Chapter 11 Constructions (Additional Questions)

To divide a line segment AB in the ratio $m:n$, where $m$ and $n$ are positive integers, using the method involving drawing a ray AX, points are marked on the ray AX at equal distances. The number of points marked on AX is equal to the sum of the ratio terms, i.e., $m+n$. The last point marked on AX is joined to the end point B of the line segment AB.

In this question, the ratio is given as $m:n$ where $m=3$ and $n=2$.

The minimum number of points required to be marked on AX is $m+n = 3+2 = 5$.

The correct option is (C) 5.

In the construction to divide a line segment AB in the ratio $m:n$, a ray AX is drawn making an acute angle with AB. Points $A_1, A_2, \dots, A_{m+n}$ are marked on AX such that $AA_1 = A_1A_2 = \dots = A_{m+n-1}A_{m+n}$.

According to the construction method, the point $A_{m+n}$ (the last point marked on AX) is joined to the endpoint B of the line segment AB.

The correct option is (B) Point B.

The construction method to divide a line segment AB in the ratio $m:n$ involves drawing a ray AX, marking $m+n$ points on it, joining the $(m+n)^{th}$ point $A_{m+n}$ to B, and then drawing a line parallel to $A_{m+n}B$ through the $m^{th}$ point $A_m$ on AX. This parallel line intersects AB at a point, say C, such that AC : CB = $m:n$.

This construction uses the Basic Proportionality Theorem (BPT), which states that if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally.

The BPT is derived from the property of similar triangles. In the construction, the triangle formed by the line segment AB, the ray AX, and the line joining $A_{m+n}$ to B (let's say $\triangle ABA_{m+n}$) is similar to the triangle formed by AC, the segment $AA_m$ on AX, and the line through $A_m$ parallel to $A_{m+n}B$ (let's say $\triangle ACA_m$). Specifically, $\triangle ACA_m \sim \triangle ABA_{m+n}$.

The ratio of corresponding sides of similar triangles is equal. This principle allows us to establish the ratio AC : CB = $m:n$.

The correct option is (B) Similar.

When a triangle is constructed similar to a given triangle ABC with a scale factor $k$, it means that the ratio of the corresponding sides of the new triangle to the original triangle is equal to $k$.

Let the given triangle be $\triangle ABC$ and the new similar triangle be $\triangle PQR$. If the scale factor is $k$, then:

Rearranging this equation, we get $PQ = k \times AB$, $QR = k \times BC$, and $RP = k \times CA$.

This shows that each side of the new triangle is $k$ times the corresponding side of the original triangle $\triangle ABC$.

The scale factor $k$ indicates the size of the new triangle relative to the original. If $k > 1$, the new triangle is an enlargement. If $0 < k < 1$, the new triangle is a reduction.

In this question, it is given that the scale factor is $k > 1$. Therefore, the sides of the new triangle will be $k$ times the corresponding sides of $\triangle ABC$.

The correct option is (B) $k$ times.

When a triangle is constructed similar to a given triangle with a scale factor $k$, the size of the new triangle relative to the original depends on the value of $k$.

If the scale factor $k > 1$, the new triangle is an enlargement, meaning it is larger than the original triangle.

If the scale factor $0 < k < 1$, the new triangle is a reduction, meaning it is smaller than the original triangle.

If the scale factor $k = 1$, the new triangle is congruent to the original triangle (same size).

In this question, the scale factor is given as $\frac{3}{4}$.

Since $\frac{3}{4} < 1$, the new triangle will be a reduction of the original triangle.

Therefore, the new triangle will be smaller than the original triangle.

The correct option is (B) Smaller than.

To construct a triangle similar to $\triangle ABC$ with a scale factor $\frac{m}{n}$ where $m < n$, the steps involve drawing a ray BX making an acute angle with BC.

On the ray BX, points $B_1, B_2, \dots, B_n$ are marked such that $BB_1 = B_1B_2 = \dots = B_{n-1}B_n$. We mark points up to the maximum of $m$ and $n$. Since the scale factor is $\frac{m}{n}$, and $m < n$, we mark $n$ points in total on BX.

The point $B_n$ (the $n^{th}$ point on BX, corresponding to the denominator of the scale factor) is joined to the vertex C of the original triangle $\triangle ABC$.

Then, a line segment $B_mC'$ is drawn parallel to $B_nC$, where $C'$ lies on BC. Finally, a line segment $A'C'$ is drawn parallel to AC, where $A'$ lies on AB. $\triangle A'BC'$ is the required similar triangle.

Therefore, the point joined to C is $B_n$.

The correct option is (B) $B_n$.

A fundamental property of circles states that the tangent at any point on a circle is perpendicular to the radius that passes through the point of contact.

To construct a tangent to a circle at a point P on the circle, we first locate the center O of the circle and draw the radius OP.

The tangent line is then constructed as a line passing through point P that makes a $90^\circ$ angle with the radius OP at P.

Therefore, the line through P that is tangent to the circle is perpendicular to the radius OP.

The correct option is (B) Perpendicular.

To construct tangents to a circle from an external point P, we follow a standard geometrical construction process. The first two steps are correctly described in the question:

1. Join the external point P to the center O of the circle.

2. Bisect the line segment PO to find its midpoint, M.

The purpose of finding the midpoint M of PO is to construct a circle whose diameter is PO. This new circle will intersect the original circle at the points where the tangents from P touch the original circle. This is based on the geometric property that the angle in a semicircle is $90^\circ$. The tangent at a point on a circle is perpendicular to the radius through that point. If A is a point of tangency on the original circle, then $\triangle$PAO must be a right-angled triangle with the right angle at A. The circle with diameter PO will pass through A.

Therefore, the next step in the construction is to draw the circle with diameter PO. Since M is the midpoint of PO, this circle has its center at M and its radius equal to MO (or MP, since $MO = MP$).

This construction step is given in option (C): Draw a circle with center M and radius MP (or MO).

This circle with center M and radius MO will intersect the original circle at two points, say A and B. Joining P to A and P to B will give the two required tangents PA and PB.

Comparing the options with the standard construction steps:

(A) Drawing a circle with center P and radius PO does not help in finding the tangent points.

(B) Drawing a circle with center O and radius OM is a circle concentric with the original circle, but with a smaller radius, and does not help in finding the tangent points from P.

(C) Drawing a circle with center M and radius MP (or MO) is precisely the next step to locate the points of tangency on the original circle.

(D) Drawing a perpendicular from P to OP would just be a line perpendicular to the line segment PO passing through P, which is not relevant to finding tangents to the original circle from P.

The correct answer is (C).

The standard method for constructing tangents from an external point P to a circle with center O involves drawing a circle with PO as its diameter. Let this new circle intersect the original circle at points A and B.

Consider the point A where the two circles intersect. Since the circle with PO as diameter passes through A, the line segment PA and OA are segments forming $\triangle$PAO within this circle. The angle $\angle$PAO is an angle subtended by the diameter PO at a point A on the circumference of the circle with diameter PO.

According to a fundamental geometric theorem, the angle subtended by a diameter at any point on the circumference of a circle is a right angle.

For the original circle with center O, OA is the radius. Since $\angle$PAO = $90^\circ$, the line segment PA is perpendicular to the radius OA at the point A on the circle. By the definition of a tangent, a line is tangent to a circle if and only if it is perpendicular to the radius through the point of contact. Thus, PA is a tangent to the original circle at point A.

Similarly, PB is a tangent to the original circle at point B because $\angle$PBO is also an angle in the semicircle of the circle with diameter PO, making $\angle$PBO = $90^\circ$, and PB is perpendicular to the radius OB at point B.

The justification for this construction method relies directly on the fact that the angle in a semicircle is a right angle.

Comparing with the options:

(A) Acute angle is less than $90^\circ$.

(B) Obtuse angle is greater than $90^\circ$ and less than $180^\circ$.

(C) Right angle is exactly $90^\circ$.

(D) Straight angle is exactly $180^\circ$.

The correct type of angle in a semicircle is a right angle.

The correct option is (C).

Let a circle have center O and radius r. The position of a point P relative to the circle is determined by its distance from the center, OP.

There are three possibilities:

• If the distance of the point from the center is less than the radius ($OP < r$), the point lies inside the circle.
• If the distance of the point from the center is equal to the radius ($OP = r$), the point lies on the circle.
• If the distance of the point from the center is greater than the radius ($OP > r$), the point lies outside the circle (i.e., it is an external point).

The question states that the distance of the point from the center is equal to the radius. Although the phrase "external point" is used, the condition $OP = r$ is explicitly given. Based on the definition above, a point whose distance from the center is equal to the radius lies on the circle.

Now, we need to determine how many tangents can be drawn from a point that lies on the circle.

A tangent to a circle is a line that intersects the circle at exactly one point (the point of tangency). Through any single point on the circumference of a circle, there exists exactly one unique line that is tangent to the circle at that point. This tangent line is perpendicular to the radius at the point of tangency.

Therefore, from a point lying on the circle, only one tangent can be drawn.

Let's evaluate the given options:

(A) 0 tangents: This is the case for a point inside the circle ($OP < r$).

(B) 1 tangent (The point is on the circle): This matches our conclusion for a point on the circle ($OP = r$).

(C) 2 tangents: This is the case for a point outside the circle ($OP > r$).

(D) Infinitely many: This is not possible for tangents from a single point to a circle.

Given that the distance from the center is equal to the radius ($OP = r$), the point is on the circle, and thus exactly one tangent can be drawn from this point.

The correct option is (B).

When constructing the division of a line segment in a given ratio $m:n$, the standard method involves drawing a ray from one endpoint of the segment, making an acute angle.

On this ray, we mark a total of $m+n$ points at equal distances.

For the ratio 2:3, we have $m=2$ and $n=3$. The total number of equal parts needed on the ray is $m+n = 2+3 = 5$.

Assertion (A): To divide a line segment in the ratio 2:3, you need to mark 5 points on the ray at equal intervals.

Based on the construction method, this is true. You mark $A_1, A_2, A_3, A_4, A_5$ on the ray such that $AA_1 = A_1A_2 = ... = A_4A_5$.

Reason (R): The total number of equal parts needed is the sum of the ratio terms.

This is also true. The construction requires marking $m+n$ equal divisions on the ray, which is the sum of the ratio terms $m$ and $n$.

The assertion states that 5 points are needed for the ratio 2:3. The reason states that the number of points needed is the sum of the ratio terms ($2+3=5$). The reason correctly explains why 5 points are needed in this specific case. Therefore, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

Let's analyze the given Assertion (A) and Reason (R) regarding the construction of similar triangles.

Assertion (A): When constructing a triangle similar to a given triangle with a scale factor greater than 1, the vertices of the new triangle will lie outside the original triangle.

When we construct a triangle similar to a given triangle $\triangle$ABC with a scale factor $k > 1$, the new triangle $\triangle$A'B'C' will be an enlargement of $\triangle$ABC. If the construction is done from a common vertex (say A=A') or with a common center of similarity, the corresponding vertices of the new triangle (B', C') will lie on the rays extending from the common point through the original vertices (B, C). Since the scale factor is greater than 1, the distances from the common point to the new vertices ($A'B'$, $A'C'$) will be greater than the distances to the original vertices ($AB$, $AC$). This implies that the new vertices (B', C') will lie outside the original triangle $\triangle$ABC. Therefore, Assertion (A) is true under typical construction methods.

Reason (R): The sides of the new triangle are larger than the corresponding sides of the original triangle.

If a triangle $\triangle$A'B'C' is similar to $\triangle$ABC with a scale factor $k$, then the ratio of corresponding sides is equal to the scale factor: $\frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{C'A'}{CA} = k$.

Given that the scale factor $k$ is greater than 1 ($k > 1$), we have:

Since $k > 1$, it follows that $A'B' > AB$, $B'C' > BC$, and $C'A' > CA$. Thus, the sides of the new triangle are indeed larger than the corresponding sides of the original triangle. Therefore, Reason (R) is true.

Now we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the vertices of the new triangle lie outside the original triangle when the scale factor is greater than 1. Reason (R) states that the sides of the new triangle are larger than the original. The fact that the sides are larger directly leads to the new triangle being an enlarged version. In the standard construction where one vertex is common or there is a common center of similarity, this enlargement means the new vertices are further away from the common point, and thus lie outside the original triangle. Therefore, Reason (R) provides a correct explanation for Assertion (A).

Both A and R are true, and R is the correct explanation of A.

The correct option is (A).

Let's match each construction type with the geometric principle it primarily utilizes:

(i) Dividing a line segment in a given ratio: This construction relies on drawing parallel lines which cut proportional segments on transversals. This is the principle behind the Basic Proportionality Theorem (BPT), also known as Thales's Theorem.

Match: (i) - (c)

(ii) Constructing a similar triangle: The construction of a triangle similar to a given triangle involves creating a triangle whose angles are equal to the corresponding angles of the original triangle (AA similarity) or whose sides are in proportion (SSS or SAS similarity). The underlying principles are the similarity criteria.

Match: (ii) - (d)

(iii) Constructing a tangent at a point on the circle: A fundamental property of a tangent to a circle is that it is perpendicular to the radius drawn to the point of tangency. The construction involves drawing the radius to the given point and then constructing a perpendicular line at that point.

Match: (iii) - (b)

(iv) Constructing tangents from an external point: The standard construction involves drawing a circle with the line segment joining the external point and the center as diameter. The points of intersection of this new circle with the original circle are the points of tangency. This works because the angle formed at the point of intersection by the line segment from the external point and the radius from the center is an angle in the semicircle of the construction circle, which is $90^\circ$. Thus, the radius is perpendicular to the line segment, confirming it's a tangent.

Match: (iv) - (a)

Putting the matches together, we get:

• (i) - (c)
• (ii) - (d)
• (iii) - (b)
• (iv) - (a)

Comparing this with the given options, option (A) matches this set of correspondences.

The correct option is (A).

The original triangle ABC has side lengths 8m, 10m, and 12m.

The sculptor is creating a scaled-down model with a scale factor of 1:4.

A scale factor of 1:4 means that each dimension of the new model will be $\frac{1}{4}$ times the corresponding dimension of the original monument.

Let the side lengths of the original triangle be $s_1 = 8$m, $s_2 = 10$m, and $s_3 = 12$m.

Let the scale factor be $k = \frac{1}{4}$.

The side lengths of the new triangular model, $s'_1, s'_2, s'_3$, are calculated by multiplying the original side lengths by the scale factor:

Calculating the new side lengths:

So, the side lengths of the triangular model will be 2m, 2.5m, and 3m.

Comparing this result with the given options:

(A) 8m, 10m, 12m - These are the original dimensions, not scaled.

(B) 2m, 2.5m, 3m - These match the calculated side lengths for a 1:4 scale factor.

(C) 32m, 40m, 48m - These correspond to a scale factor of 4:1 (enlargement).

(D) 4m, 5m, 6m - These correspond to a scale factor of 1:2.

The correct side lengths for the triangular model are 2m, 2.5m, and 3m.

The correct option is (B).

From the case study in Question 14, the original monument is a triangle and the model is a similar triangle (scaled version) with a scale factor of 1:4.

Let the original triangle be $\triangle ABC$ and the model triangle be $\triangle A'B'C'$. The scale factor of the model relative to the original is given as 1:4. This means that the ratio of corresponding linear dimensions (like side lengths) of the model to the original is $\frac{1}{4}$.

For similar figures, the ratio of their areas is equal to the square of the ratio of their corresponding linear dimensions (the scale factor squared).

Rearranging the equation to express the Area of the model in terms of the Area of the original monument:

Alternatively, this can be written as:

Comparing this result with the given options:

(A) Area of model = Area of monument / 4

(B) Area of model = Area of monument / 16

(C) Area of model = 4 $\times$ Area of monument

(D) Area of model = 16 $\times$ Area of monument

The correct comparison is that the Area of the model is the Area of the original monument divided by 16.

The correct option is (B).

The question asks which tools are typically used for geometric constructions in this chapter. Based on the types of constructions covered (like dividing a line segment, constructing similar triangles, and constructing tangents to a circle), we use specific tools to perform the geometrical operations.

Let's examine each tool listed:

• (A) Ruler (straightedge): A ruler is used as a straightedge to draw straight lines connecting two points. This is a fundamental operation in all geometric constructions. Thus, a ruler is typically used.
• (B) Compass: A compass is used to draw circles or arcs with a given center and radius. It is essential for transferring lengths and finding points that are a specific distance from another point. Constructions involving circles (like tangents) or transferring lengths rely heavily on the compass. Thus, a compass is typically used.
• (C) Protractor: A protractor is used for measuring or drawing angles of a specific degree measure. Geometric constructions in this context typically focus on *constructing* specific angles (like $90^\circ$ for perpendiculars, or copying angles) using only compass and straightedge techniques or by relying on geometric properties, rather than measuring angles with a protractor. Thus, a protractor is generally not typically used for the constructions themselves, although it might be used to verify the result.
• (D) Set squares: Set squares (often used in pairs) are useful for drawing parallel and perpendicular lines accurately and efficiently. Constructions like dividing a line segment in a ratio often involve drawing parallel lines, and tangent constructions involve perpendicular lines. While these can theoretically be constructed using only compass and straightedge, set squares are commonly used as practical aids in educational settings to perform these steps more quickly and accurately. Thus, set squares are typically used.

Considering the common methods taught for these constructions in school geometry, the tools typically employed are the Ruler (straightedge), Compass, and Set squares.

Since the instruction is "Select all that apply", the correct options are those representing the tools typically used.

The correct options are:

(A)

(B)

(D)

The construction method for dividing a line segment (say AB) in a ratio $m:n$ involves drawing a ray AX from A, marking $m+n$ points ($A_1, A_2, ..., A_{m+n}$) at equal distances on AX, joining $A_{m+n}$ to B, and then drawing a line through $A_m$ parallel to $A_{m+n}B$ that intersects AB at point C.

In $\triangle AA_{m+n}B$, the line segment $A_mC$ is drawn parallel to the side $A_{m+n}B$. This line segment $A_mC$ intersects the other two sides of the triangle, AA$_{m+n}$ and AB, at points $A_m$ and C respectively.

According to the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Applying BPT to $\triangle AA_{m+n}B$ with line $A_mC \parallel A_{m+n}B$, we get:

By construction, the points $A_1, A_2, ..., A_{m+n}$ are marked such that $AA_1 = A_1A_2 = ... = A_{m+n-1}A_{m+n}$. If each equal segment has length 'k', then $AA_m = m \times k$ and $A_m A_{m+n} = (m+n - m) \times k = n \times k$.

Therefore, from BPT, we have:

This shows that point C divides the line segment AB in the ratio $m:n$. The theorem that guarantees this proportionality is the Basic Proportionality Theorem (BPT).

Let's look at the options:

(A) Pythagoras Theorem: Applies to right-angled triangles and relates the square of the hypotenuse to the sum of the squares of the other two sides.

(B) Converse of Pythagoras Theorem: Used to prove if a triangle is right-angled.

(C) Basic Proportionality Theorem (Thales Theorem): States that a line parallel to one side of a triangle divides the other two sides proportionally. This is exactly the principle used.

(D) Midpoint Theorem: A special case of BPT where the line segment connects the midpoints of two sides of a triangle.

The theorem that directly guarantees the proportionality needed for dividing a line segment in a given ratio using parallel lines is the Basic Proportionality Theorem.

The correct option is (C).

Given:

Radius of the circle, $R = 6$ cm.

Distance of the point P from the center O, $d = OP = 10$ cm.

We need to determine the number of tangents that can be drawn from point P to the circle.

The number of tangents from a point to a circle depends on the location of the point relative to the circle:

• If the point lies inside the circle, its distance from the center is less than the radius ($d < R$). In this case, no tangent can be drawn from the point.
• If the point lies on the circle, its distance from the center is equal to the radius ($d = R$). In this case, exactly one tangent can be drawn at the point.
• If the point lies outside the circle (i.e., it is an external point), its distance from the center is greater than the radius ($d > R$). In this case, exactly two tangents can be drawn from the point.

In this problem, the distance of the point P from the center is $d = 10$ cm and the radius of the circle is $R = 6$ cm.

Comparing the distance and the radius, we have $10 > 6$.

So, $d > R$.

Since the distance of the point from the center is greater than the radius, the point P lies outside the circle.

According to the property mentioned above, from an external point, exactly two tangents can be drawn to the circle.

Therefore, 2 tangents can be drawn from point P to the circle.

Comparing our result with the given options:

(A) 0: Incorrect ($d > R$).

(B) 1: Incorrect ($d = R$).

(C) 2: Correct ($d > R$).

(D) Infinitely many: Incorrect.

The correct option is (C).

The question describes the construction of a tangent to a circle at a given point on the circle. It also states the principle behind this construction: drawing a perpendicular at the point P to the radius OP.

This method is based on a fundamental theorem in circle geometry which describes the relationship between a radius and a tangent at the point where the tangent touches the circle.

The theorem states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

In other words, the angle formed between the radius and the tangent line at the point where they meet on the circle is always $90^\circ$.

Therefore, the property utilized in this construction is that the angle between the radius and the tangent at the point of contact is always a right angle, which is $90^\circ$.

Comparing this with the given options:

(A) $0^\circ$: Incorrect.

(B) $45^\circ$: Incorrect.

(C) $90^\circ$: Correct.

(D) $180^\circ$: Incorrect.

The correct option is (C).

We are asked about the construction of a triangle similar to a given triangle $\triangle ABC$ with a scale factor of $\frac{5}{3}$. The scale factor is greater than 1, which means the resulting similar triangle will be an enlargement of the original triangle.

The standard construction method for a similar triangle $\triangle A'BC'$ (where A' lies on the ray AB extended, and C' lies on the ray BC extended) with vertex B common involves the following steps:

1. Draw a ray BX from B, making an acute angle with BC.

2. On the ray BX, mark points $B_1, B_2, \dots, B_n$ at equal distances, where $n$ is the greater of the numerator and the denominator of the scale factor. The scale factor is $\frac{5}{3}$, so the greater value is 5. Thus, we mark 5 points: $B_1, B_2, B_3, B_4, B_5$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

3. Join the point corresponding to the denominator of the scale factor to the vertex C of the original triangle. The scale factor is $\frac{5}{3}$, and the denominator is 3. The ray BX starts from B along BC. So, we join the point $B_3$ to C.

4. Draw a line through the point corresponding to the numerator of the scale factor ($B_5$) parallel to $B_3C$. This line intersects the ray BC extended at point C'.

5. Draw a line through C' parallel to AC, intersecting the ray BA extended at point A'.

Then, $\triangle A'BC'$ is the required triangle similar to $\triangle ABC$ with scale factor $\frac{5}{3}$.

In the given question, the scale factor is $\frac{5}{3}$. The ray BX is drawn. Points $B_1, B_2, \dots, B_5$ are marked.

According to the construction method, the point corresponding to the denominator (3) on the ray BX is joined to the vertex C.

The point corresponding to the denominator 3 is $B_3$.

Thus, the point that is joined to C is $B_3$.

Comparing with the given options:

(A) $B_3$: This is the point corresponding to the denominator 3.

(B) $B_5$: This is the point corresponding to the numerator 5.

(C) $B_1$: This is the point corresponding to 1.

(D) B: This is the common vertex.

The correct point to be joined to C is $B_3$.

The correct option is (A).

Let the two similar triangles be $\triangle_1$ and $\triangle_2$.

We are given that the ratio of the areas of these two similar triangles is 4:25.

A property of similar triangles states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Furthermore, this property extends to other corresponding linear dimensions such as altitudes, medians, and angle bisectors.

So, the ratio of the areas of two similar triangles is also equal to the square of the ratio of their corresponding altitudes.

Let $h_1$ be the length of an altitude of $\triangle_1$, and $h_2$ be the length of the corresponding altitude of $\triangle_2$.

According to the property:

Substituting the given ratio of areas into this equation:

To find the ratio of the altitudes $\frac{h_1}{h_2}$, we need to take the square root of both sides of the equation:

Thus, the ratio of their corresponding altitudes is 2:5.

Comparing this result with the given options:

(A) 4:25: This is the ratio of areas, not altitudes.

(B) 2:5: This matches our calculated ratio of altitudes.

(C) 5:2: This is the inverse ratio.

(D) 16:625: This is the square of 4:25, which is the ratio of the areas squared.

The correct ratio of corresponding altitudes is 2:5.

The correct option is (B).

The construction of a triangle similar to a given triangle with a scale factor $m/n$ typically involves drawing a ray from one vertex of the original triangle, marking points at equal intervals along this ray, and then using parallel lines to find the vertices of the new similar triangle.

Consider a triangle $\triangle ABC$. To construct a similar triangle $\triangle A'BC'$ with scale factor $m/n$ (say, $m > n$), sharing vertex B, we draw a ray BX making an acute angle with BC. We mark $m$ points $B_1, B_2, \dots, B_m$ on BX such that $BB_1 = B_1B_2 = \dots = B_{m-1}B_m$. We join $B_n$ to C. Then we draw a line through $B_m$ parallel to $B_nC$, intersecting the ray BC extended at C'. Finally, we draw a line through C' parallel to AC, intersecting the ray BA extended at A'.

In $\triangle BB_m C'$, the line segment $B_nC$ is parallel to $B_mC'$. By the Basic Proportionality Theorem (BPT), if a line is parallel to one side of a triangle and intersects the other two sides (or their extensions), it divides the two sides proportionally.

Applying BPT to $\triangle BB_m C'$ with $B_nC \parallel B_mC'$:

By construction, $\frac{BB_n}{BB_m} = \frac{n \times \text{equal unit}}{m \times \text{equal unit}} = \frac{n}{m}$.

So, $\frac{BC}{BC'} = \frac{n}{m}$, which means $\frac{BC'}{BC} = \frac{m}{n}$. This shows that the side BC is scaled by the factor $m/n$.

Similarly, the construction of $A'C' \parallel AC$ ensures that $\triangle A'BC'$ is similar to $\triangle ABC$ and the ratio of corresponding sides is $m/n$. This step also implicitly uses properties related to parallel lines cut by transversals, which are linked to the concept of proportionality embodied by BPT or similar triangle properties derived from BPT.

The core principle that allows us to divide lines proportionally using parallel lines in this construction is the Basic Proportionality Theorem.

Let's consider the options:

(A) Pythagoras Theorem: Deals with the sides of a right-angled triangle.

(B) Converse of Pythagoras Theorem: Used to check if a triangle is right-angled.

(C) Basic Proportionality Theorem: Deals with proportional division of sides by a line parallel to one side of a triangle.

(D) Angle Sum Property: States that the sum of angles in a triangle is $180^\circ$.

The justification for this construction method primarily relies on the Basic Proportionality Theorem (BPT).

The correct option is (C).

Given:

Total length of the line segment PQ = 14 cm.

The line segment is divided in the ratio 4:3.

Let the line segment PQ be divided at a point R into two parts, PR and RQ, such that the ratio of their lengths is $PR : RQ = 4 : 3$.

The sum of the ratio terms is $4 + 3 = 7$.

This means the line segment PQ is divided into 7 equal parts in terms of the ratio.

The total length of the line segment is 14 cm, which corresponds to the total of 7 parts.

The length of one part is the total length divided by the sum of the ratio terms:

The ratio of the two parts is 4:3. This means one part (PR) consists of 4 units of length, and the other part (RQ) consists of 3 units of length, where each unit is the 'length of one part' calculated above.

Length of the first part (corresponding to the ratio 4) = $4 \times \text{Length of one part} = 4 \times 2 \text{ cm} = 8 \text{ cm}$.

Length of the second part (corresponding to the ratio 3) = $3 \times \text{Length of one part} = 3 \times 2 \text{ cm} = 6 \text{ cm}$.

The lengths of the two parts are 8 cm and 6 cm.

The larger part is the one with the greater length, which is 8 cm.

This calculation corresponds to the result obtained by the geometric construction method, which divides the line segment into a total number of equal subdivisions equal to the sum of the ratio terms and then uses parallel lines to project the required division point onto the original segment.

Comparing the calculated lengths with the given options:

(A) 6 cm: This is the smaller part.

(B) 8 cm: This is the larger part.

(C) 10 cm: Incorrect.

(D) 7 cm: Incorrect.

The length of the larger part is 8 cm.

The correct option is (B).

Let's analyze each statement about tangents and secants to a circle.

(A) A line intersecting a circle at two points is a secant.

This is the standard definition of a secant line with respect to a circle. A secant is a line that passes through two distinct points of a circle.

This statement is True.

(B) A tangent is a special case of a secant where the two intersection points coincide.

A secant intersects a circle at two *distinct* points. A tangent intersects a circle at exactly one point (the point of contact). While a tangent can be viewed as the limit of a secant as the two intersection points approach each other and merge, by definition, a secant requires two distinct intersection points. A line intersecting at a single point is classified as a tangent, not a secant. Therefore, a tangent is not a special case of a secant based on the definitions.

This statement is False.

(C) The tangent at a point on the circle is unique.

For any given point on the circumference of a circle, there is only one line that is perpendicular to the radius at that point. This unique line is the tangent to the circle at that point. Thus, the tangent at a specific point on the circle is indeed unique.

This statement is True.

(D) From a point inside the circle, no tangent can be drawn.

A tangent line touches the circle at exactly one point. If a line passes through a point located inside the circle, any line passing through that interior point and extending outwards will necessarily intersect the circle at two points unless it doesn't intersect the circle at all (which isn't a tangent). Therefore, it is impossible to draw a line through an interior point that is tangent to the circle.

This statement is True.

We are looking for the FALSE statement. Based on our analysis, statement (B) is false.

The correct option is (B).

To divide a line segment AB in a given ratio $m:n$ using the construction method, we follow these steps:

1. Draw a ray AX from point A, making an acute angle with the line segment AB.

2. On the ray AX, mark points $A_1, A_2, \dots, A_{m+n}$ such that the distances $AA_1, A_1A_2, \dots, A_{m+n-1}A_{m+n}$ are all equal.

3. Join the point $A_{m+n}$ to the endpoint B of the line segment AB.

4. Through the point $A_m$, draw a line parallel to $A_{m+n}B$, intersecting AB at point C. The point C divides AB in the ratio $m:n$, i.e., $\frac{AC}{CB} = \frac{m}{n}$.

In this question, the ratio is 3:5. So, we have $m=3$ and $n=5$.

The total number of equal parts required on the ray AX is $m+n = 3+5 = 8$.

The points marked on the ray AX are given as $A_1, A_2, \dots, A_8$. This means we have marked 8 equal divisions on the ray, which corresponds to $A_{m+n}$ where $m+n=8$. So, the point is $A_8$.

According to step 3 of the construction, the point corresponding to the total number of divisions on the ray (which is $A_{m+n}$) is joined to the endpoint B of the line segment.

In this case, $m+n=8$, so the point $A_8$ is joined to B.

Comparing this with the given options:

(A) $A_3$: This corresponds to the first part of the ratio (m=3), used for drawing the parallel line.

(B) $A_5$: This corresponds to the second part of the ratio (n=5), but this point is not directly joined to B in the standard construction unless the ratio is flipped or the other endpoint is used.

(C) $A_8$: This corresponds to the sum of the ratio terms (m+n=8).

(D) $A_1$: This corresponds to the first division point.

The point joined to B is the point on the ray corresponding to the total number of parts, which is $A_8$.

The correct option is (C).

Given:

A circle with center O.

Two radii OA and OB.

The angle between the two radii is $\angle AOB = 100^\circ$.

To Find:

The angle between the tangents drawn at points A and B.

Let the tangents at points A and B intersect at point P. We are looking for the measure of $\angle APB$.

Consider the quadrilateral PAOB formed by the two radii OA and OB, and the two tangents PA and PB.

We know that the tangent at any point on a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OA is perpendicular to the tangent PA at point A.

Similarly, the radius OB is perpendicular to the tangent PB at point B.

The sum of the interior angles of a quadrilateral is always $360^\circ$.

In quadrilateral PAOB, the sum of angles is:

Substitute the known values into the equation:

Now, solve for $\angle APB$:

Thus, the angle between the tangents at the ends of the radii is $80^\circ$.

Comparing this result with the given options:

(A) $80^\circ$: This matches our calculated angle.

(B) $100^\circ$: This is the angle between the radii.

(C) $50^\circ$: Incorrect.

(D) $40^\circ$: Incorrect.

The correct option is (A).

Let's examine the typical construction method for each option to see if drawing a perpendicular is a required step.

(A) Dividing a line segment in a given ratio: This construction involves drawing a ray from one endpoint, marking equal segments on the ray, joining the last marked point to the other endpoint of the segment, and then drawing a line through the appropriate marked point on the ray parallel to the joining line. While parallel lines can be constructed using perpendiculars to a transversal, the core method relies on parallel lines and the Basic Proportionality Theorem, not directly on constructing a perpendicular line to the original segment.

(B) Constructing a similar triangle: Similar to dividing a line segment, constructing a similar triangle using the standard method (with a common vertex) involves drawing a ray, marking points, joining a point to a vertex of the original triangle, and then drawing a parallel line through another marked point to one side of the original triangle. This construction also relies on parallel lines and similarity principles (like AA or BPT), not typically on drawing a perpendicular as a primary step.

(C) Constructing a tangent at a point on the circle: To construct a tangent at a given point P on a circle with center O, we draw the radius OP. A fundamental property of a tangent is that it is perpendicular to the radius at the point of contact. Therefore, the construction requires drawing a line that is perpendicular to the radius OP at the point P. This directly involves the construction of a perpendicular line at a point on a given line segment (or ray, in this case, the radius).

(D) All of the above: Since constructions (A) and (B) do not fundamentally require drawing a perpendicular as part of their core method (they require parallel lines), this option is incorrect.

Based on the standard geometric construction techniques, constructing a tangent at a point on the circle explicitly requires drawing a perpendicular line.

The correct option is (C).

Given:

Side lengths of the original triangular sketch are 6 cm, 8 cm, and 10 cm.

Scale factor for the mural relative to the sketch is 2:1.

To Find:

The perimeter of the triangle in the mural.

Solution:

First, let's find the perimeter of the original triangular sketch. The perimeter is the sum of the lengths of its sides.

The scale factor for the mural is 2:1, which can be written as $\frac{2}{1} = 2$. This means that every linear dimension of the mural triangle is 2 times the corresponding linear dimension of the sketch triangle.

Since the mural triangle is similar to the sketch triangle with a scale factor of 2, the side lengths of the mural triangle will be 2 times the corresponding side lengths of the sketch triangle.

Side lengths of the mural triangle:

Now, let's find the perimeter of the mural triangle by summing its side lengths.

Alternate Solution:

A property of similar figures is that the ratio of their perimeters is equal to the scale factor of similarity.

Scale factor (mural to sketch) = 2:1 = 2.

Perimeter of sketch = 24 cm (calculated above).

Both methods yield the same result.

Comparing the result with the given options:

(A) 24 cm: This is the perimeter of the original sketch.

(B) 48 cm: This matches the calculated perimeter of the mural.

(C) 12 cm: Incorrect.

(D) $24 \times 4$ cm = 96 cm: This would correspond to a scale factor of 4, or (scale factor)^2 = 4.

The correct option is (B).

Given:

Side lengths of the original triangular sketch are 6 cm, 8 cm, and 10 cm.

Scale factor for the mural relative to the sketch is 2:1.

To Find:

The area of the triangle in the mural.

Solution:

First, let's determine if the original triangular sketch is a right-angled triangle. The side lengths are 6 cm, 8 cm, and 10 cm. Let's check if they satisfy the Pythagorean theorem ($a^2 + b^2 = c^2$).

Since $6^2 + 8^2 = 10^2$, the triangle is a right-angled triangle with legs 6 cm and 8 cm.

The area of the original triangular sketch can be calculated using the formula for the area of a right triangle:

The mural triangle is similar to the sketch triangle with a scale factor of 2:1, which means the linear dimensions are scaled by a factor of $k = \frac{2}{1} = 2$.

For similar figures, the ratio of their areas is equal to the square of the scale factor.

Now, we can find the area of the mural triangle:

Alternate Solution:

The side lengths of the mural triangle are the side lengths of the sketch multiplied by the scale factor 2.

Side lengths of mural: $2 \times 6 = 12$ cm, $2 \times 8 = 16$ cm, $2 \times 10 = 20$ cm.

This is also a right-angled triangle with legs 12 cm and 16 cm.

Area of mural = $\frac{1}{2} \times 12\text{ cm} \times 16\text{ cm}$

Area of mural = $\frac{1}{2} \times 192\text{ cm}^2$

Area of mural = $96\text{ cm}^2$

Both methods give the same result. The area of the triangle in the mural is $96\text{ cm}^2$.

Comparing the result with the given options:

(A) $24 \text{ cm}^2$: Area of sketch.

(B) $48 \text{ cm}^2$: Perimeter of mural (from Q28).

(C) $96 \text{ cm}^2$: Matches the calculated area.

(D) $192 \text{ cm}^2$: $2 \times 96\text{ cm}^2$. Incorrect.

The correct option is (C).

The standard construction method for drawing tangents from an external point P to a circle with center O involves drawing a circle that has the line segment PO as its diameter.

Let the original circle have center O and some radius. Let P be the external point. We join P and O.

We then find the midpoint M of the line segment PO and draw a circle with center M and radius MO (or MP). This new circle has PO as its diameter.

Let this new circle intersect the original circle at two points, say A and B.

Consider the point of intersection A. A lies on the original circle, and A also lies on the circle with diameter PO. In the circle with diameter PO, the angle subtended by the diameter PO at any point on the circumference is a right angle.

For the original circle, OA is the radius drawn to the point A. Since $\angle PAO = 90^\circ$, the line segment PA is perpendicular to the radius OA at the point A on the circle.

According to the theorem about tangents, a line is tangent to a circle at a point if and only if it is perpendicular to the radius through that point.

Therefore, PA is a tangent to the original circle at the point A.

Similarly, for the point of intersection B, $\angle PBO = 90^\circ$ (angle in a semicircle of the circle with diameter PO). Since OB is the radius of the original circle, PB is perpendicular to the radius OB at point B. Therefore, PB is a tangent to the original circle at the point B.

The points A and B where the circle with diameter OP intersects the given circle are precisely the points on the given circle where the tangents from P touch the circle. These points are known as the points of contact of the tangents.

Comparing this with the given options:

(A) Center of the given circle (O) is usually inside the circle with diameter OP, not an intersection point with the original circle unless P is on the original circle (which is not the case for an external point).

(B) External point P is an endpoint of the diameter OP of the new circle, not an intersection point with the original circle unless P is on the original circle.

(C) Points of contact of the tangents. As shown by the geometric principle ($\angle$ in a semicircle leading to radius $\perp$ tangent), the intersection points are the points where the tangents touch the original circle.

(D) Midpoint of OP (M) is the center of the new circle, not an intersection point with the original circle.

The correct option is (C).

Let the two similar triangles be $\triangle_1$ and $\triangle_2$.

We are given that the ratio of their corresponding sides is 3:5.

For similar figures, the ratio of their perimeters is equal to the ratio of their corresponding linear dimensions, which includes the ratio of their corresponding sides.

Let $P_1$ be the perimeter of $\triangle_1$, and $P_2$ be the perimeter of $\triangle_2$.

According to the property:

Substituting the given ratio of sides into this equation:

Thus, the ratio of their perimeters is 3:5.

Comparing this result with the given options:

(A) 3:5: This matches our calculated ratio of perimeters.

(B) 5:3: This is the inverse ratio.

(C) 9:25: This is the square of the ratio of sides, which is the ratio of areas ($(\frac{3}{5})^2 = \frac{9}{25}$).

(D) 25:9: This is the square of the inverse ratio of sides.

The correct ratio of corresponding perimeters is 3:5.

The correct option is (A).

Two circles are said to be concentric if they have the same center but different radii.

Let the two concentric circles have center O and radii $r_1$ and $r_2$, where we assume $r_1 < r_2$ without loss of generality.

A tangent to a circle is a line that touches the circle at exactly one point. A key property of a tangent is that it is perpendicular to the radius at the point of contact. Equivalently, the perpendicular distance from the center of the circle to the tangent line is equal to the radius of the circle.

For a line to be a common tangent to two circles, it must be tangent to both circles simultaneously.

Consider a line L. If L is tangent to the smaller circle (with radius $r_1$), then the perpendicular distance from the center O to the line L must be equal to $r_1$.

If the same line L is also tangent to the larger circle (with radius $r_2$), then the perpendicular distance from the center O to the line L must also be equal to $r_2$.

For a line to be a common tangent, both conditions must hold simultaneously. That is, the distance from the center O to the line L must be equal to both $r_1$ and $r_2$.

This implies that $r_1 = r_2$.

However, the definition of two distinct concentric circles requires their radii to be different ($r_1 \neq r_2$). Therefore, it is impossible for the perpendicular distance from the center to a line to be equal to two different values simultaneously.

This means there is no line that can be tangent to both concentric circles.

Thus, the number of common tangents that can be drawn to two concentric circles is 0.

Comparing this result with the given options:

(A) 0: This matches our conclusion.

(B) 1: Incorrect.

(C) 2: Incorrect.

(D) Infinitely many: Incorrect.

The correct option is (A).

We are given the task of constructing a triangle similar to $\triangle PQR$ with a scale factor of $\frac{7}{4}$. The scale factor is $\frac{m}{n} = \frac{7}{4}$, where $m=7$ and $n=4$. Since the scale factor is greater than 1 ($7/4 > 1$), the resulting similar triangle will be an enlargement of the original triangle.

The standard construction method involves drawing a ray PX from one vertex (say P) making an acute angle with the side PR (or PQ). On the ray PX, we mark a number of points equal to the greater of the numerator and the denominator of the scale factor. In this case, $\max(7, 4) = 7$. The question confirms that 7 points ($P_1, P_2, \dots, P_7$) are marked on the ray PX such that $PP_1 = P_1P_2 = \dots = P_6P_7$ (at equal distances).

The next step in the construction depends on whether the scale factor is less than 1 or greater than 1, and which vertex the ray is drawn from. In the standard method where the new triangle shares vertex P and the ray PX makes an acute angle with PR, we follow these steps after marking the points:

1. Join the point corresponding to the denominator of the scale factor on the ray PX to the vertex R of the original triangle.

2. Draw a line through the point corresponding to the numerator of the scale factor on the ray PX, parallel to the line segment drawn in step 1. This parallel line intersects the ray PR extended at R', the corresponding vertex of the new triangle.

3. Draw a line through R' parallel to QR, intersecting the ray PQ extended at Q'. Then $\triangle PQ'R'$ is the required similar triangle.

In this question, the scale factor is $\frac{7}{4}$, the numerator is 7, and the denominator is 4. The ray is PX. The points $P_1, P_2, \dots, P_7$ are marked.

According to step 1, we join the point corresponding to the denominator (4) on the ray PX to the vertex R.

The point corresponding to the denominator 4 is $P_4$.

Therefore, the point that is joined to R is $P_4$.

Comparing this with the given options:

(A) $P_4$: This is the point corresponding to the denominator 4.

(B) $P_7$: This is the point corresponding to the numerator 7.

(C) $P_1$: Incorrect.

(D) P: Incorrect.

The correct option is (A).

Given:

A circle with center O.

An external point P.

Tangents PA and PB are drawn from P to the circle, where A and B are the points of contact.

The angle between the tangents is $\angle APB = 70^\circ$.

To Find:

The measure of $\angle AOP$, where A is a point of contact.

Solution:

We know that the line segment joining the external point P to the center O ($\text{PO}$) bisects the angle between the two tangents drawn from P.

Substitute the given value of $\angle APB$:

We also know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

So, the radius OA is perpendicular to the tangent PA at A.

Now consider the right-angled triangle $\triangle OAP$. The sum of the angles in a triangle is $180^\circ$.

In $\triangle OAP$, the angles are $\angle AOP$, $\angle OAP$, and $\angle APO$.

Substitute the known values into the equation:

Solve for $\angle AOP$:

Thus, the measure of $\angle AOP$ is $55^\circ$.

Alternate Approach:

Consider the quadrilateral PAOB. The sum of interior angles of a quadrilateral is $360^\circ$.

We know $\angle APB = 70^\circ$, $\angle PAO = 90^\circ$, and $\angle PBO = 90^\circ$.

We know that $\text{PO}$ bisects $\angle AOB$ as well (or $\triangle OAP \cong \triangle OBP$ by RHS congruence, implying $\angle AOP = \angle BOP$).

Both methods confirm that $\angle AOP = 55^\circ$.

Comparing the result with the given options:

(A) $70^\circ$: Incorrect.

(B) $35^\circ$: This is $\angle APO$, not $\angle AOP$.

(C) $110^\circ$: This is $\angle AOB$, not $\angle AOP$.

(D) $55^\circ$: This matches our calculated angle.

The correct option is (D).

Given:

Total length of the road segment = 21 km.

The road segment is to be divided into two parts in the ratio 3:4.

To Find:

The length of the shorter part of the road segment.

Solution:

Let the line segment representing the road be divided into two parts in the ratio $m:n = 3:4$.

The sum of the ratio terms is $3 + 4 = 7$. This means that the total length of the road segment is considered as 7 equal conceptual parts based on the ratio.

The total length of the road segment is 21 km, which corresponds to these 7 parts.

The length of one conceptual part is calculated by dividing the total length by the sum of the ratio terms:

The two parts of the road segment correspond to the ratio terms 3 and 4.

The length of the first part (corresponding to the ratio 3) is $3 \times \text{Length of one part}$.

The length of the second part (corresponding to the ratio 4) is $4 \times \text{Length of one part}$.

The lengths of the two parts are 9 km and 12 km.

We need to find the length of the shorter part. Comparing 9 km and 12 km, the shorter length is 9 km.

The construction method for dividing a line segment in a ratio $m:n$ geometrically achieves this exact partitioning based on the Basic Proportionality Theorem.

Comparing our result with the given options:

(A) 7 km: Incorrect.

(B) 9 km: This matches the length of the shorter part.

(C) 12 km: This is the length of the longer part.

(D) 3 km: This is the length of one conceptual unit based on the sum of the ratio terms, but not the length of one of the two main parts.

The correct option is (B).

We are considering two circles that intersect at two distinct points. This means the circles cross over each other. Let the centers of the circles be $O_1$ and $O_2$ and their radii be $r_1$ and $r_2$. When two circles intersect at two distinct points, the distance between their centers is less than the sum of their radii but greater than the absolute difference of their radii, i.e., $|r_1 - r_2| < O_1O_2 < r_1 + r_2$.

A common tangent is a line that is tangent to both circles. There are two types of common tangents:

• Direct common tangents: These tangents lie on the same side of the line connecting the centers of the two circles.
• Transverse common tangents: These tangents lie on opposite sides of the line connecting the centers of the two circles and intersect the line segment connecting the centers.

Let's consider the case of two circles intersecting at two distinct points.

The region where the two circles intersect is common to both circles. Any line passing through this common region will intersect both circles at least within this region, making it a secant, not a tangent.

Consider the **transverse common tangents**. These tangents would pass through the space between the centers $O_1$ and $O_2$. However, for intersecting circles, this space includes the area of intersection. A line passing through the area of intersection cannot be a tangent to either circle (as it would intersect them at more than one point or pass through the interior). Therefore, no transverse common tangents can be drawn to two intersecting circles.

Consider the **direct common tangents**. These tangents do not pass through the space between the centers; they are on the "outside". For two intersecting circles, we can draw lines that touch the outer edge of both circles without passing through the intersection region. There are exactly two such direct common tangents, one on each side relative to the line connecting the centers.

Thus, for two intersecting circles (not touching), there are 0 transverse common tangents and 2 direct common tangents.

Total number of common tangents = Number of direct common tangents + Number of transverse common tangents

Total number of common tangents = $2 + 0 = 2$.

Comparing this result with the given options:

(A) 1: Incorrect.

(B) 2: Correct.

(C) 3: Incorrect.

(D) 4: Incorrect (4 common tangents exist when circles are disjoint and outside each other).

The correct option is (B).

Let's analyze each construction to see if it typically involves drawing arcs from points on a line segment.

(A) Dividing a line segment in a given ratio: This construction involves drawing a ray AX from one endpoint A, marking points $A_1, A_2, \dots, A_n$ at equal intervals on this ray (which is part of a line), and then drawing parallel lines. The equal intervals are typically marked using a compass by drawing arcs of the same radius from the previous point on the ray. Thus, arcs are drawn from points ($A, A_1, A_2, \dots$) on the ray AX.

(B) Constructing a similar triangle: This construction is based on dividing a line segment in a given ratio and drawing parallel lines. It involves drawing a ray from a vertex, marking points on the ray (points on a line segment), joining a point to a vertex of the original triangle, and drawing a parallel line. Drawing parallel lines often involves copying angles or using perpendiculars, both of which use arcs drawn from points on lines or line segments. Also, marking equal divisions on the ray uses arcs from points on the ray.

(C) Constructing a tangent at a point on the circle (by drawing a perpendicular): To draw a tangent at a point P on a circle, we draw the radius OP and construct a line perpendicular to OP at P. Constructing a perpendicular at a point P on a line involves drawing arcs from P to intersect the line on both sides, and then drawing arcs of a larger radius from these two intersection points (which are points on the line segment OP or its extension) to find the direction of the perpendicular. This method directly involves drawing arcs from points on the line segment (or line) OP.

(D) Constructing tangents from an external point: This construction involves joining the external point P to the center O, finding the midpoint M of OP, and drawing a circle with center M and radius MO. The points where this circle intersects the given circle are the points of tangency. While finding the midpoint M of OP (a line segment) typically involves drawing arcs from the endpoints P and O, the subsequent crucial step to find the tangent points is the intersection of the circle centered at M with the original circle. We are not drawing arcs *from points on the line segment OP* to directly find the tangent points A and B in the same way as finding a perpendicular or a parallel line.

Based on the analysis, constructions (A), (B), and (C) directly involve using arcs drawn from points located on a line segment (or a ray/line) as a core step to determine the required line or point. Construction (D), while using arcs to find the midpoint of OP, finds the tangent points through the intersection of two circles, one of which is centered at the midpoint M of OP and has OP as its diameter. The tangent points are defined by this intersection, not directly by drawing arcs from points on the segment OP to locate them.

Therefore, the construction that does NOT involve drawing arcs *from points on a line segment* in the same fundamental way as the others is constructing tangents from an external point.

The correct option is (D).

Given:

Two tangents are drawn from an external point to a circle.

The lengths of the two tangents are 10 cm and $x$ cm.

To Find:

The value of $x$.

Solution:

We use a fundamental theorem regarding tangents drawn from an external point to a circle. This theorem states:

Theorem: The lengths of tangents drawn from an external point to a circle are equal.

In this problem, we have an external point from which two tangents are drawn to a circle. Let the external point be P, and let the points of contact of the tangents be A and B. The lengths of the tangents are PA and PB.

According to the theorem, the lengths of these two tangents must be equal.

Therefore, the value of $x$ must be equal to 10 cm.

Comparing this result with the given options:

(A) Equal to 10 cm: This matches our conclusion.

(B) Greater than 10 cm: Incorrect, as tangent lengths must be equal.

(C) Less than 10 cm: Incorrect, as tangent lengths must be equal.

(D) Can be any value: Incorrect, as the value is fixed by the theorem.

The correct option is (A).

Let's review the typical steps for each construction method listed in the options to determine which one involves finding the midpoint of a line segment.

(A) Dividing a line segment in a given ratio (general case $m:n$): This construction typically involves drawing a ray from one endpoint, marking $m+n$ equal divisions on the ray using a compass, joining the $(m+n)^{th}$ division mark to the other endpoint of the segment, and then drawing a line parallel to this joining line through the $m^{th}$ division mark. This method divides the segment in the ratio $m:n$ but does not require finding the midpoint of the original segment or any other segment as a standard step, unless the ratio is specifically 1:1 (dividing the segment into two equal halves, which is finding the midpoint). However, the question refers to the general case.

(B) Constructing a similar triangle: This construction is often based on the principle used for dividing a line segment in a ratio or using angle properties. It typically involves drawing a ray, marking divisions based on the scale factor, and drawing parallel lines. Finding a midpoint is not a standard required step in the general construction of a similar triangle.

(C) Constructing a tangent from an external point: The standard method for constructing tangents from an external point P to a circle with center O is as follows:

1. Join the external point P to the center O.

2. Find the midpoint M of the line segment PO by performing a perpendicular bisector construction on PO (which yields the midpoint). Alternatively, measure the length of PO and mark the halfway point.

3. With M as the center and radius MO (or MP), draw a circle.

4. The points where this circle intersects the original circle are the points of tangency.

In this construction, finding the midpoint of the line segment OP is a necessary and explicit step.

(D) Constructing a tangent at a point on the circle: To construct a tangent at a point P on the circle with center O, we draw the radius OP and construct a line perpendicular to OP at P. Constructing a perpendicular at a point on a line segment (or line) involves drawing arcs from the point and then from points equidistant from the point on the line. This method constructs a perpendicular line but does not involve finding the midpoint of the radius OP or any other segment as its primary step.

Based on the analysis of the standard construction methods, the construction of tangents from an external point explicitly requires finding the midpoint of the line segment joining the external point and the center of the circle.

The correct option is (C).

Given:

Two similar triangular plots.

Ratio of corresponding sides of the smaller plot to the larger plot = 3:5.

Perimeter of the smaller plot = 60 metres.

To Find:

The perimeter of the larger plot.

Solution:

For two similar figures, the ratio of their corresponding perimeters is equal to the ratio of their corresponding sides.

Let $P_s$ be the perimeter of the smaller plot and $P_l$ be the perimeter of the larger plot.

Let the ratio of corresponding sides (smaller to larger) be $k = \frac{3}{5}$.

According to the property of similar figures:

We are given that $P_s = 60$ metres. Substitute this value into the equation:

Now, we can solve for $P_l$ by cross-multiplying:

The perimeter of the larger plot is 100 metres.

Comparing this result with the given options:

(A) 60 metres: Perimeter of the smaller plot.

(B) 100 metres: Matches our calculated perimeter for the larger plot.

(C) 36 metres: This would correspond to a scale factor of 3/5 applied to the smaller plot (60 * 3/5 = 36), which is incorrect for finding the larger plot's perimeter.

(D) 150 metres: This would correspond to a scale factor of 5/3 applied to the smaller plot (60 * 5/3 = 100) - wait, the ratio is smaller to larger is 3:5. So larger to smaller is 5:3. Let's double-check the calculation. $P_l = 60 \times (5/3) = 20 \times 5 = 100$. The calculation is correct.

The correct option is (B).

Given:

A line segment of length $7.6 \text{ cm}$.

Ratio to divide the line segment is $5:8$.

To Divide:

The given line segment in the ratio $5:8$.

Construction Steps:

1. Draw a line segment AB of length $7.6 \text{ cm}$.

2. Draw a ray AX making an acute angle with AB.

3. Locate $(5+8) = 13$ points, say $A_1, A_2, ..., A_{13}$, on the ray AX such that $AA_1 = A_1A_2 = ... = A_{12}A_{13}$.

4. Join $BA_{13}$.

5. Through the point $A_5$ (since the first part of the ratio is 5), draw a line parallel to $BA_{13}$ intersecting AB at point C.

6. The point C divides the line segment AB in the ratio $5:8$. Thus, AC : CB = $5:8$.

Measurement:

The point C divides the line segment AB such that AC : CB = $5:8$.

The total length of AB is $7.6 \text{ cm}$.

The sum of the parts of the ratio is $5+8=13$.

Length of the first part, AC, should be $\frac{5}{13}$ of the total length.

AC = $\frac{5}{13} \times 7.6 \text{ cm} = \frac{38}{13} \text{ cm}$.

Calculating the value:

$38 \div 13 \approx 2.92 \text{ cm}$.

Length of the second part, CB, should be $\frac{8}{13}$ of the total length.

CB = $\frac{8}{13} \times 7.6 \text{ cm} = \frac{60.8}{13} \text{ cm}$.

Calculating the value:

$60.8 \div 13 \approx 4.68 \text{ cm}$.

Upon measuring the two parts, you should find that AC $\approx 2.9 \text{ cm}$ and CB $\approx 4.7 \text{ cm}$ (or more precisely, AC $\approx 2.92 \text{ cm}$ and CB $\approx 4.68 \text{ cm}$), which sum up to the total length $2.9 + 4.7 = 7.6 \text{ cm}$.

Given:

A triangle ABC with side lengths AB = $5 \text{ cm}$, BC = $6 \text{ cm}$, and AC = $7 \text{ cm}$.

A scale factor of $\frac{2}{3}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{2}{3}$ of the corresponding sides of $\triangle \text{ABC}$.

Construction Steps:

1. Construct $\triangle \text{ABC}$ with the given side lengths. Draw a line segment BC of length $6 \text{ cm}$. With B as center and radius $5 \text{ cm}$, draw an arc. With C as center and radius $7 \text{ cm}$, draw another arc intersecting the previous arc at A. Join AB and AC to form $\triangle \text{ABC}$.

2. Draw a ray BX extending downwards from B, making an acute angle with BC.

3. Since the scale factor is $\frac{2}{3}$ (which is less than 1), locate 3 points (the greater of the numerator and denominator), say $B_1, B_2, B_3$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3$.

4. Join the point $B_3$ (corresponding to the denominator of the ratio) to C.

5. Through the point $B_2$ (corresponding to the numerator of the ratio), draw a line parallel to $B_3C$ by making an angle equal to $\angle BB_3C$ at $B_2$. Let this line intersect BC at C'.

6. Through C', draw a line parallel to AC by making an angle equal to $\angle BCA$ at C'. Let this line intersect AB at A'.

7. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line through C' is parallel to AC.

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals AB and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle \text{BB}_3\text{C}$, the line $B_2C'$ is parallel to $B_3C$.

By the Basic Proportionality Theorem (BPT), $\frac{\text{BB}_2}{\text{BB}_3} = \frac{\text{BC'}}{\text{BC}}$.

By construction, $\frac{\text{BB}_2}{\text{BB}_3} = \frac{2}{3}$.

Therefore, $\frac{\text{BC'}}{\text{BC}} = \frac{2}{3}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{2}{3}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{2}{3}$ of the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with center O and radius $3 \text{ cm}$.

An external point P at a distance of $7 \text{ cm}$ from the center O.

To Construct:

The pair of tangents from point P to the circle and measure their lengths.

Construction Steps:

1. Draw a circle with center O and radius $3 \text{ cm}$.

2. Mark a point P at a distance of $7 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Measurement:

Measure the length of the tangent segments PT and PT' using a ruler. You will find that PT $\approx 6.3 \text{ cm}$ and PT' $\approx 6.3 \text{ cm}$. The lengths of the tangents from an external point to a circle are equal.

Verification:

We know that the radius through the point of contact is perpendicular to the tangent at that point. Therefore, $\angle \text{OTP} = 90^\circ$ and $\angle \text{OT'P} = 90^\circ$.

Consider the right-angled triangle $\triangle \text{OTP}$.

By the Pythagorean theorem, we have:

$OP^2 = OT^2 + PT^2$

Given $OP = 7 \text{ cm}$ and $OT = 3 \text{ cm}$ (radius).

$7^2 = 3^2 + PT^2$

$49 = 9 + PT^2$

$PT^2 = 49 - 9$

$PT^2 = 40$

$PT = \sqrt{40} \text{ cm}$

$PT = \sqrt{4 \times 10} \text{ cm} = 2\sqrt{10} \text{ cm}$.

Since $\sqrt{10} \approx 3.162$,

$PT \approx 2 \times 3.162 \text{ cm} \approx 6.324 \text{ cm}$.

The measured length is close to the calculated value, verifying the construction.

Given:

A line segment MN of length $6 \text{ cm}$.

Ratio to divide the line segment internally is $3:2$.

To Divide:

The line segment MN internally in the ratio $3:2$.

Construction Steps:

1. Draw a line segment MN of length $6 \text{ cm}$.

2. Draw a ray MX extending downwards from M, making an acute angle with MN.

3. Locate $(3+2) = 5$ points, say $M_1, M_2, M_3, M_4, M_5$, on the ray MX such that $MM_1 = M_1M_2 = M_2M_3 = M_3M_4 = M_4M_5$. (The number of points is the sum of the ratio parts).

4. Join the point $M_5$ (corresponding to the sum of the ratio parts) to N.

5. Through the point $M_3$ (corresponding to the first part of the ratio, 3), draw a line parallel to $M_5N$. To do this, construct an angle equal to $\angle MM_5N$ at $M_3$. Let this line intersect MN at point P.

6. The point P divides the line segment MN internally in the ratio $3:2$. Thus, MP : PN = $3:2$.

7. Upon measurement, you will find that MP $\approx 3.6 \text{ cm}$ and PN $\approx 2.4 \text{ cm}$.

Justification:

By construction, $M_3P$ is parallel to $M_5N$.

In $\triangle MM_5N$, by the Basic Proportionality Theorem (BPT), since $M_3P \parallel M_5N$, we have:

$\frac{MM_3}{M_3M_5} = \frac{MP}{PN}$

By construction, the points $M_1, M_2, ..., M_5$ are marked such that $MM_1 = M_1M_2 = ... = M_4M_5$.

So, $MM_3$ consists of 3 equal parts and $M_3M_5$ consists of 2 equal parts ($M_3M_4$ and $M_4M_5$).

Therefore, $\frac{MM_3}{M_3M_5} = \frac{3}{2}$.

From the BPT, $\frac{MP}{PN} = \frac{MM_3}{M_3M_5} = \frac{3}{2}$.

Thus, the point P divides MN in the ratio $3:2$.

Given:

A triangle PQR with side lengths PQ = $4 \text{ cm}$, QR = $5 \text{ cm}$, and PR = $6 \text{ cm}$.

A scale factor of $\frac{5}{4}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{PQR}$ whose sides are $\frac{5}{4}$ times the corresponding sides of $\triangle \text{PQR}$.

Construction Steps:

1. Construct $\triangle \text{PQR}$ with the given side lengths. Draw a line segment QR of length $5 \text{ cm}$. With Q as center and radius $4 \text{ cm}$, draw an arc. With R as center and radius $6 \text{ cm}$, draw another arc intersecting the previous arc at P. Join PQ and PR to form $\triangle \text{PQR}$.

2. Draw a ray QX extending downwards from Q, making an acute angle with QR.

3. Since the scale factor is $\frac{5}{4}$, locate 5 points (the greater of the numerator and denominator), say $Q_1, Q_2, Q_3, Q_4, Q_5$, on the ray QX such that $QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5$.

4. Join the point $Q_4$ (corresponding to the denominator of the ratio) to R.

5. Through the point $Q_5$ (corresponding to the numerator of the ratio), draw a line parallel to $Q_4R$ by making an angle equal to $\angle QQ_4R$ at $Q_5$. Let this line intersect the extended line segment QR at R'.

6. Through R', draw a line parallel to PR by making an angle equal to $\angle QRP$ at R'. Let this line intersect the extended line segment QP at P'.

7. $\triangle \text{P'QR'}$ is the required triangle similar to $\triangle \text{PQR}$.

Justification:

By construction, the line through R' is parallel to PR ($P'R' \parallel PR$).

Therefore, $\triangle \text{P'QR'}$ is similar to $\triangle \text{PQR}$ by the AA similarity criterion ( $\angle \text{Q}$ is common to both triangles and $\angle \text{QP'R'} = \angle \text{QPR}$ and $\angle \text{QR'P'} = \angle \text{QRP}$ as corresponding angles between parallel lines P'R' and PR with transversals QP and QR respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{QP'}}{\text{QP}} = \frac{\text{QR'}}{\text{QR}} = \frac{\text{P'R'}}{\text{PR}}$.

In $\triangle QQ_5R'$, the line $Q_4R$ is parallel to $Q_5R'$.

By the Basic Proportionality Theorem (BPT), $\frac{QQ_4}{QQ_5} = \frac{QR}{QR'}$.

By construction, $\frac{QQ_4}{QQ_5} = \frac{4}{5}$.

Therefore, $\frac{QR}{QR'} = \frac{4}{5}$, which implies $\frac{QR'}{QR} = \frac{5}{4}$.

Hence, $\frac{\text{QP'}}{\text{QP}} = \frac{\text{QR'}}{\text{QR}} = \frac{\text{P'R'}}{\text{PR}} = \frac{5}{4}$.

Thus, the constructed triangle $\triangle \text{P'QR'}$ has sides which are $\frac{5}{4}$ times the corresponding sides of $\triangle \text{PQR}$.

Given:

A circle with center O and radius $4 \text{ cm}$.

A point P on the circle.

To Construct:

A tangent to the circle at the point P.

Construction Steps:

1. Draw a circle with center O and radius $4 \text{ cm}$.

2. Mark any point P on the circumference of the circle.

3. Join the center O to the point P. This line segment OP is the radius through the point of contact.

4. Draw a line perpendicular to the radius OP at the point P. To do this, extend OP to a point Q such that O, P, and Q are collinear.

5. With P as center and any suitable radius, draw arcs intersecting the line segment OQ (extended) at two points, say A and B.

6. With A as center and a radius greater than AP, draw an arc above (or below) the line OQ.

7. With B as center and the same radius, draw another arc intersecting the previous arc at a point R.

8. Join PR. The line PR is the required tangent to the circle at the point P.

Justification:

By construction, the line PR is perpendicular to the radius OP at the point P on the circle.

We know that a line drawn perpendicular to the radius through the point of contact is a tangent to the circle.

Therefore, PR is the tangent to the circle at P.

Given:

A line segment of length $9 \text{ cm}$.

Ratio to divide the line segment is $4:3$.

To Divide:

The given line segment in the ratio $4:3$ and measure the two parts.

Construction Steps:

1. Draw a line segment AB of length $9 \text{ cm}$.

2. Draw a ray AX making an acute angle with AB.

3. Locate $(4+3) = 7$ points, say $A_1, A_2, ..., A_{7}$, on the ray AX such that $AA_1 = A_1A_2 = ... = A_{6}A_{7}$.

4. Join $BA_{7}$.

5. Through the point $A_4$ (since the first part of the ratio is 4), draw a line parallel to $BA_{7}$ intersecting AB at point C. To do this, construct an angle equal to $\angle AA_{7}B$ at $A_4$.

6. The point C divides the line segment AB in the ratio $4:3$. Thus, AC : CB = $4:3$.

Measurement:

The point C divides the line segment AB such that AC : CB = $4:3$.

The total length of AB is $9 \text{ cm}$.

The sum of the parts of the ratio is $4+3=7$.

Length of the first part, AC, should be $\frac{4}{7}$ of the total length.

AC = $\frac{4}{7} \times 9 \text{ cm} = \frac{36}{7} \text{ cm}$.

Calculating the value:

AC $\approx 5.14 \text{ cm}$.

Length of the second part, CB, should be $\frac{3}{7}$ of the total length.

CB = $\frac{3}{7} \times 9 \text{ cm} = \frac{27}{7} \text{ cm}$.

Calculating the value:

CB $\approx 3.86 \text{ cm}$.

Upon measuring the two parts, you will find that AC $\approx 5.1 \text{ cm}$ (or $5.2 \text{ cm}$) and CB $\approx 3.9 \text{ cm}$ (or $3.8 \text{ cm}$), which sum up to the total length $9 \text{ cm}$.

Given:

An isosceles triangle with base $8 \text{ cm}$ and altitude $4 \text{ cm}$.

A scale factor of $1\frac{1}{2} = \frac{3}{2}$ for the sides of the similar triangle.

To Construct:

An isosceles triangle with given dimensions, and a triangle similar to it whose sides are $\frac{3}{2}$ times the corresponding sides of the original triangle.

Construction Steps:

1. Construct the isosceles triangle. Draw a line segment BC of length $8 \text{ cm}$.

2. Draw the perpendicular bisector of BC. Let M be the midpoint of BC. M is $4 \text{ cm}$ from B and C.

3. Along the perpendicular bisector from M, mark a point A such that AM = $4 \text{ cm}$ (the altitude). AM is perpendicular to BC.

4. Join AB and AC. $\triangle \text{ABC}$ is the required isosceles triangle with base BC = $8 \text{ cm}$ and altitude AM = $4 \text{ cm}$.

5. Draw a ray BX extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{3}{2}$, locate 3 points (the greater of the numerator and denominator), say $B_1, B_2, B_3$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3$.

7. Join the point $B_2$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_3$ (corresponding to the numerator of the ratio), draw a line parallel to $B_2C$ by making an angle equal to $\angle BB_2C$ at $B_3$. Let this line intersect the extended line segment BC at C'.

9. Through C', draw a line parallel to AC by making an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line through C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_3C'$, the line $B_2C$ is parallel to $B_3C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_2}{BB_3} = \frac{BC}{BC'}$.

By construction, $\frac{BB_2}{BB_3} = \frac{2}{3}$.

Therefore, $\frac{BC}{BC'} = \frac{2}{3}$, which implies $\frac{BC'}{BC} = \frac{3}{2}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{3}{2}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{3}{2}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with center O and radius $2.5 \text{ cm}$.

An external point P at a distance of $6 \text{ cm}$ from the center O.

To Construct:

The pair of tangents from point P to the circle.

Construction Steps:

1. Draw a circle with center O and radius $2.5 \text{ cm}$.

2. Mark a point P at a distance of $6 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Justification:

The circle with center M and radius MO passes through O and P.

The points of intersection T and T' lie on both circles.

Consider the line segment OT and PT. OT is the radius of the original circle.

The angle $\angle \text{OTP}$ is an angle in the semicircle of the circle centered at M.

Therefore, $\angle \text{OTP} = 90^\circ$.

Since OT is the radius and PT is perpendicular to OT at the point of contact T, PT is a tangent to the circle.

Similarly, $\angle \text{OT'P} = 90^\circ$, and PT' is also a tangent to the circle.

Given:

A line segment AB of length $8 \text{ cm}$.

Ratio to divide the line segment internally is $2:5$.

To Divide:

The line segment AB internally in the ratio $2:5$ at point C, and verify the ratio by measurement.

Construction Steps:

1. Draw a line segment AB of length $8 \text{ cm}$.

2. Draw a ray AX extending downwards from A, making an acute angle with AB.

3. Locate $(2+5) = 7$ points, say $A_1, A_2, ..., A_{7}$, on the ray AX such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$. (The number of points is the sum of the ratio parts).

4. Join the point $A_{7}$ (corresponding to the sum of the ratio parts) to B.

5. Through the point $A_2$ (corresponding to the first part of the ratio, 2), draw a line parallel to $A_{7}B$. To do this, construct an angle equal to $\angle AA_{7}B$ at $A_2$. Let this line intersect AB at point C.

6. The point C divides the line segment AB internally in the ratio $2:5$. Thus, AC : CB = $2:5$.

Measurement and Verification:

The point C divides the line segment AB such that AC : CB = $2:5$.

The total length of AB is $8 \text{ cm}$.

The sum of the parts of the ratio is $2+5=7$.

Length of the first part, AC, should be $\frac{2}{7}$ of the total length.

AC = $\frac{2}{7} \times 8 \text{ cm} = \frac{16}{7} \text{ cm}$.

Calculating the value:

AC $\approx 2.29 \text{ cm}$.

Length of the second part, CB, should be $\frac{5}{7}$ of the total length.

CB = $\frac{5}{7} \times 8 \text{ cm} = \frac{40}{7} \text{ cm}$.

Calculating the value:

CB $\approx 5.71 \text{ cm}$.

Upon measuring the two parts, you will find that AC is approximately $2.3 \text{ cm}$ and CB is approximately $5.7 \text{ cm}$.

The sum of the measured lengths is $2.3 + 5.7 = 8.0 \text{ cm}$, which is the original length of AB.

The ratio of the measured lengths is AC : CB $\approx 2.3 : 5.7$. While not exactly $2:5$ due to measurement inaccuracies, it is close. The calculated ratio $\frac{2.3}{5.7} \approx 0.4035$ and the required ratio $\frac{2}{5} = 0.4$. The measured lengths are approximately in the ratio $2:5$, verifying the construction.

Justification:

By construction, $A_2C$ is parallel to $A_{7}B$.

In $\triangle AA_{7}B$, by the Basic Proportionality Theorem (BPT), since $A_2C \parallel A_{7}B$, we have:

$\frac{AA_2}{A_2A_{7}} = \frac{AC}{CB}$

By construction, the points $A_1, A_2, ..., A_{7}$ are marked such that $AA_1 = A_1A_2 = ... = A_6A_{7}$.

So, $AA_2$ consists of 2 equal parts and $A_2A_{7}$ consists of 5 equal parts ($A_2A_3, A_3A_4, A_4A_5, A_5A_6, A_6A_{7}$).

Therefore, $\frac{AA_2}{A_2A_{7}} = \frac{2}{5}$.

From the BPT, $\frac{AC}{CB} = \frac{AA_2}{A_2A_{7}} = \frac{2}{5}$.

Thus, the point C divides AB in the ratio $2:5$.

Given:

A triangle ABC with BC = $7 \text{ cm}$, $\angle \text{B} = 45^\circ$, and $\angle \text{A} = 105^\circ$.

A scale factor of $\frac{4}{3}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle \text{ABC}$.

Calculation of $\angle \text{C}$ in $\triangle \text{ABC}$:

The sum of angles in a triangle is $180^\circ$.

$\angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ$

$105^\circ + 45^\circ + \angle \text{C} = 180^\circ$

$150^\circ + \angle \text{C} = 180^\circ$

$\angle \text{C} = 180^\circ - 150^\circ$

$\angle \text{C} = 30^\circ$.

Construction Steps:

1. Construct $\triangle \text{ABC}$ with the given measures. Draw a line segment BC of length $7 \text{ cm}$.

2. At point B, construct an angle of $45^\circ$ using a protractor or compass and ruler. Draw a ray BY.

3. At point C, construct an angle of $30^\circ$ using a protractor or compass and ruler. Draw a ray CZ.

4. The rays BY and CZ intersect at point A, forming $\triangle \text{ABC}$. (Check that $\angle \text{A}$ is $105^\circ$).

5. Draw a ray BX extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{4}{3}$, locate 4 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

7. Join the point $B_3$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_4$ (corresponding to the numerator of the ratio), draw a line parallel to $B_3C$. To do this, construct an angle equal to $\angle BB_3C$ at $B_4$. Let this line intersect the extended line segment BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line through C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_4C'$, the line $B_3C$ is parallel to $B_4C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_3}{BB_4} = \frac{BC}{BC'}$.

By construction, $\frac{BB_3}{BB_4} = \frac{3}{4}$.

Therefore, $\frac{BC}{BC'} = \frac{3}{4}$, which implies $\frac{BC'}{BC} = \frac{4}{3}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{4}{3}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{4}{3}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with radius $5 \text{ cm}$.

A pair of tangents inclined to each other at an angle of $60^\circ$.

To Construct:

A pair of tangents to the given circle such that the angle between them is $60^\circ$.

Relationship between the angle between tangents and the angle at the center:

Let O be the center of the circle and P be the external point from which the tangents are drawn. Let T and T' be the points of contact of the tangents from P.

The angle between the tangents is $\angle \text{TPT'} = 60^\circ$.

In the quadrilateral OTPT', we have $\angle \text{OTP} = 90^\circ$ and $\angle \text{OT'P} = 90^\circ$ (radius is perpendicular to the tangent at the point of contact).

The sum of angles in a quadrilateral is $360^\circ$.

So, $\angle \text{TOT'} + \angle \text{OTP} + \angle \text{OT'P} + \angle \text{TPT'} = 360^\circ$

$\angle \text{TOT'} + 90^\circ + 90^\circ + 60^\circ = 360^\circ$

$\angle \text{TOT'} + 240^\circ = 360^\circ$

$\angle \text{TOT'} = 360^\circ - 240^\circ = 120^\circ$.

Thus, the angle between the radii to the points of contact must be $120^\circ$.

Construction Steps:

1. Draw a circle with center O and radius $5 \text{ cm}$.

2. Draw any radius of the circle, say OT.

3. At the center O, construct an angle $\angle \text{TOT'} = 120^\circ$ such that OT' is another radius of the circle.

4. At point T, construct a line perpendicular to OT. To do this, draw an arc with center T, mark points on OT, and construct a $90^\circ$ angle at T.

5. At point T', construct a line perpendicular to OT'. Similarly, construct a $90^\circ$ angle at T'.

6. Let the two perpendicular lines intersect at point P.

7. PT and PT' are the required tangents inclined at an angle of $60^\circ$ to each other.

Justification:

By construction, $\angle \text{OTP} = 90^\circ$ and $\angle \text{OT'P} = 90^\circ$. Since these angles are formed by the lines PT and PT' with the radii at the points on the circle, PT and PT' are tangents to the circle.

Also by construction, $\angle \text{TOT'} = 120^\circ$.

In quadrilateral OTPT', the sum of angles is $360^\circ$.

$\angle \text{TPT'} = 360^\circ - (\angle \text{OTP} + \angle \text{OT'P} + \angle \text{TOT'})$

$\angle \text{TPT'} = 360^\circ - (90^\circ + 90^\circ + 120^\circ)$

$\angle \text{TPT'} = 360^\circ - 300^\circ$

$\angle \text{TPT'} = 60^\circ$.

Thus, the tangents are inclined to each other at an angle of $60^\circ$, as required.

Given:

A line segment of length $10 \text{ cm}$.

To divide it into 5 equal parts.

To Divide:

The given line segment into 5 equal parts using construction.

Construction Steps:

1. Draw a line segment AB of length $10 \text{ cm}$.

2. Draw a ray AX extending downwards from A, making an acute angle with AB.

3. Locate 5 points, say $A_1, A_2, A_3, A_4, A_5$, on the ray AX such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$. (You can use any convenient length for these divisions, but they must be equal).

4. Join the point $A_5$ to B.

5. Through the points $A_1, A_2, A_3, A_4$, draw lines parallel to $A_5B$, intersecting the line segment AB at points $C_1, C_2, C_3, C_4$ respectively. To do this, construct angles equal to $\angle AA_5B$ at $A_1, A_2, A_3, A_4$.

6. The points $C_1, C_2, C_3, C_4$ divide the line segment AB into 5 equal parts: $AC_1 = C_1C_2 = C_2C_3 = C_3C_4 = C_4B$.

Justification:

By construction, $A_1C_1 \parallel A_2C_2 \parallel A_3C_3 \parallel A_4C_4 \parallel A_5B$.

We have transversal AX intersecting these parallel lines, and the segments on AX are equal ($AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$).

According to the theorem which states that if a transversal makes equal intercepts on three or more parallel lines, then any other transversal intersecting them will also make equal intercepts, the points $C_1, C_2, C_3, C_4$ divide the line segment AB into 5 equal parts.

Thus, $AC_1 = C_1C_2 = C_2C_3 = C_3C_4 = C_4B$.

The length of each part is $\frac{10}{5} = 2 \text{ cm}$.

Upon measurement, you will find that $AC_1, C_1C_2, C_2C_3, C_3C_4$, and $C_4B$ are each approximately $2 \text{ cm}$ long.

Given:

A triangle ABC with BC = $6 \text{ cm}$, AB = $5 \text{ cm}$, and $\angle \text{B} = 60^\circ$.

A similar triangle PQR such that $\text{PQ} = \frac{3}{5} \text{AB}$. This implies the scale factor of $\triangle \text{PQR}$ with respect to $\triangle \text{ABC}$ is $\frac{3}{5}$.

To Construct:

A triangle PQR similar to $\triangle \text{ABC}$ such that the ratio of corresponding sides is $\frac{\text{PQ}}{\text{AB}} = \frac{3}{5}$.

(Note: Since $\triangle \text{PQR} \sim \triangle \text{ABC}$ and PQ corresponds to AB, QR to BC, PR to AC, the required triangle PQR will have its vertex Q at B, vertex P on AB, and vertex R on BC).

Construction Steps:

1. Construct $\triangle \text{ABC}$ with the given measures. Draw a line segment BC of length $6 \text{ cm}$.

2. At point B, construct an angle of $60^\circ$ using a compass and ruler. Draw a ray BY.

3. With B as center and radius $5 \text{ cm}$, draw an arc intersecting the ray BY at point A.

4. Join AC to form $\triangle \text{ABC}$.

5. Draw a ray BX extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{3}{5}$, locate 5 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4, B_5$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

7. Join the point $B_5$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_3$ (corresponding to the numerator of the ratio), draw a line parallel to $B_5C$. To do this, construct an angle equal to $\angle BB_5C$ at $B_3$. Let this line intersect BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect AB at A'.

10. $\triangle \text{A'BC'}$ is the required triangle. Here, vertex Q is B, vertex P is A', and vertex R is C'. Thus, $\triangle \text{A'BC'}$ corresponds to $\triangle \text{PQR}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_5C$, the line $B_3C'$ is parallel to $B_5C$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_3}{BB_5} = \frac{BC'}{BC}$.

By construction, $\frac{BB_3}{BB_5} = \frac{3}{5}$.

Therefore, $\frac{BC'}{BC} = \frac{3}{5}$.

Since $\triangle \text{A'BC'} \sim \triangle \text{ABC}$, the ratio of all corresponding sides is the same.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{3}{5}$.

As P corresponds to A', Q to B, and R to C', we have $\text{PQ} = \text{A'B}$, $\text{QR} = \text{BC'}$, and $\text{PR} = \text{A'C'}$.

Thus, $\frac{\text{PQ}}{\text{AB}} = \frac{\text{QR}}{\text{BC}} = \frac{\text{PR}}{\text{AC}} = \frac{3}{5}$.

This confirms that the constructed triangle $\triangle \text{A'BC'}$ (i.e., $\triangle \text{PQR}$) has sides which are $\frac{3}{5}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with diameter $8 \text{ cm}$.

A point P at a distance of $5 \text{ cm}$ from the circumference of the circle.

To Construct:

The pair of tangents from point P to the circle.

Calculation of distance of P from the center:

The radius of the circle is half of the diameter.

Radius = $\frac{\text{Diameter}}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm}$.

The distance of the point P from the center O is the sum of the radius and the distance from the circumference.

Distance OP = Radius + Distance from circumference

Distance OP = $4 \text{ cm} + 5 \text{ cm} = 9 \text{ cm}$.

So, the point P is $9 \text{ cm}$ away from the center O.

Construction Steps:

1. Draw a circle with center O and radius $4 \text{ cm}$.

2. Mark a point P at a distance of $9 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle (centered at O) at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Justification:

The circle with center M and radius MO passes through O and P.

The points of intersection T and T' lie on both circles.

Consider the line segment OT and PT. OT is the radius of the original circle.

The angle $\angle \text{OTP}$ is an angle inscribed in the semicircle of the circle centered at M (since OP is the diameter of this circle).

Therefore, $\angle \text{OTP} = 90^\circ$.

Since OT is the radius and PT is perpendicular to OT at the point of contact T, PT is a tangent to the circle by the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Similarly, $\angle \text{OT'P} = 90^\circ$, and PT' is also a tangent to the circle.

Given:

A line segment of length $7 \text{ cm}$.

Ratio to divide the line segment is $3:4$.

To Divide:

The given line segment in the ratio $3:4$ and justify the construction.

Construction Steps:

1. Draw a line segment AB of length $7 \text{ cm}$.

2. Draw a ray AX making an acute angle with AB.

3. Locate $(3+4) = 7$ points, say $A_1, A_2, A_3, A_4, A_5, A_6, A_{7}$, on the ray AX such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$. (You can use any convenient length for these divisions, but they must be equal).

4. Join the point $A_{7}$ (corresponding to the sum of the ratio parts) to B.

5. Through the point $A_3$ (corresponding to the first part of the ratio, 3), draw a line parallel to $A_{7}B$ intersecting AB at point C. To do this, construct an angle equal to $\angle AA_{7}B$ at $A_3$.

6. The point C divides the line segment AB in the ratio $3:4$. Thus, AC : CB = $3:4$.

Justification:

By construction, $A_3C$ is parallel to $A_{7}B$.

In $\triangle AA_{7}B$, by the Basic Proportionality Theorem (BPT), since $A_3C \parallel A_{7}B$, we have:

$\frac{AA_3}{A_3A_{7}} = \frac{AC}{CB}$

By construction, the points $A_1, A_2, ..., A_{7}$ are marked such that $AA_1 = A_1A_2 = ... = A_6A_{7}$. Let the length of each small segment be $k$.

Then $AA_3 = 3 \times k$ and $A_3A_{7} = A_3A_4 + A_4A_5 + A_5A_6 + A_6A_7 = k + k + k + k = 4k$.

Therefore, $\frac{AA_3}{A_3A_{7}} = \frac{3k}{4k} = \frac{3}{4}$.

From the BPT, $\frac{AC}{CB} = \frac{AA_3}{A_3A_{7}} = \frac{3}{4}$.

Thus, the point C divides the line segment AB internally in the ratio $3:4$.

Given:

A triangle with side lengths $5 \text{ cm}$, $6 \text{ cm}$, and $7 \text{ cm}$.

A scale factor of $\frac{7}{5}$ for the sides of the similar triangle.

To Construct:

A triangle with the given side lengths, and a triangle similar to it whose sides are $\frac{7}{5}$ times the corresponding sides of the first triangle.

Construction Steps:

1. Construct the first triangle. Draw a line segment BC of length $6 \text{ cm}$. With B as center and radius $5 \text{ cm}$, draw an arc. With C as center and radius $7 \text{ cm}$, draw another arc intersecting the previous arc at A. Join AB and AC to form $\triangle \text{ABC}$.

2. Draw a ray BX extending downwards from B, making an acute angle with BC.

3. Since the scale factor is $\frac{7}{5}$, locate 7 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4, B_5, B_6, B_7$, on the ray BX such that $BB_1 = B_1B_2 = ... = B_6B_7$.

4. Join the point $B_5$ (corresponding to the denominator of the ratio) to C.

5. Through the point $B_7$ (corresponding to the numerator of the ratio), draw a line parallel to $B_5C$. To do this, construct an angle equal to $\angle BB_5C$ at $B_7$. Let this line intersect the extended line segment BC at C'.

6. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

7. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line through C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_7C'$, the line $B_5C$ is parallel to $B_7C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_5}{BB_7} = \frac{BC}{BC'}$.

By construction, $\frac{BB_5}{BB_7} = \frac{5}{7}$.

Therefore, $\frac{BC}{BC'} = \frac{5}{7}$, which implies $\frac{BC'}{BC} = \frac{7}{5}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{7}{5}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{7}{5}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with center O and radius $3.5 \text{ cm}$.

To Construct:

A pair of tangents to the circle which are parallel to each other.

Construction Steps:

1. Draw a circle with center O and radius $3.5 \text{ cm}$.

2. Draw any diameter of the circle. Let the endpoints of the diameter be P and Q.

3. At point P, construct a line perpendicular to the diameter PQ. To do this, draw an arc with center P, mark points on the extension of PQ, and construct a $90^\circ$ angle at P.

4. At point Q, construct a line perpendicular to the diameter PQ. Similarly, construct a $90^\circ$ angle at Q.

5. Let the line through P be $l$ and the line through Q be $m$. The lines $l$ and $m$ are the required parallel tangents to the circle.

Justification:

By construction, the line $l$ is perpendicular to the radius OP at the point P on the circle. According to the theorem, a line drawn perpendicular to the radius through the point of contact is a tangent to the circle. Hence, line $l$ is a tangent to the circle at P.

Similarly, the line $m$ is perpendicular to the radius OQ at the point Q on the circle. Hence, line $m$ is a tangent to the circle at Q.

Since both lines $l$ and $m$ are perpendicular to the same line segment (diameter PQ), they are parallel to each other. This follows from the property that two lines perpendicular to the same line are parallel.

Thus, the constructed lines $l$ and $m$ are a pair of parallel tangents to the given circle.

Given:

An angle of $60^\circ$.

To bisect this angle.

To divide one of the bisected angles (which is $30^\circ$) into the ratio $1:2$.

To Construct:

An angle of $60^\circ$, its bisector, and a ray that divides one of the $30^\circ$ angles in the ratio $1:2$.

Construction Steps:

Part 1: Constructing $60^\circ$ and its bisector

1. Draw a ray OA.

2. With O as center and any convenient radius, draw an arc intersecting OA at point B.

3. With B as center and the same radius, draw an arc intersecting the first arc at point C.

4. Join OC. Then, $\angle \text{AOC} = 60^\circ$.

5. To bisect $\angle \text{AOC}$: With B and C as centers and a radius greater than half the distance BC, draw arcs intersecting each other at point D.

6. Join OD. The ray OD is the angle bisector of $\angle \text{AOC}$. Thus, $\angle \text{AOD} = \angle \text{DOC} = \frac{1}{2} \times 60^\circ = 30^\circ$.

Part 2: Dividing the bisected angle ($30^\circ$) in the ratio $1:2$

We will divide one of the $30^\circ$ angles, say $\angle \text{AOD}$, in the ratio $1:2$. The sum of the ratio parts is $1+2 = 3$.

7. With O as center and any convenient radius (it can be the same as before or different), draw an arc that intersects OA at point P and OD at point Q.

8. Draw the line segment PQ (this is the chord subtending the $30^\circ$ angle). Note that dividing the chord in a ratio does not strictly divide the angle in the same ratio, but this method is used to approximate angular division when exact construction is not possible.

9. Draw a ray PK making an acute angle with the line segment PQ.

10. On the ray PK, locate $1+2 = 3$ points, say $K_1, K_2, K_3$, such that $PK_1 = K_1K_2 = K_2K_3$.

11. Join the point $K_3$ to Q.

12. Through the point $K_1$ (corresponding to the first part of the ratio, 1), draw a line parallel to $K_3Q$. To do this, construct an angle equal to $\angle PK_3Q$ at $K_1$. Let this line intersect the line segment PQ at point R.

13. Join OR. The ray OR approximately divides the angle $\angle \text{AOD}$ in the ratio $1:2$. This means $\angle \text{AOR} : \angle \text{ROD} \approx 1:2$. (The exact angles would be $10^\circ$ and $20^\circ$, which are not constructible with elementary compass and ruler methods).

Justification:

Part 1: The construction of the $60^\circ$ angle and its bisector are standard constructions based on the properties of equilateral triangles and rhombuses formed by the arcs, which ensure the angles are correct.

Part 2: By construction, $K_1R$ is parallel to $K_3Q$. In $\triangle PK_3Q$, by the Basic Proportionality Theorem (BPT), since $K_1R \parallel K_3Q$, we have:

$\frac{PK_1}{K_1K_3} = \frac{PR}{RQ}$

By construction, $PK_1$ consists of 1 part and $K_1K_3$ consists of 2 equal parts ($K_1K_2, K_2K_3$).

Therefore, $\frac{PK_1}{K_1K_3} = \frac{1}{2}$.

From the BPT, $\frac{PR}{RQ} = \frac{1}{2}$.

Thus, the point R divides the line segment (chord) PQ in the ratio $1:2$. While this construction divides the chord in the desired ratio, it provides only an approximation for dividing the angle $\angle \text{POQ}$ in the same ratio for larger angles. The exact angular division of $30^\circ$ into $10^\circ$ and $20^\circ$ is not achievable with standard compass and ruler constructions.

Given:

A triangle ABC with BC = $8 \text{ cm}$, $\angle \text{B} = 45^\circ$, and $\angle \text{C} = 30^\circ$.

A scale factor of $\frac{3}{4}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{3}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Calculation of $\angle \text{A}$ in $\triangle \text{ABC}$:

The sum of angles in a triangle is $180^\circ$.

$\angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ$

$\angle \text{A} + 45^\circ + 30^\circ = 180^\circ$

$\angle \text{A} + 75^\circ = 180^\circ$

$\angle \text{A} = 180^\circ - 75^\circ = 105^\circ$.

Construction Steps:

1. Construct $\triangle \text{ABC}$ with the given measures. Draw a line segment BC of length $8 \text{ cm}$.

2. At point B, construct an angle of $45^\circ$ using a protractor or compass and ruler. Draw a ray BX.

3. At point C, construct an angle of $30^\circ$ using a protractor or compass and ruler. Draw a ray CY.

4. The rays BX and CY intersect at point A, forming $\triangle \text{ABC}$. (You can verify that $\angle \text{A} = 105^\circ$).

5. Draw a ray BZ extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{3}{4}$, locate 4 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4$, on the ray BZ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

7. Join the point $B_4$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_3$ (corresponding to the numerator of the ratio), draw a line parallel to $B_4C$. To do this, construct an angle equal to $\angle BB_4C$ at $B_3$. Let this line intersect BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect AB at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_4C$, the line $B_3C'$ is parallel to $B_4C$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_3}{BB_4} = \frac{BC'}{BC}$.

By construction, $\frac{BB_3}{BB_4} = \frac{3}{4}$.

Therefore, $\frac{BC'}{BC} = \frac{3}{4}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{3}{4}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{3}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with center O and radius $4 \text{ cm}$.

An external point P at a distance of $9 \text{ cm}$ from the center O.

To Construct:

The pair of tangents from point P to the circle and state the relationship between their lengths.

Construction Steps:

1. Draw a circle with center O and radius $4 \text{ cm}$.

2. Mark a point P at a distance of $9 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle (centered at O) at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Justification:

The circle with center M and radius MO passes through O and P.

The points of intersection T and T' lie on both circles.

Consider the line segment OT and PT. OT is the radius of the original circle.

The angle $\angle \text{OTP}$ is an angle inscribed in the semicircle of the circle centered at M (since OP is the diameter of this circle).

Therefore, $\angle \text{OTP} = 90^\circ$.

Since OT is the radius and PT is perpendicular to OT at the point of contact T, PT is a tangent to the circle by the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Similarly, $\angle \text{OT'P} = 90^\circ$, and PT' is also a tangent to the circle.

Relationship between the lengths of the two tangents:

The relationship between the lengths of the two tangents drawn from an external point to a circle is that they are equal in length.

This can be verified by measurement (PT = PT') or proven using congruence of triangles $\triangle \text{OTP}$ and $\triangle \text{OT'P}$ (Both are right-angled triangles, OP is common hypotenuse, OT = OT' are radii. By RHS congruence, $\triangle \text{OTP} \cong \triangle \text{OT'P}$, so PT = PT' by CPCT).

Given:

A line segment of any given length.

To divide it internally in the ratio $m:n$, where $m$ and $n$ are positive integers such that $m+n=8$. (For example, the ratio could be $1:7$, $2:6$, $3:5$, $4:4$, $5:3$, $6:2$, or $7:1$).

To Divide:

The given line segment in the ratio $m:n$, where $m+n=8$.

Construction Steps:

1. Draw a line segment AB of any desired length.

2. Draw a ray AX extending downwards from A, making an acute angle with AB.

3. Since $m+n=8$, locate $m+n=8$ points on the ray AX. Let these points be $A_1, A_2, ..., A_8$, such that $AA_1 = A_1A_2 = A_2A_3 = ... = A_7A_8$. (You can use any convenient length for these divisions, but they must be equal).

4. Join the point $A_8$ (corresponding to the sum $m+n=8$) to B.

5. Through the point $A_m$ (corresponding to the first part of the ratio, $m$), draw a line parallel to $A_8B$. To do this, construct an angle equal to $\angle AA_8B$ at $A_m$. Let this line intersect the line segment AB at point C.

6. The point C divides the line segment AB internally in the ratio $m:n$. Thus, AC : CB = $m:n$.

Justification:

By construction, $A_mC$ is parallel to $A_8B$.

In $\triangle AA_8B$, by the Basic Proportionality Theorem (BPT), since $A_mC \parallel A_8B$, we have:

$\frac{AA_m}{A_mA_8} = \frac{AC}{CB}$

By construction, the points $A_1, A_2, ..., A_8$ are marked on AX such that $AA_1 = A_1A_2 = ... = A_7A_8$. Let the length of each small segment be $k$.

Then $AA_m = m \times k$.

The segment $A_mA_8$ consists of the remaining $8 - m = n$ equal segments ($A_mA_{m+1}, ..., A_7A_8$).

So, $A_mA_8 = n \times k$.

Therefore, $\frac{AA_m}{A_mA_8} = \frac{mk}{nk} = \frac{m}{n}$.

From the BPT, $\frac{AC}{CB} = \frac{AA_m}{A_mA_8} = \frac{m}{n}$.

Thus, the point C divides the line segment AB internally in the ratio $m:n$, where $m+n=8$.

Given:

A circle with center O and radius $6 \text{ cm}$.

An external point P at a distance of $10 \text{ cm}$ from the center O.

To Construct:

The pair of tangents from point P to the circle.

Measure the lengths of the tangents.

Write the steps of construction and justify the construction.

Construction Steps:

1. Draw a circle with center O and radius $6 \text{ cm}$.

2. Mark a point P at a distance of $10 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle (centered at O) at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Measurement:

Measure the length of the tangent segments PT and PT' using a ruler.

You will find that PT = PT' = $8 \text{ cm}$.

Justification:

By construction, the circle with center M and radius MO passes through O and P. The points of intersection T and T' lie on both circles.

Consider the line segment OT and PT. OT is the radius of the original circle.

The angle $\angle \text{OTP}$ is an angle inscribed in the semicircle of the circle centered at M (since OP is the diameter of this circle).

Therefore, $\angle \text{OTP} = 90^\circ$.

Since OT is the radius and PT is perpendicular to OT at the point of contact T, PT is a tangent to the circle by the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Similarly, $\angle \text{OT'P} = 90^\circ$, and PT' is also a tangent to the circle.

We can also verify the length using the Pythagorean theorem in the right-angled triangle $\triangle \text{OTP}$:

$OP^2 = OT^2 + PT^2$

$10^2 = 6^2 + PT^2$

$100 = 36 + PT^2$

$PT^2 = 100 - 36$

$PT^2 = 64$

$PT = \sqrt{64} = 8 \text{ cm}$.

Similarly, PT' = $8 \text{ cm}$. The measured length matches the calculated length.

Given:

A triangle ABC with sides AB = $5 \text{ cm}$, BC = $7 \text{ cm}$, and $\angle \text{B} = 60^\circ$.

A scale factor of $\frac{3}{4}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{3}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Construction Steps:

1. Construct $\triangle \text{ABC}$. Draw a line segment BC of length $7 \text{ cm}$.

2. At point B, construct an angle of $60^\circ$ using a compass and ruler. Draw a ray BX.

3. With B as center and radius $5 \text{ cm}$, draw an arc intersecting the ray BX at point A.

4. Join AC to form $\triangle \text{ABC}$.

5. Draw a ray BY extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{3}{4}$, locate 4 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4$, on the ray BY such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

7. Join the point $B_4$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_3$ (corresponding to the numerator of the ratio), draw a line parallel to $B_4C$. To do this, construct an angle equal to $\angle BB_4C$ at $B_3$. Let this line intersect BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect AB at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_4C$, the line $B_3C'$ is parallel to $B_4C$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_3}{BB_4} = \frac{BC'}{BC}$.

By construction, the points $B_1, B_2, B_3, B_4$ are marked such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

So, $\frac{BB_3}{BB_4} = \frac{3}{4}$.

Therefore, $\frac{BC'}{BC} = \frac{3}{4}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{3}{4}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{3}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

Two concentric circles with center O.

Radius of the smaller circle ($r$) = $3 \text{ cm}$.

Radius of the larger circle ($R$) = $5 \text{ cm}$.

A point P on the larger circle.

To Construct:

A pair of tangents from point P on the larger circle to the smaller circle.

Measure the length of the tangent.

Write the steps of construction and justify the construction.

Construction Steps:

1. Draw two concentric circles with center O and radii $3 \text{ cm}$ and $5 \text{ cm}$.

2. Take any point P on the circumference of the larger circle.

3. Join O to P. OP is the radius of the larger circle, so $OP = 5 \text{ cm}$.

4. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

5. With M as center and radius MO (or MP), draw a circle. This circle will intersect the smaller circle (centered at O) at two distinct points.

6. Let the points of intersection be T and T'.

7. Join PT and PT'. These are the required tangents from P to the smaller circle.

Measurement:

Measure the length of the tangent segments PT and PT' using a ruler.

You will find that PT = PT' = $4 \text{ cm}$.

Justification:

By construction, the circle with center M and radius MO passes through O and P. The points of intersection T and T' lie on both circles.

Consider the line segment OT and PT. OT is the radius of the smaller circle ($3 \text{ cm}$).

The angle $\angle \text{OTP}$ is an angle inscribed in the semicircle of the circle centered at M (since OP is the diameter of this circle).

Therefore, $\angle \text{OTP} = 90^\circ$.

Since OT is the radius of the smaller circle and PT is perpendicular to OT at the point of contact T, PT is a tangent to the smaller circle by the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Similarly, $\angle \text{OT'P} = 90^\circ$, and PT' is also a tangent to the smaller circle.

We can also verify the length using the Pythagorean theorem in the right-angled triangle $\triangle \text{OTP}$:

$OP^2 = OT^2 + PT^2$

$5^2 = 3^2 + PT^2$

$25 = 9 + PT^2$

$PT^2 = 25 - 9$

$PT^2 = 16$

$PT = \sqrt{16} = 4 \text{ cm}$.

Similarly, PT' = $4 \text{ cm}$. The measured length matches the calculated length, and the lengths of the tangents from an external point to a circle are equal.

Given:

A triangle with side lengths $4 \text{ cm}$, $5 \text{ cm}$, and $6 \text{ cm}$.

A scale factor of $\frac{2}{3}$ for the sides of the similar triangle.

To Construct:

A triangle with the given side lengths, and a triangle similar to it whose sides are $\frac{2}{3}$ times the corresponding sides of the first triangle. Write the steps of construction and justification.

Construction Steps:

1. Construct the first triangle. Draw a line segment BC of length $6 \text{ cm}$. With B as center and radius $4 \text{ cm}$, draw an arc. With C as center and radius $5 \text{ cm}$, draw another arc intersecting the previous arc at A. Join AB and AC to form $\triangle \text{ABC}$.

2. Draw a ray BX extending downwards from B, making an acute angle with BC.

3. Since the scale factor is $\frac{2}{3}$, locate 3 points (the greater of the numerator and denominator), say $B_1, B_2, B_3$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3$.

4. Join the point $B_3$ (corresponding to the denominator of the ratio) to C.

5. Through the point $B_2$ (corresponding to the numerator of the ratio), draw a line parallel to $B_3C$. To do this, construct an angle equal to $\angle BB_3C$ at $B_2$. Let this line intersect BC at C'.

6. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect AB at A'.

7. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_3C$, the line $B_2C'$ is parallel to $B_3C$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_2}{BB_3} = \frac{BC'}{BC}$.

By construction, the points $B_1, B_2, B_3$ are marked such that $BB_1 = B_1B_2 = B_2B_3$.

So, $\frac{BB_2}{BB_3} = \frac{2}{3}$.

Therefore, $\frac{BC'}{BC} = \frac{2}{3}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{2}{3}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{2}{3}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with center O and radius $3.5 \text{ cm}$.

An external point P such that OP = $7 \text{ cm}$.

To Construct:

The pair of tangents from point P to the circle.

Write the steps of construction.

Verify the length of the tangents using the Pythagorean theorem.

Construction Steps:

1. Draw a circle with center O and radius $3.5 \text{ cm}$.

2. Mark a point P at a distance of $7 \text{ cm}$ from O. Join OP.

3. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. With O as center and radius greater than half of OP, draw arcs on both sides of OP. With P as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of these arcs. This line is the perpendicular bisector of OP, and it intersects OP at its midpoint. Let this midpoint be M.

4. With M as center and radius MO (or MP), draw a circle. This circle will intersect the original circle (centered at O) at two distinct points.

5. Let the points of intersection be T and T'.

6. Join PT and PT'. These are the required tangents from P to the circle.

Verification using Pythagorean Theorem:

Let T be one of the points of contact of the tangent from P to the circle. Join OT. OT is the radius of the circle.

We know that the radius through the point of contact is perpendicular to the tangent at that point. Therefore, $\angle \text{OTP} = 90^\circ$.

Consider the right-angled triangle $\triangle \text{OTP}$.

By the Pythagorean theorem, we have:

$OP^2 = OT^2 + PT^2$

We are given $OP = 7 \text{ cm}$ and $OT = 3.5 \text{ cm}$ (radius).

Substituting these values into the equation:

$7^2 = (3.5)^2 + PT^2$

$49 = 12.25 + PT^2$

$PT^2 = 49 - 12.25$

$PT^2 = 36.75$

$PT = \sqrt{36.75} \text{ cm}$

$PT = \sqrt{\frac{3675}{100}} = \frac{\sqrt{3675}}{10} \text{ cm}$

Prime factorisation of 3675:

$\begin{array}{c|cc} 3 & 3675 \\ \hline 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $3675 = 3 \times 5^2 \times 7^2$.

$PT = \frac{\sqrt{3 \times 5^2 \times 7^2}}{10} = \frac{5 \times 7 \sqrt{3}}{10} = \frac{35\sqrt{3}}{10} = \frac{7\sqrt{3}}{2} \text{ cm}$.

Using the approximate value $\sqrt{3} \approx 1.732$,

$PT \approx \frac{7 \times 1.732}{2} = \frac{12.124}{2} \approx 6.06 \text{ cm}$.

Upon measuring the constructed tangent length PT, you will find it is approximately $6.1 \text{ cm}$, which is close to the calculated value, thus verifying the construction.

Given:

A triangle ABC with BC = $5 \text{ cm}$, $\angle \text{B} = 60^\circ$, and AB = $6 \text{ cm}$.

A scale factor of $\frac{5}{3}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{5}{3}$ times the corresponding sides of $\triangle \text{ABC}$.

Construction Steps:

1. Construct $\triangle \text{ABC}$. Draw a line segment BC of length $5 \text{ cm}$.

2. At point B, construct an angle of $60^\circ$ using a compass and ruler. Draw a ray BX.

3. With B as center and radius $6 \text{ cm}$, draw an arc intersecting the ray BX at point A.

4. Join AC to form $\triangle \text{ABC}$.

5. Draw a ray BY extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{5}{3}$, locate 5 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4, B_5$, on the ray BY such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

7. Join the point $B_3$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_5$ (corresponding to the numerator of the ratio), draw a line parallel to $B_3C$. To do this, construct an angle equal to $\angle BB_3C$ at $B_5$. Let this line intersect the extended line segment BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_5C'$, the line $B_3C$ is parallel to $B_5C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_3}{BB_5} = \frac{BC}{BC'}$.

By construction, the points $B_1, B_2, B_3, B_4, B_5$ are marked such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

So, $\frac{BB_3}{BB_5} = \frac{3}{5}$.

Therefore, $\frac{BC}{BC'} = \frac{3}{5}$, which implies $\frac{BC'}{BC} = \frac{5}{3}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{5}{3}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{5}{3}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A circle with radius $5 \text{ cm}$.

A pair of tangents inclined to each other at an angle of $50^\circ$.

To Construct:

A pair of tangents to the given circle such that the angle between them is $50^\circ$.

Relationship between the angle between tangents and the angle at the center:

Let O be the center of the circle and P be the external point from which the tangents are drawn. Let T and T' be the points of contact of the tangents from P.

The angle between the tangents is $\angle \text{TPT'} = 50^\circ$.

In the quadrilateral OTPT', we have $\angle \text{OTP} = 90^\circ$ and $\angle \text{OT'P} = 90^\circ$ (radius is perpendicular to the tangent at the point of contact).

The sum of angles in a quadrilateral is $360^\circ$.

So, $\angle \text{TOT'} + \angle \text{OTP} + \angle \text{OT'P} + \angle \text{TPT'} = 360^\circ$

$\angle \text{TOT'} + 90^\circ + 90^\circ + 50^\circ = 360^\circ$

$\angle \text{TOT'} + 230^\circ = 360^\circ$

$\angle \text{TOT'} = 360^\circ - 230^\circ = 130^\circ$.

Thus, the angle between the radii to the points of contact must be $130^\circ$.

Construction Steps:

1. Draw a circle with center O and radius $5 \text{ cm}$.

2. Draw any radius of the circle, say OT.

3. At the center O, construct an angle $\angle \text{TOT'} = 130^\circ$ using a protractor, such that OT' is another radius of the circle.

4. At point T, construct a line perpendicular to OT. To do this, extend OT outwards to a point Q. With T as center and any suitable radius, draw arcs intersecting OQ at two points. With these two points as centers and a radius greater than half the distance between them, draw arcs intersecting each other. Join T to the intersection point of these arcs. This line is perpendicular to OT at T.

5. At point T', construct a line perpendicular to OT'. Similarly, construct a $90^\circ$ angle at T'.

6. Let the two perpendicular lines intersect at point P.

7. PT and PT' are the required tangents inclined at an angle of $50^\circ$ to each other.

Given:

An isosceles triangle ABC with base BC = $7 \text{ cm}$ and altitude from A to BC is $5 \text{ cm}$.

A scale factor of $\frac{7}{4}$ for the sides of the similar triangle.

To Construct:

An isosceles triangle with given dimensions, and a triangle similar to it whose sides are $\frac{7}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Construction Steps:

1. Construct the isosceles triangle ABC. Draw a line segment BC of length $7 \text{ cm}$.

2. Draw the perpendicular bisector of BC. Let M be the midpoint of BC. To find the midpoint and perpendicular bisector, with B as center and radius greater than half of BC, draw arcs above and below BC. With C as center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection. This line is the perpendicular bisector and intersects BC at M.

3. Along the perpendicular bisector from M, mark a point A such that AM = $5 \text{ cm}$ (the altitude). AM is perpendicular to BC.

4. Join AB and AC. $\triangle \text{ABC}$ is the required isosceles triangle with base BC = $7 \text{ cm}$ and altitude AM = $5 \text{ cm}$. (Since A is on the perpendicular bisector of BC, AB = AC, making it an isosceles triangle).

5. Draw a ray BX extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{7}{4}$, locate 7 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4, B_5, B_6, B_7$, on the ray BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

7. Join the point $B_4$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_7$ (corresponding to the numerator of the ratio), draw a line parallel to $B_4C$. To do this, construct an angle equal to $\angle BB_4C$ at $B_7$. Let this line intersect the extended line segment BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_7C'$, the line $B_4C$ is parallel to $B_7C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_4}{BB_7} = \frac{BC}{BC'}$.

By construction, the points $B_1, B_2, ..., B_7$ are marked such that $BB_1 = B_1B_2 = ... = B_6B_7$.

So, $\frac{BB_4}{BB_7} = \frac{4}{7}$.

Therefore, $\frac{BC}{BC'} = \frac{4}{7}$, which implies $\frac{BC'}{BC} = \frac{7}{4}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{7}{4}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{7}{4}$ times the corresponding sides of $\triangle \text{ABC}$.

Given:

A right triangle ABC with AB = $6 \text{ cm}$, BC = $8 \text{ cm}$, and $\angle \text{B} = 90^\circ$.

BD is perpendicular from B on AC.

A circle passing through points B, C, and D.

To Construct:

The tangents from point A to the circle passing through B, C, and D.

Understanding the circle through B, C, D:

Since $\angle \text{BDC} = 90^\circ$ (because BD $\perp$ AC), the circle passing through B, C, and D must have BC as its diameter. This is because the angle subtended by the diameter at any point on the circumference is $90^\circ$. Both $\angle \text{BDC}$ and $\angle \text{BBC}$ (angle at circumference is $0^\circ$) subtend the arc BC. Thus, BC is the diameter of the circle passing through B, C, and D. The center of this circle is the midpoint of BC.

Construction Steps:

1. Draw a line segment BC of length $8 \text{ cm}$.

2. At point B, construct a right angle ($\angle \text{CBY}$).

3. With B as center and radius $6 \text{ cm}$, draw an arc intersecting the ray BY at point A. Join AC. $\triangle \text{ABC}$ is constructed.

4. Find the midpoint M of the line segment BC by drawing its perpendicular bisector. M is the center of the circle passing through B, C, and D.

5. With M as center and radius MB (or MC), draw a circle. This is the circle passing through B, C, and D.

6. Point A is the external point from which tangents are to be drawn. Join AM.

7. Find the midpoint N of the line segment AM by drawing its perpendicular bisector.

8. With N as center and radius AN (or NM), draw a circle.

9. This circle (centered at N with diameter AM) intersects the circle with center M (the circle through B, C, D) at two points. One point is B (since $\angle \text{ABM} = 90^\circ$, AB is perpendicular to the radius MB at B, so AB is a tangent). Let the other point of intersection be T.

10. Join AB and AT. AB and AT are the required tangents from A to the circle through B, C, D.

Justification:

The circle through B, C, D has BC as its diameter and M (midpoint of BC) as its center.

By construction, $\angle \text{ABC} = 90^\circ$. Since MB is the radius of the circle through B, C, D, and AB is a line segment perpendicular to the radius MB at the point on the circle B, AB is a tangent to the circle at B.

The other tangent is AT. The construction for finding tangents from an external point A to the circle centered at M involves drawing a circle with diameter AM. Any angle subtended by the diameter AM on the circumference of the circle centered at N is $90^\circ$. Thus, $\angle \text{ATM} = 90^\circ$. Since MT is the radius of the circle centered at M (the circle through B, C, D) and AT is perpendicular to MT at the point of contact T, AT is a tangent to the circle.

The lengths of the tangents from an external point to a circle are equal. Thus, the length of tangent AT should be equal to the length of tangent AB, which is $6 \text{ cm}$.

Given:

A line segment of length $7.6 \text{ cm}$.

Ratio to divide the line segment internally is $5:8$.

To Divide:

The given line segment in the ratio $5:8$ using construction.

Construction Steps:

1. Draw a line segment AB of length $7.6 \text{ cm}$.

2. Draw a ray AX extending downwards from A, making an acute angle with AB.

3. Locate $(5+8) = 13$ points on the ray AX, say $A_1, A_2, A_3, ..., A_{13}$, such that $AA_1 = A_1A_2 = A_2A_3 = ... = A_{12}A_{13}$. You can use a compass to mark these points with equal divisions along the ray AX.

4. Join the point $A_{13}$ (corresponding to the sum of the ratio parts) to B.

5. Through the point $A_5$ (corresponding to the first part of the ratio, 5), draw a line parallel to $A_{13}B$. To do this, construct an angle at $A_5$ equal to $\angle AA_{13}B$. Let this line intersect the line segment AB at point C.

6. The point C divides the line segment AB internally in the ratio $5:8$. Thus, AC : CB = $5:8$.

Justification:

By construction, the line $A_5C$ is parallel to $A_{13}B$.

Consider the triangle $\triangle AA_{13}B$. The line segment $A_5C$ is drawn parallel to the side $A_{13}B$ and intersects the other two sides AA$_{13}$ and AB at points $A_5$ and C respectively.

According to the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Applying the BPT to $\triangle AA_{13}B$ with line $A_5C \parallel A_{13}B$, we get:

$\frac{AA_5}{A_5A_{13}} = \frac{AC}{CB}$

By our construction, the points $A_1, A_2, ..., A_{13}$ are marked on the ray AX such that $AA_1 = A_1A_2 = ... = A_{12}A_{13}$. Let the length of each of these small segments be $k$.

Then, $AA_5$ consists of 5 such equal segments, so $AA_5 = 5k$.

And, $A_5A_{13}$ consists of the remaining $13 - 5 = 8$ equal segments ($A_5A_6, A_6A_7, ..., A_{12}A_{13}$). So, $A_5A_{13} = 8k$.

Therefore, the ratio $\frac{AA_5}{A_5A_{13}} = \frac{5k}{8k} = \frac{5}{8}$.

Substituting this back into the BPT equation:

$\frac{AC}{CB} = \frac{AA_5}{A_5A_{13}} = \frac{5}{8}$.

Thus, the point C divides the line segment AB internally in the ratio $5:8$, which justifies the construction.

Given:

A circle with center O and radius $4 \text{ cm}$.

Two points P and Q on an extended diameter, such that OP = OQ = $9 \text{ cm}$. P and Q are on opposite sides of the center O.

To Construct:

The pair of tangents from point P to the circle and the pair of tangents from point Q to the circle.

Construction Steps:

1. Draw a circle with center O and radius $4 \text{ cm}$.

2. Draw a line passing through the center O. Let this line be the extended diameter.

3. On this line, mark two points P and Q on opposite sides of O such that OP = $9 \text{ cm}$ and OQ = $9 \text{ cm}$.

4. For point P:

a. Find the midpoint of the line segment OP. To do this, draw the perpendicular bisector of OP. Let the midpoint be M.

b. With M as center and radius MO (or MP), draw a circle.

c. This circle intersects the original circle (centered at O) at two distinct points. Let these points be T1 and T2.

d. Join PT1 and PT2. These are the two tangents from P to the circle.

5. For point Q:

a. Find the midpoint of the line segment OQ. To do this, draw the perpendicular bisector of OQ. Let the midpoint be N.

b. With N as center and NO (or NQ), draw a circle.

c. This circle intersects the original circle (centered at O) at two distinct points. Let these points be T3 and T4.

d. Join QT3 and QT4. These are the two tangents from Q to the circle.

6. PT1, PT2, QT3, and QT4 are the four required tangents.

Justification:

For the tangents from P: The circle with center M and diameter OP intersects the original circle at T1 and T2. The angle $\angle \text{OT1P}$ is an angle in a semicircle of the circle centered at M, so $\angle \text{OT1P} = 90^\circ$. Since OT1 is a radius of the original circle and PT1 is perpendicular to OT1 at the point of contact T1, PT1 is a tangent to the circle. Similarly, PT2 is a tangent.

For the tangents from Q: The circle with center N and diameter OQ intersects the original circle at T3 and T4. The angle $\angle \text{OT3Q}$ is an angle in a semicircle of the circle centered at N, so $\angle \text{OT3Q} = 90^\circ$. Since OT3 is a radius of the original circle and QT3 is perpendicular to OT3 at the point of contact T3, QT3 is a tangent to the circle. Similarly, QT4 is a tangent.

Given:

A triangle ABC with BC = $6 \text{ cm}$, $\angle \text{B} = 35^\circ$, and $\angle \text{C} = 75^\circ$.

A scale factor of $\frac{7}{5}$ for the sides of the similar triangle.

To Construct:

A triangle similar to $\triangle \text{ABC}$ whose sides are $\frac{7}{5}$ times the corresponding sides of $\triangle \text{ABC}$.

Calculation of $\angle \text{A}$ in $\triangle \text{ABC}$:

The sum of angles in a triangle is $180^\circ$.

$\angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ$

$\angle \text{A} + 35^\circ + 75^\circ = 180^\circ$

$\angle \text{A} + 110^\circ = 180^\circ$

$\angle \text{A} = 180^\circ - 110^\circ = 70^\circ$.

Construction Steps:

1. Construct $\triangle \text{ABC}$ with the given measures. Draw a line segment BC of length $6 \text{ cm}$.

2. At point B, construct an angle of $35^\circ$ using a protractor. Draw a ray BX.

3. At point C, construct an angle of $75^\circ$ using a protractor. Draw a ray CY.

4. The rays BX and CY intersect at point A, forming $\triangle \text{ABC}$. (You can verify that $\angle \text{A} = 70^\circ$).

5. Draw a ray BZ extending downwards from B, making an acute angle with BC.

6. Since the scale factor is $\frac{7}{5}$, locate 7 points (the greater of the numerator and denominator), say $B_1, B_2, B_3, B_4, B_5, B_6, B_7$, on the ray BZ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

7. Join the point $B_5$ (corresponding to the denominator of the ratio) to C.

8. Through the point $B_7$ (corresponding to the numerator of the ratio), draw a line parallel to $B_5C$. To do this, construct an angle equal to $\angle BB_5C$ at $B_7$. Let this line intersect the extended line segment BC at C'.

9. Through C', draw a line parallel to AC. To do this, construct an angle equal to $\angle BCA$ at C'. Let this line intersect the extended line segment BA at A'.

10. $\triangle \text{A'BC'}$ is the required triangle similar to $\triangle \text{ABC}$.

Justification:

By construction, the line A'C' is parallel to AC ($A'C' \parallel AC$).

Therefore, $\triangle \text{A'BC'}$ is similar to $\triangle \text{ABC}$ by the AA similarity criterion ( $\angle \text{B}$ is common to both triangles and $\angle \text{BA'C'} = \angle \text{BAC}$ and $\angle \text{BC'A'} = \angle \text{BCA}$ as corresponding angles between parallel lines A'C' and AC with transversals BA and BC respectively).

Since the triangles are similar, the ratio of their corresponding sides is equal.

$\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}}$.

In $\triangle BB_7C'$, the line $B_5C$ is parallel to $B_7C'$.

By the Basic Proportionality Theorem (BPT), $\frac{BB_5}{BB_7} = \frac{BC}{BC'}$.

By construction, the points $B_1, B_2, ..., B_7$ are marked such that $BB_1 = B_1B_2 = ... = B_6B_7$.

So, $\frac{BB_5}{BB_7} = \frac{5}{7}$.

Therefore, $\frac{BC}{BC'} = \frac{5}{7}$, which implies $\frac{BC'}{BC} = \frac{7}{5}$.

Hence, $\frac{\text{A'B}}{\text{AB}} = \frac{\text{BC'}}{\text{BC}} = \frac{\text{A'C'}}{\text{AC}} = \frac{7}{5}$.

Thus, the constructed triangle $\triangle \text{A'BC'}$ has sides which are $\frac{7}{5}$ times the corresponding sides of $\triangle \text{ABC}$.