Chapter 10 Constructions

To divide a line segment AB in the ratio $p:q$, we use the following construction method:

1. Draw a ray AX such that $\angle$BAX is an acute angle.

2. Mark points $A_1, A_2, A_3, \ldots$ on the ray AX at equal distances from A. Let these equal distances be denoted by $k$. So, $AA_1 = A_1A_2 = A_2A_3 = \ldots = k$.

3. To divide the line segment AB in the ratio $p:q$, we need to locate a point C on AB such that AC : CB = $p:q$.

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In the described method, we typically mark $p+q$ points on the ray AX, say $A_1, A_2, \ldots, A_{p+q}$.

We then join the point $A_{p+q}$ to B.

Next, we draw a line through the point $A_p$ (the $p^{\text{th}}$ point on AX) parallel to $A_{p+q}$B, which intersects AB at point C.

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According to the Basic Proportionality Theorem (also known as Thales's 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.

In triangle $AA_{p+q}$B, the line segment $A_p$C is parallel to $A_{p+q}$B. Therefore, by BPT, we have:

Since the points $A_1, A_2, \ldots, A_{p+q}$ are marked at equal distances $k$, we have:

Substituting these into the ratio equation:

Thus, the point C divides the line segment AB in the ratio $p:q$.

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To perform this construction, we need to reach the point $A_{p+q}$ on the ray AX. This requires marking at least $p+q$ points at equal distances along AX.

The minimum number of points needed is $p+q$.

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The correct option is (B) $p+q$.

Therefore, the minimum number of these points is $p+q$.

The final answer is $\boxed{p+q}$.

Let the circle have center O.

Let the two radii be OP and OQ.

Let the tangents drawn at points P and Q intersect at point T.

We are given that the angle between the tangents is $35^\circ$, so $\angle$PTQ = $35^\circ$.

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We know that the radius through the point of contact is perpendicular to the tangent.

Therefore, the angle between the radius OP and the tangent PT is $90^\circ$.

Similarly, the angle between the radius OQ and the tangent QT is $90^\circ$.

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Consider the quadrilateral formed by the points O, P, T, and Q, which is OPTQ.

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

Substitute the known values:

Solve for $\angle$POQ:

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The angle between the two radii is $145^\circ$.

This corresponds to option (D).

The final answer is $\boxed{145^\circ}$.

To divide a line segment AB in the ratio $m:n$, the standard construction method involves drawing a ray AX making an acute angle with AB.

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Points are then marked on the ray AX at equal distances.

The number of points marked on AX is the sum of the two parts of the ratio, i.e., $m+n$.

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In this question, the line segment AB is to be divided in the ratio $5:7$.

Here, the ratio is $p:q$ where $p=5$ and $q=7$.

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According to the construction method, the minimum number of points to be marked on the ray AX at equal distances is $p+q$.

Minimum number of points = $5 + 7 = 12$.

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We mark points $A_1, A_2, \ldots, A_{12}$ on ray AX such that $AA_1 = A_1A_2 = \ldots = A_{11}A_{12}$.

Then, we join $A_{12}$ to B.

Finally, we draw a line through $A_5$ parallel to $A_{12}$B, intersecting AB at C. The point C divides AB in the ratio $5:7$.

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Therefore, the minimum number of points required is 12.

This corresponds to option (D).

The final answer is 12.

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

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

2. Mark points $A_1, A_2, A_3, \ldots$ on ray AX at equal distances.

3. Join the point $A_{m+n}$ to B.

4. Draw a line through $A_m$ parallel to $A_{m+n}$B to intersect AB at C. Then AC : CB = $m:n$.

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In this question, the line segment AB is to be divided in the ratio $4:7$.

Here, $m=4$ and $n=7$.

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The total number of equal divisions required on the ray AX is $m+n$.

$m+n = 4+7 = 11$.

So, we mark points $A_1, A_2, \ldots, A_{11}$ on the ray AX at equal distances.

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According to the construction method, the point B is joined to the $(m+n)^{\text{th}}$ point on ray AX.

Since $m+n = 11$, the point B is joined to $A_{11}$.

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This corresponds to option (B).

The final answer is A₁₁.

To divide a line segment AB in the ratio $m:n$ using the alternate method with two parallel rays, we follow these steps:

1. Draw a ray AX from A making an acute angle $\angle$BAX.

2. Draw a ray BY parallel to AX from B such that the points X and Y are on opposite sides of AB.

3. Mark points $A_1, A_2, \ldots, A_m$ on AX and $B_1, B_2, \ldots, B_n$ on BY such that $AA_1 = A_1A_2 = \ldots = A_{m-1}A_m$ and $BB_1 = B_1B_2 = \ldots = B_{n-1}B_n$. The equal distances marked on AX and BY must be the same.

4. Join $A_m$ to $B_n$. The point where the line segment $A_mB_n$ intersects AB is the required point C such that AC : CB = $m:n$.

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In this question, the line segment AB is to be divided in the ratio $5:6$.

Here, the ratio is $m:n$ where $m=5$ and $n=6$.

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According to the alternate construction method, we mark $m=5$ points on ray AX, i.e., $A_1, A_2, A_3, A_4, A_5$.

We mark $n=6$ points on ray BY, i.e., $B_1, B_2, B_3, B_4, B_5, B_6$.

The points to be joined are $A_m$ and $B_n$, which are $A_5$ and $B_6$.

Joining $A_5$ and $B_6$ gives a line segment that intersects AB at a point C such that AC : CB = $5:6$.

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Therefore, the points joined are $A_5$ and $B_6$.

This corresponds to option (A).

The final answer is A₅ and B₆.

To construct a triangle similar to a given $\Delta$ABC with its sides equal to a fraction $\frac{m}{n}$ of the corresponding sides of $\Delta$ABC, where $m$ and $n$ are positive integers, we follow a standard construction method.

Given the scale factor is $\frac{3}{7}$, we have $m=3$ and $n=7$. Since the scale factor $\frac{3}{7} < 1$, the resulting similar triangle will be smaller than the given $\Delta$ABC.

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The steps for construction when the scale factor is $\frac{m}{n}$ and $m < n$ are as follows:

1. Draw a ray BX from B such that $\angle$CBX is an acute angle and X is on the opposite side of A with respect to BC.

2. Locate $n$ points, $B_1, B_2, \ldots, B_n$, on the ray BX at equal distances from B. The number of points corresponds to the denominator of the scale factor.

3. Join the point $B_n$ (corresponding to the denominator) to the vertex C.

4. Draw a line through the point $B_m$ (corresponding to the numerator) parallel to $B_n$C, intersecting BC at C'.

5. Draw a line through C' parallel to AC, intersecting AB at A'.

Then $\Delta$A'BC' is the required triangle similar to $\Delta$ABC with scale factor $\frac{m}{n}$.

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In this problem, the scale factor is $\frac{3}{7}$. So $m=3$ and $n=7$.

We draw a ray BX and locate points $B_1, B_2, \ldots, B_7$ on BX at equal distances.

The next step is to join the point $B_n$, which is $B_7$, to the vertex C.

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The point to be joined to C is $B_7$.

This corresponds to option (C).

The final answer is B₇ to C.

To construct a triangle similar to a given $\Delta$ABC with its sides $\frac{m}{n}$ of the corresponding sides of $\Delta$ABC, we draw a ray BX such that $\angle$CBX is an acute angle.

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Points are then located at equal distances on the ray BX.

The minimum number of points to be located on the ray BX at equal distances is the greater of the numerator and the denominator of the scale factor $\frac{m}{n}$. This is because we need to draw a line through the $m^{\text{th}}$ point parallel to the line joining the $n^{\text{th}}$ point to C.

In other words, the minimum number of points is $\max(m, n)$.

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In this question, the scale factor is $\frac{8}{5}$.

Here, the numerator is $m=8$ and the denominator is $n=5$.

The minimum number of points to be located on ray BX is $\max(8, 5)$.

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We mark points $B_1, B_2, \ldots, B_8$ on ray BX at equal distances.

This corresponds to option (B).

The final answer is 8.

Let the circle have center O.

Let the two radii be OP and OQ.

Let the tangents drawn at points P and Q intersect at point T.

We are given that the angle between the tangents is $60^\circ$, so $\angle$PTQ = $60^\circ$.

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We know that the radius through the point of contact is perpendicular to the tangent.

Therefore, the angle between the radius OP and the tangent PT is $90^\circ$.

Similarly, the angle between the radius OQ and the tangent QT is $90^\circ$.

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Consider the quadrilateral formed by the points O, P, T, and Q, which is OPTQ.

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

Substitute the known values:

Solve for $\angle$POQ:

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The angle between the two radii is $120^\circ$.

This corresponds to option (D).

The final answer is $\boxed{120^\circ}$.

The statement is False.

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Geometrical construction by ruler and compass allows us to perform basic arithmetic operations (addition, subtraction, multiplication, division) and extract square roots on given lengths.

When we divide a line segment in a ratio $m:n$ using the standard construction methods, we are essentially applying the Basic Proportionality Theorem.

In the construction method where we draw a ray AX and mark points $A_1, A_2, \ldots, A_k$ at equal distances, we join the $k^{\text{th}}$ point to one endpoint of the segment and draw a parallel line through another marked point.

For the ratio $\frac{\text{AC}}{\text{CB}} = \frac{m}{n}$, the lengths along the ray AX are in the ratio $\frac{\text{AA}_m}{\text{A}_m\text{A}_n} = \frac{m \times \text{unit length}}{n \times \text{unit length}} = \frac{m}{n}$.

This construction relies on taking an integer number of equal parts on the ray AX. Therefore, the ratio in which the line segment is divided must be a ratio of integers, i.e., a rational number.

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The given ratio is $(2 + \sqrt{3}) : (2 - \sqrt{3})$.

This ratio is $\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$.

To check if this ratio is rational, we can rationalize the denominator:

The value $7 + 4\sqrt{3}$ is an irrational number because $\sqrt{3}$ is irrational, and the set of irrational numbers is closed under multiplication by a non-zero rational number and addition/subtraction with a rational number.

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Since the ratio $(2 + \sqrt{3}) : (2 - \sqrt{3})$ is an irrational ratio, it is not possible to divide a line segment in this ratio using standard geometric constructions (compass and straightedge), which are limited to constructing lengths and ratios that are expressible as rational numbers or involve square roots of constructible numbers.

The statement is True.

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The given ratio is $\sqrt{3} : \frac{1}{\sqrt{3}}$.

We can simplify this ratio by multiplying both parts by $\sqrt{3}$:

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The given ratio $\sqrt{3} : \frac{1}{\sqrt{3}}$ simplifies to the ratio $3:1$.

Geometrical construction using ruler and compass allows us to divide a line segment in any ratio $m:n$, where $m$ and $n$ are positive integers.

The ratio $3:1$ is a ratio of positive integers (where $m=3$ and $n=1$).

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Since the given ratio is equivalent to the rational ratio $3:1$, it is possible to divide a line segment in this ratio using standard geometrical construction methods based on the Basic Proportionality Theorem.

The statement is False.

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To construct a triangle similar to a given $\Delta$ABC with its sides $\frac{m}{n}$ of the corresponding sides of $\Delta$ABC, where $m=7$ and $n=3$, we use the following steps (since $m > n$, the similar triangle will be larger):

1. Draw a ray BX making an acute angle with BC, with X on the opposite side of A relative to BC.

2. Locate the greater of $m$ and $n$, which is 7, points $B_1, B_2, \ldots, B_7$ on BX at equal distances from B.

3. Join the point $B_n$ to C. In this case, $n=3$, so we join $B_3$ to C.

4. Draw a line through the point $B_m$ (which is $B_7$) parallel to $B_n$C (which is $B_3$C).

This line intersects BC produced at a point C'.

5. Draw a line through C' parallel to AC, intersecting BA produced at A'.

Then $\Delta$A'BC' is the required similar triangle.

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The question states that points $B_1, \ldots, B_7$ are located, which is correct ($max(7, 3) = 7$).

It states that $B_3$ is joined to C, which is correct ($n=3$).

However, it states that a line segment $B_6$C' is drawn parallel to $B_3$C.

According to the correct construction method for a scale factor $\frac{m}{n}$ with $m>n$, the line parallel to $B_n$C should be drawn through $B_m$.

In this case, $m=7$, so the line parallel to $B_3$C should be drawn through $B_7$, not $B_6$.

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Therefore, the step described in the question, drawing the parallel line through $B_6$, is incorrect for achieving a scale factor of $\frac{7}{3}$.

The statement is False.

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To construct a pair of tangents from a point P to a circle, the point P must be located outside the circle.

A point P is outside the circle if its distance from the center of the circle is greater than the radius of the circle.

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In this question, the radius of the circle is given as 3.5 cm.

The distance of the point P from the centre is given as 3 cm.

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Comparing the distance of the point P from the centre with the radius:

Since $3 \text{ cm} < 3.5 \text{ cm}$, the distance of point P from the centre is less than the radius.

This means that the point P is located inside the circle.

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It is not possible to draw any tangent to a circle from a point lying inside the circle.

Therefore, a pair of tangents cannot be constructed from a point P situated at a distance of 3 cm from the centre to a circle of radius 3.5 cm.

The statement is True.

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When a pair of tangents is drawn from an external point to a circle, the quadrilateral formed by the center of the circle, the two points of contact, and the external point has a specific property regarding its angles.

Let O be the center of the circle, T be the external point, and P and Q be the points of contact where the tangents from T touch the circle.

The quadrilateral formed is OPTQ.

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

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

In quadrilateral OPTQ:

Substitute the known angles:

This shows that the angle between the radii joining the center to the points of contact ($\angle$POQ) and the angle between the tangents at these points ($\angle$PTQ) are supplementary.

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The question asks if a pair of tangents can be constructed inclined at an angle of $170^\circ$. This means $\angle$PTQ = $170^\circ$.

Using the relationship from equation (i):

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To construct tangents inclined at $170^\circ$, we need to draw two radii with an angle of $10^\circ$ between them.

It is possible to construct two radii of a circle with an angle of $10^\circ$ between them at the center. We can draw one radius, say OP, and then construct an angle of $10^\circ$ at O and draw the other radius OQ.

Once the two radii with an angle of $10^\circ$ are drawn, we can draw lines perpendicular to these radii at their endpoints P and Q. These perpendicular lines will be the tangents, and they will intersect at an external point T such that $\angle$PTQ = $170^\circ$.

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Since the required angle between the radii ($10^\circ$) is a valid and constructible angle, it is possible to construct tangents inclined at an angle of $170^\circ$.

Given: $\Delta$ABC is an equilateral triangle (AB = BC = CA, $\angle$A = $\angle$B = $\angle$C = $60^\circ$).

Construction: A triangle $\Delta$A'BC' is constructed similar to $\Delta$ABC with a scale factor of $\frac{3}{5}$.

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By construction, $\Delta$A'BC' is similar to $\Delta$ABC.

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A fundamental property of similar triangles is that their corresponding angles are equal.

Since $\Delta$ABC is an equilateral triangle, each of its angles is $60^\circ$.

Because $\Delta$A'BC' is similar to $\Delta$ABC, their corresponding angles are equal:

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Since all three interior angles of $\Delta$A'BC' are $60^\circ$, $\Delta$A'BC' is an equilateral triangle.

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Alternatively, for similar triangles, the ratio of corresponding sides is equal to the scale factor:

Since $\Delta$ABC is equilateral, AB = BC = CA.

Therefore, A'B = $\frac{3}{5}$AB, BC' = $\frac{3}{5}$BC, and C'A' = $\frac{3}{5}$CA.

Substituting AB = BC = CA, we get A'B = BC' = C'A'.

Since all sides of $\Delta$A'BC' are equal, it is an equilateral triangle.

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The new triangle is also an equilateral triangle.

The final answer is Yes.

Given:

A line segment AB of length 7 cm.

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To Find:

A point P on AB which divides it in the ratio $3:5$.

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Construction:

1. Draw a line segment AB of length 7 cm.

2. Draw a ray AX from A making an acute angle $\angle$BAX with AB.

3. The ratio is $3:5$. The sum of the parts is $3+5=8$. Locate $8$ points $A_1, A_2, \ldots, A_8$ on the ray AX such that the distances $AA_1 = A_1A_2 = \ldots = A_7A_8$ are equal.

4. Join the point $A_8$ (corresponding to the total number of parts) to the endpoint B.

5. Through the point $A_3$ (corresponding to the first part of the ratio, 3), draw a line segment $A_3$P parallel to $A_8$B, intersecting the line segment AB at point P.

6. The point P is the required point that divides AB in the ratio $3:5$.

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Justification:

By construction, we have $A_3$P parallel to $A_8$B.

Consider the triangle $\Delta AA_8$B. Since $A_3$P $\parallel$ $A_8$B and P lies on AB, and $A_3$ lies on $AA_8$, we can apply the Basic Proportionality Theorem (BPT).

According to BPT, if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

In $\Delta AA_8$B, $A_3$P $\parallel$ $A_8$B, so we have:

By construction, the points $A_1, A_2, \ldots, A_8$ are marked at equal distances on AX. Let the equal distance be $k$.

Then, $AA_3 = AA_1 + A_1A_2 + A_2A_3 = 3k$.

And, $A_3A_8 = A_3A_4 + A_4A_5 + A_5A_6 + A_6A_7 + A_7A_8 = 5k$.

So, the ratio $\frac{\text{AA}_3}{\text{A}_3\text{A}_8} = \frac{3k}{5k} = \frac{3}{5}$.

Substituting this back into the BPT ratio:

Thus, the point P divides the line segment AB in the ratio $3:5$.

Given:

A right triangle ABC where BC = 12 cm, AB = 5 cm, and $\angle$B = $90^\circ$.

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To Construct:

A triangle similar to $\Delta$ABC with a scale factor of $\frac{2}{3}$.

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Construction:

1. Draw the right triangle ABC with BC = 12 cm, AB = 5 cm, and $\angle$B = $90^\circ$.

2. Below the base BC, draw a ray BX such that $\angle$CBX is an acute angle.

3. The scale factor is $\frac{2}{3}$. The denominator is 3. Locate 3 points, $B_1, B_2, B_3$, on the ray BX at equal distances from B.

4. Join the point $B_3$ (corresponding to the denominator 3) to the vertex C.

5. The numerator of the scale factor is 2. Through the point $B_2$ (corresponding to the numerator 2), draw a line segment $B_2$C' parallel to $B_3$C, where C' lies on BC.

6. Through the point C' on BC, draw a line segment A'C' parallel to AC, where A' lies on AB.

Then $\Delta$A'BC' is the required triangle similar to $\Delta$ABC with scale factor $\frac{2}{3}$.

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Justification:

By construction, $B_2$C' is parallel to $B_3$C in $\Delta BB_3$C.

Using the Basic Proportionality Theorem (BPT) in $\Delta BB_3$C, we have:

Since $B_1, B_2, B_3$ are located at equal distances on BX,

Therefore,

Now, consider $\Delta$BA'C'. By construction, A'C' is parallel to AC.

In $\Delta$BAC, since A'C' $\parallel$ AC, we have:

Also, $\angle$B is common to both $\Delta$ABC and $\Delta$A'BC'.

By AAA similarity criterion, $\Delta$A'BC' is similar to $\Delta$ABC.

For similar triangles, the ratio of corresponding sides is equal to the scale factor:

Thus, the sides of $\Delta$A'BC' are $\frac{2}{3}$ of the corresponding sides of $\Delta$ABC.

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Is the new triangle also a right triangle?

Yes, the new triangle $\Delta$A'BC' is also a right triangle.

Since $\Delta$A'BC' is similar to $\Delta$ABC, their corresponding angles are equal.

Given that $\Delta$ABC is a right triangle with $\angle$B = $90^\circ$.

The corresponding angle in $\Delta$A'BC' is $\angle$A'BC', which is the same as $\angle$B.

Since one angle of $\Delta$A'BC' is $90^\circ$, it is a right triangle.

The final answer is Yes.

Given:

A triangle ABC with side lengths BC = 6 cm, CA = 5 cm, and AB = 4 cm.

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To Construct:

A triangle similar to $\Delta$ABC with a scale factor of $\frac{5}{3}$.

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Construction:

1. Draw the triangle ABC with given side lengths: Draw a line segment BC = 6 cm. With B as center and radius 4 cm, draw an arc. With C as center and radius 5 cm, draw another arc intersecting the first arc at point A. Join AB and AC to form $\Delta$ABC.

2. Below the base BC, draw a ray BX such that $\angle$CBX is an acute angle.

3. The scale factor is $\frac{5}{3}$. The greater of the numerator (5) and the denominator (3) is 5. Locate 5 points, $B_1, B_2, B_3, B_4, B_5$, on the ray BX at equal distances from B.

4. The denominator of the scale factor is 3. Join the point $B_3$ (corresponding to the denominator) to the vertex C.

5. The numerator of the scale factor is 5. Through the point $B_5$ (corresponding to the numerator), draw a line segment $B_5$C' parallel to $B_3$C, intersecting the ray BC produced at point C'.

6. Through the point C' on BC produced, draw a line segment A'C' parallel to AC, intersecting the ray BA produced at point A'.

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

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Justification:

By construction, we have $B_5$C' parallel to $B_3$C.

Consider the triangle $\Delta BB_5$C. Since $B_3$C $\parallel$ $B_5$C' and C lies on BC produced, we can apply the Basic Proportionality Theorem (BPT) or similarity of triangles.

In $\Delta BB_3$C and $\Delta BB_5$C', $\angle B_3$BC = $\angle B_5$BC (common angle) and $\angle BB_3$C = $\angle BB_5$C' (corresponding angles since $B_3$C $\parallel$ $B_5$C').

Thus, by AA similarity criterion, $\Delta BB_3$C $\sim$ $\Delta BB_5$C'.

The ratio of corresponding sides is equal:

By construction, the points $B_1, B_2, \ldots, B_5$ are marked at equal distances on BX. Let the equal distance be $k$.

Then, $BB_3 = 3k$ and $BB_5 = 5k$.

So, the ratio $\frac{\text{BB}_3}{\text{BB}_5} = \frac{3k}{5k} = \frac{3}{5}$.

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

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Now, consider $\Delta$BA'C'. By construction, A'C' is parallel to AC.

In $\Delta$BAC and $\Delta$BA'C', $\angle$ABC = $\angle$A'BC' (common angle) and $\angle$BCA = $\angle$BC'A' (corresponding angles since AC $\parallel$ A'C').

Thus, by AA similarity criterion, $\Delta$A'BC' is similar to $\Delta$ABC.

For similar triangles, the ratio of corresponding sides is equal:

We already showed that $\frac{\text{BC'}}{\text{BC}} = \frac{5}{3}$.

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

Thus, the sides of $\Delta$A'BC' are $\frac{5}{3}$ times the corresponding sides of $\Delta$ABC.

Given:

A circle with centre O and radius $r = 4$ cm.

An external point P at a distance $d = 6$ cm from the centre O.

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To Construct:

A pair of tangents from point P to the circle.

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Construction:

1. Draw a circle with centre O and radius 4 cm.

2. Mark a point P outside the circle such that the distance OP = 6 cm.

3. Join the points O and P.

4. Find the midpoint M of the line segment OP. To find the midpoint, draw the perpendicular bisector of OP. With O and P as centres, and a radius greater than half of OP, draw arcs on both sides of OP. Join the intersection points of the arcs. This line intersects OP at its midpoint M.

5. With M as centre and radius OM (or MP), draw a circle.

6. This circle with diameter OP intersects the given circle (with centre O) at two distinct points. Let these points be T and T'.

7. Join the points P to T and P to T'. These line segments PT and PT' are the required tangents from P to the given circle.

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Justification:

Join OT and OT'.

Consider the circle drawn with OP as diameter. The point T lies on the circumference of this circle.

The angle $\angle$OTP is the angle subtended by the diameter OP at a point T on the circumference of the circle with diameter OP.

According to Thales's theorem (or converse of angle in a semicircle), the angle in a semicircle is a right angle.

Since OT is the radius of the given circle and $\angle$OTP = $90^\circ$, it means that the line segment PT is perpendicular to the radius OT at the point T on the circle.

We know that a line drawn perpendicular to a radius through its endpoint on the circle is a tangent to the circle.

Therefore, PT is a tangent to the circle.

Similarly, in the circle with diameter OP, $\angle$OT'P is also an angle in a semicircle, so $\angle$OT'P = $90^\circ$. Since OT' is the radius of the given circle and PT' is perpendicular to OT' at the point T', PT' is also a tangent to the circle.

Thus, PT and PT' are the two tangents from point P to the given circle.

Given:

A rhombus ABCD with AB = 4 cm and $\angle$ABC = $60^\circ$.

The rhombus is divided into two triangles $\Delta$ABC and $\Delta$ADC by the diagonal AC.

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To Construct:

A triangle AB'C' similar to $\Delta$ABC with a scale factor of $\frac{2}{3}$.

A line segment C'D' parallel to CD where D' lies on AD.

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To Determine:

Whether the quadrilateral AB'C'D' is a rhombus.

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Construction:

1. Draw the rhombus ABCD. Since AB = 4 cm and $\angle$ABC = $60^\circ$, and it's a rhombus (all sides equal), we have AB = BC = CD = DA = 4 cm. Also, adjacent angles are supplementary, so $\angle$BAD = $180^\circ - 60^\circ = 120^\circ$. Opposite angles are equal, so $\angle$ADC = $60^\circ$ and $\angle$BCD = $120^\circ$. Since $\Delta$ABC has AB = BC = 4 cm and $\angle$ABC = $60^\circ$, it is an equilateral triangle. Similarly, since AD = CD = 4 cm and $\angle$ADC = $60^\circ$, $\Delta$ADC is an equilateral triangle. The diagonal AC has length 4 cm.

2. Draw the diagonal AC, dividing the rhombus into $\Delta$ABC and $\Delta$ADC.

3. To construct $\Delta$AB'C' similar to $\Delta$ABC with scale factor $\frac{2}{3}$ such that A is the common vertex, we need to find points B' on AB and C' on AC such that $\frac{\text{AB'}}{\text{AB}} = \frac{\text{AC'}}{\text{AC}} = \frac{2}{3}$.

Draw a ray AX from A making an acute angle with AB.

Locate 3 points, $A_1, A_2, A_3$, on AX at equal distances from A.

Join $A_3$ to B.

Through the point $A_2$, draw a line parallel to $A_3$B intersecting AB at B'. Then $\frac{\text{AB'}}{\text{AB}} = \frac{\text{AA}_2}{\text{AA}_3} = \frac{2}{3}$.

Similarly, draw a ray AY from A making an acute angle with AC.

Locate 3 points, $Y_1, Y_2, Y_3$, on AY at equal distances from A.

Join $Y_3$ to C.

Through the point $Y_2$, draw a line parallel to $Y_3$C intersecting AC at C'. Then $\frac{\text{AC'}}{\text{AC}} = \frac{\text{AY}_2}{\text{AY}_3} = \frac{2}{3}$.

$\Delta$AB'C' is the required triangle similar to $\Delta$ABC with scale factor $\frac{2}{3}$.

4. Draw a line segment C'D' parallel to CD such that D' lies on AD. Since C' is on AC, draw a line through C' parallel to CD, intersecting AD at D'.

5. The quadrilateral formed is AB'C'D'.

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Is AB'C'D' a rhombus? Give reasons.

Yes, AB'C'D' is a rhombus.

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Reasons:

From the construction of B' on AB, we have $\frac{\text{AB'}}{\text{AB}} = \frac{2}{3}$.

From the construction of C' on AC and drawing C'D' parallel to CD with D' on AD, consider $\Delta$ACD. Since C' is on AC, D' is on AD, and C'D' $\parallel$ CD, by the Basic Proportionality Theorem (BPT), we have:

From the construction of C' on AC, we have $\frac{\text{AC'}}{\text{AC}} = \frac{2}{3}$.

Therefore, $\frac{\text{AD'}}{\text{AD}} = \frac{2}{3}$ and $\frac{\text{C'D'}}{\text{CD}} = \frac{2}{3}$.

Now consider the side B'C'. Since $\Delta$AB'C' is similar to $\Delta$ABC with scale factor $\frac{2}{3}$, the ratio of corresponding sides is $\frac{2}{3}$.

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

The side lengths of the quadrilateral AB'C'D' are:

Since all four sides of the quadrilateral AB'C'D' are equal in length, it is a rhombus.

Given:

Line segments AB and AC such that AB = 5 cm, AC = 7 cm, and the angle between them is $\angle$BAC = $60^\circ$.

Point P on AB such that AP = $\frac{3}{4}$AB.

Point Q on AC such that AQ = $\frac{1}{4}$AC.

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To Find:

The length of the line segment PQ by measurement.

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Construction:

1. Draw a line segment AB of length 5 cm.

2. At point A, construct an angle of $60^\circ$. Draw a ray AY representing the direction of AC.

3. On the ray AY, mark a point C such that AC = 7 cm.

4. Draw a ray AX from A making an acute angle with AB, on the opposite side of BC.

5. The denominators of the fractions $\frac{3}{4}$ and $\frac{1}{4}$ are both 4. Locate 4 points, $A_1, A_2, A_3, A_4$, on the ray AX at equal distances from A.

6. To locate point P on AB such that AP = $\frac{3}{4}$AB, join the point $A_4$ to B. Through the point $A_3$ (since the numerator for AP is 3), draw a line segment $A_3$P parallel to $A_4$B, intersecting AB at point P.

7. To locate point Q on AC such that AQ = $\frac{1}{4}$AC, join the point $A_4$ to C. Through the point $A_1$ (since the numerator for AQ is 1), draw a line segment $A_1$Q parallel to $A_4$C, intersecting AC at point Q.

8. Join the points P and Q.

9. Measure the length of the line segment PQ using a ruler.

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Justification of locating P and Q:

By construction, $A_3$P $\parallel$ $A_4$B in $\Delta AA_4$B. By Basic Proportionality Theorem, $\frac{\text{AP}}{\text{PB}} = \frac{\text{AA}_3}{\text{A}_3\text{A}_4}$. Since $AA_3 = 3 \times AA_1$ and $A_3A_4 = 1 \times AA_1$, $\frac{\text{AA}_3}{\text{A}_3\text{A}_4} = \frac{3}{1}$. Thus, $\frac{\text{AP}}{\text{PB}} = \frac{3}{1}$, which means AP = 3PB. Since AB = AP + PB, AB = 3PB + PB = 4PB, so PB = $\frac{1}{4}$AB and AP = $\frac{3}{4}$AB. This confirms the position of P.

By construction, $A_1$Q $\parallel$ $A_4$C in $\Delta AA_4$C. By Basic Proportionality Theorem, $\frac{\text{AQ}}{\text{QC}} = \frac{\text{AA}_1}{\text{A}_1\text{A}_4}$. Since $AA_1 = 1 \times AA_1$ and $A_1A_4 = 3 \times AA_1$, $\frac{\text{AA}_1}{\text{A}_1\text{A}_4} = \frac{1}{3}$. Thus, $\frac{\text{AQ}}{\text{QC}} = \frac{1}{3}$, which means AQ = $\frac{1}{3}$QC. Since AC = AQ + QC, AC = AQ + 3AQ = 4AQ, so AQ = $\frac{1}{4}$AC. This confirms the position of Q.

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Measurement:

Upon accurately performing the construction and measuring the length of PQ with a ruler, you will find that:

PQ $\approx$ 3.25 cm

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Verification (Calculation using Law of Cosines):

AP = $\frac{3}{4} \times \text{AB} = \frac{3}{4} \times 5 = \frac{15}{4}$ cm = 3.75 cm.

AQ = $\frac{1}{4} \times \text{AC} = \frac{1}{4} \times 7 = \frac{7}{4}$ cm = 1.75 cm.

The angle between AP and AQ is $\angle$PAQ = $\angle$BAC = $60^\circ$.

In $\Delta$APQ, using the Law of Cosines:

The measured length PQ should be approximately 3.25 cm.

Given:

A parallelogram ABCD with BC = 5 cm, AB = 3 cm and $\angle$ABC = $60^\circ$.

The parallelogram is divided into triangles BCD and ABD by the diagonal BD.

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To Construct:

A triangle BD'C' similar to $\Delta$BDC with scale factor $\frac{4}{3}$.

A line segment D'A' parallel to DA where A' lies on the extended side BA.

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To Determine:

Whether the quadrilateral A'BC'D' is a parallelogram.

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Construction:

1. Draw a line segment BC = 5 cm.

2. At point B, construct an angle of $60^\circ$ ($\angle$CBX = $60^\circ$). Mark a point A on the ray BX such that AB = 3 cm.

3. Draw a line through A parallel to BC. Draw a line through C parallel to AB. Let these lines intersect at D. ABCD is the required parallelogram.

4. Draw the diagonal BD.

5. To construct $\Delta$BD'C' similar to $\Delta$BDC with scale factor $\frac{4}{3}$ (where B is the common vertex):

Draw a ray BY from B making an acute angle with BD, on the opposite side of BC.

Locate 4 points, $Y_1, Y_2, Y_3, Y_4$, on the ray BY at equal distances from B. ($BY_1 = Y_1Y_2 = Y_2Y_3 = Y_3Y_4$).

Join the point $Y_3$ (corresponding to the denominator 3) to the point D.

Through the point $Y_4$ (corresponding to the numerator 4), draw a line parallel to $Y_3$D, intersecting the ray BD extended at point D'.

Through the point D', draw a line parallel to DC, intersecting the ray BC extended at point C'.

Then $\Delta$BD'C' is the required triangle similar to $\Delta$BDC with scale factor $\frac{4}{3}$.

6. Extend the side BA beyond A to form a ray. Through the point D', draw a line parallel to DA, intersecting the extended side BA at point A'.

7. The quadrilateral formed is A'BC'D'.

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Is A'BC'D' a parallelogram? Give reasons.

Yes, A'BC'D' is a parallelogram.

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Reasons:

1. By construction, $\Delta$BD'C' is similar to $\Delta$BDC with B as the center of similarity and scale factor $\frac{4}{3}$. This implies that C' lies on the ray BC such that BC' = $\frac{4}{3}$ BC, D' lies on the ray BD such that BD' = $\frac{4}{3}$ BD, and the corresponding side D'C' is parallel to DC.

2. Since ABCD is a parallelogram, the opposite sides are parallel. Thus, DC $\parallel$ AB.

3. From (1) and (2), we have D'C' $\parallel$ DC and DC $\parallel$ AB. Therefore, D'C' $\parallel$ AB. Since the point A' lies on the extended side BA (which is the line containing AB), we can say that D'C' is parallel to A'B.

4. Since ABCD is a parallelogram, the opposite sides are parallel. Thus, DA $\parallel$ BC.

5. By construction, D'A' is drawn parallel to DA.

6. From (4) and (5), we have D'A' $\parallel$ DA and DA $\parallel$ BC. Therefore, D'A' $\parallel$ BC. Since the point C' lies on the extended side BC (which is the line containing BC), we can say that D'A' is parallel to BC'.

7. A quadrilateral is defined as a parallelogram if both pairs of its opposite sides are parallel.

8. From (3) and (6), we have shown that D'C' $\parallel$ A'B and D'A' $\parallel$ BC'.

Therefore, the quadrilateral A'BC'D' is a parallelogram.

Given:

Two concentric circles with centre O.

Radius of the inner circle, $r_1 = 3$ cm.

Radius of the outer circle, $r_2 = 5$ cm.

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To Construct:

A pair of tangents to the inner circle from a point on the outer circle.

Measure the length of a tangent and verify it by calculation.

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Construction:

1. Draw two concentric circles with centre O and radii 3 cm and 5 cm.

2. Take any point P on the circumference of the outer circle (radius 5 cm). Join OP.

3. Find the midpoint M of the line segment OP. To find the midpoint, draw the perpendicular bisector of OP. With O and P as centres, and a radius greater than half of OP, draw arcs on both sides of OP. Join the intersection points of the arcs. This line intersects OP at its midpoint M.

4. With M as centre and radius OM (or MP), draw a circle.

5. This circle with diameter OP intersects the inner circle (with centre O and radius 3 cm) at two distinct points. Let these points be T and T'.

6. Join the points P to T and P to T'. These line segments PT and PT' are the required tangents from the point P on the outer circle to the inner circle.

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Justification:

Join OT and OT'.

Consider the circle drawn with OP as diameter. The point T lies on the circumference of this circle.

The angle $\angle$OTP is the angle subtended by the diameter OP at a point T on the circumference of the circle with diameter OP.

According to Thales's theorem (or converse of angle in a semicircle), the angle in a semicircle is a right angle.

Since OT is the radius of the inner circle and $\angle$OTP = $90^\circ$, it means that the line segment PT is perpendicular to the radius OT at the point T on the inner circle.

We know that a line drawn perpendicular to a radius through its endpoint on the circle is a tangent to the circle.

Therefore, PT is a tangent to the inner circle.

Similarly, PT' is also a tangent to the inner circle.

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Measurement:

Measure the length of the tangent segment PT (or PT') using a ruler. Upon accurate construction, the measured length should be approximately 4 cm.

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Verification by calculation:

In the right-angled triangle $\Delta$OTP (since $\angle$OTP = $90^\circ$), O is the center of the circles, P is a point on the outer circle, and T is the point of contact on the inner circle.

OP is the radius of the outer circle, so OP = 5 cm.

OT is the radius of the inner circle, so OT = 3 cm.

PT is the tangent segment whose length we need to find.

Using the Pythagorean theorem in $\Delta$OTP:

Substitute the values of OP and OT:

The calculated length of the tangent is 4 cm. This matches the expected measured length, verifying the construction.

Given:

An isosceles triangle ABC with AB = 6 cm, AC = 6 cm, and BC = 5 cm.

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To Construct:

A triangle PQR similar to $\Delta$ABC such that PQ = 8 cm.

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Calculation of Scale Factor and Sides of $\Delta$PQR:

Since $\Delta$PQR is similar to $\Delta$ABC, their corresponding sides are proportional.

Given that PQ corresponds to AB (based on the order PQR $\sim$ ABC implied by the problem statement asking for PQ = 8 cm, which corresponds to AB = 6 cm), the scale factor is the ratio of the lengths of corresponding sides:

Since the scale factor is $\frac{4}{3} > 1$, the constructed triangle will be larger than $\Delta$ABC.

The sides of $\Delta$PQR will be $\frac{4}{3}$ times the corresponding sides of $\Delta$ABC:

So, we need to construct a triangle with sides PQ = 8 cm, QR = $\frac{20}{3}$ cm ($\approx 6.67$ cm), and PR = 8 cm.

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Construction:

1. Draw the triangle ABC with AB = 6 cm, AC = 6 cm, and BC = 5 cm. (Draw BC = 5 cm, then draw arcs from B and C with radius 6 cm intersecting at A).

2. Draw a ray BX from B making an acute angle with BC, on the opposite side of vertex A.

3. The scale factor is $\frac{4}{3}$. The greater of the numerator (4) and the denominator (3) is 4. Locate 4 points, $B_1, B_2, B_3, B_4$, on the ray BX at equal distances from B.

4. Join the point $B_3$ (corresponding to the denominator 3) to the vertex C.

5. Through the point $B_4$ (corresponding to the numerator 4), draw a line parallel to $B_3$C. To do this, construct $\angle BB_4$C' equal to $\angle BB_3$C. Extend the line segment BC to meet this parallel line at C'.

6. Through the point C' on BC produced, draw a line parallel to AC. To do this, construct $\angle$BC'A' equal to $\angle$BCA. Extend the line segment BA to meet this parallel line at A'.

7. The triangle A'BC' is the required triangle similar to $\Delta$ABC with scale factor $\frac{4}{3}$. Relabel the vertices as P, Q, R such that P corresponds to A', Q corresponds to B, and R corresponds to C'. So, $\Delta$PQR is $\Delta$A'BC'.

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Justification:

1. By construction, we drew $B_4$C' $\parallel$ $B_3$C. Applying the Basic Proportionality Theorem (BPT) in $\Delta BB_4$C', considering the line segment $B_3$C cutting sides $BB_4$ and BC', and $B_3$C $\parallel$ $B_4$C', the ratio of segments is equal: $\frac{\text{BB}_3}{\text{BB}_4} = \frac{\text{BC}}{\text{BC'}}$.

Since the points $B_1, B_2, B_3, B_4$ are marked at equal distances, we have $\frac{\text{BB}_3}{\text{BB}_4} = \frac{3}{4}$.

Therefore,

This implies $\text{BC'} = \frac{4}{3}\text{BC}$.

2. By construction, we drew A'C' $\parallel$ AC. Consider $\Delta$BA'C' and $\Delta$BAC.

$\angle$A'BC' = $\angle$ABC (Common angle)

$\angle$BC'A' = $\angle$BCA (Corresponding angles since A'C' $\parallel$ AC)

Therefore, by AA similarity criterion, $\Delta$BA'C' $\sim$ $\Delta$BAC.

3. For similar triangles, the ratio of corresponding sides is equal:

From step 1, we know $\frac{\text{BC'}}{\text{BC}} = \frac{4}{3}$.

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

This means the sides of $\Delta$A'BC' are $\frac{4}{3}$ times the corresponding sides of $\Delta$ABC.

4. Specifically, A'B = $\frac{4}{3}$ AB = $\frac{4}{3} \times 6$ cm = 8 cm.

When we relabel $\Delta$A'BC' as $\Delta$PQR with P $\leftrightarrow$ A', Q $\leftrightarrow$ B, R $\leftrightarrow$ C', the side PQ corresponds to A'B, and PQ = A'B = 8 cm.

Thus, the constructed triangle $\Delta$PQR (which is $\Delta$A'BC') is similar to $\Delta$ABC with a scale factor of $\frac{4}{3}$ and has the side PQ = 8 cm as required.

Given:

A triangle ABC with side lengths AB = 5 cm, BC = 6 cm, and angle $\angle$ABC = $60^\circ$.

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To Construct:

A triangle similar to $\Delta$ABC with a scale factor of $\frac{5}{7}$.

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Construction:

1. Draw a line segment BC of length 6 cm.

2. At point B, construct an angle of $60^\circ$. Draw a ray BX such that $\angle$CBX = $60^\circ$.

3. On the ray BX, cut off a line segment BA equal to 5 cm.

4. Join points A and C to form the triangle ABC.

5. From B, draw a ray BY making an acute angle with BC, on the opposite side of vertex A.

6. The scale factor is $\frac{5}{7}$. The denominator is 7. Locate 7 points, $B_1, B_2, B_3, B_4, B_5, B_6, B_7$, on the ray BY at equal distances from B.

7. Join the point $B_7$ (corresponding to the denominator 7) to the vertex C.

8. The numerator of the scale factor is 5. Through the point $B_5$ (corresponding to the numerator 5), draw a line segment A'C' parallel to $B_7$C. To do this, construct $\angle$BB$_5$C' equal to $\angle$BB$_7$C. This line through $B_5$ intersects BC at a point C'.

9. Through the point C' on BC, draw a line segment A'C' parallel to AC. To do this, construct $\angle$BC'A' equal to $\angle$BCA. This line through C' intersects AB at a point A'.

10. The triangle A'BC' is the required triangle similar to $\Delta$ABC with scale factor $\frac{5}{7}$.

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Justification:

1. By construction, we drew $B_5$C' parallel to $B_7$C.

Consider the triangle $\Delta BB_7$C. Since $B_5$C' $\parallel$ $B_7$C and C' lies on BC and $B_5$ lies on $BB_7$, by the Basic Proportionality Theorem (BPT), we have:

By construction, the points $B_1, B_2, \ldots, B_7$ are marked at equal distances on BY. Let the equal distance be $k$.

Then, $BB_5 = 5k$ and $BB_7 = 7k$.

So, the ratio $\frac{\text{BB}_5}{\text{BB}_7} = \frac{5k}{7k} = \frac{5}{7}$.

Therefore,

2. By construction, we drew A'C' parallel to AC.

Consider the triangle $\Delta$ABC and $\Delta$A'BC'.

$\angle$ABC = $\angle$A'BC' (Common angle)

$\angle$BCA = $\angle$BC'A' (Corresponding angles since A'C' $\parallel$ AC)

Therefore, by AA similarity criterion, $\Delta$A'BC' is similar to $\Delta$ABC.

3. For similar triangles, the ratio of corresponding sides is equal:

From equation (i), we know $\frac{\text{BC'}}{\text{BC}} = \frac{5}{7}$.

Therefore,

This shows that the sides of the constructed triangle $\Delta$A'BC' are $\frac{5}{7}$ times the corresponding sides of the original triangle $\Delta$ABC, which is the required scale factor.

Given:

A circle with radius $r = 4$ cm.

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

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To Construct:

A pair of tangents to the circle inclined at $60^\circ$.

Justify the construction.

Measure and verify the distance between the centre and the point of intersection of tangents.

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Construction:

1. Draw a circle with centre O and radius 4 cm.

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

3. At the centre O, construct an angle $\angle$AOB such that $\angle$AOB = $180^\circ - 60^\circ = 120^\circ$. Draw the radius OB.

4. At point A, draw a line perpendicular to OA. This line will be a tangent to the circle at A. (Construct a $90^\circ$ angle at A with OA as one arm).

5. At point B, draw a line perpendicular to OB. This line will be a tangent to the circle at B. (Construct a $90^\circ$ angle at B with OB as one arm).

6. Let the two perpendicular lines (tangents) intersect at point P.

7. PA and PB are the required tangents to the circle inclined at an angle of $60^\circ$.

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Justification:

We are constructing tangents at the endpoints of two radii OA and OB such that the angle between the radii is $120^\circ$.

Since PA is constructed perpendicular to the radius OA at the point A on the circle, PA is a tangent to the circle at A.

Similarly, since PB is constructed perpendicular to the radius OB at the point B on the circle, PB is a tangent to the circle at B.

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

We constructed $\angle$AOB = $120^\circ$, and by the properties of tangents, $\angle$OAP = $90^\circ$ and $\angle$OBP = $90^\circ$.

Substitute these values into the equation:

Thus, the angle between the tangents PA and PB is $60^\circ$, as required.

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Measurement of the distance OP:

Measure the length of the line segment OP using a ruler. Upon accurate construction, the measured distance OP should be approximately 8 cm.

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Verification by actual calculation:

Consider the triangle OAP. PA and PB are tangents from an external point P. The line segment OP connects the center to the external point. OP bisects the angle between the tangents and the angle between the radii at the center.

Consider the right-angled triangle $\Delta$OAP (since $\angle$OAP = $90^\circ$).

We have the radius OA = 4 cm, $\angle$OAP = $90^\circ$, and $\angle$APO = $30^\circ$. We want to find the length of the hypotenuse OP.

Using the sine function:

We know that $\sin(30^\circ) = \frac{1}{2}$.

Cross-multiplying:

The calculated distance between the center of the circle and the point of intersection of tangents is 8 cm. This matches the expected measured length, verifying the construction.

Given:

A triangle ABC with side lengths AB = 4 cm, BC = 6 cm, and AC = 9 cm.

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To Construct:

A triangle similar to $\Delta$ABC with a scale factor of $\frac{3}{2}$.

Justify the construction.

Determine if the two triangles are congruent.

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Construction:

1. Draw a line segment BC of length 6 cm.

2. With B as centre and radius 4 cm, draw an arc. With C as centre and radius 9 cm, draw another arc intersecting the first arc at point A. Join AB and AC to form $\Delta$ABC.

3. From B, draw a ray BX making an acute angle with BC, on the opposite side of vertex A.

4. The scale factor is $\frac{3}{2}$. The greater of the numerator (3) and the denominator (2) is 3. Locate 3 points, $B_1, B_2, B_3$, on the ray BX at equal distances from B.

5. The denominator of the scale factor is 2. Join the point $B_2$ (corresponding to the denominator 2) to the vertex C.

6. The numerator of the scale factor is 3. Through the point $B_3$ (corresponding to the numerator 3), draw a line segment $B_3$C' parallel to $B_2$C. To do this, construct $\angle BB_3$C' equal to $\angle BB_2$C. Extend the line segment BC to meet this parallel line at C'.

7. Through the point C' on BC produced, draw a line segment A'C' parallel to AC. To do this, construct $\angle$BC'A' equal to $\angle$BCA. Extend the line segment BA to meet this parallel line at A'.

8. The triangle A'BC' is the required triangle similar to $\Delta$ABC with scale factor $\frac{3}{2}$.

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Justification:

1. By construction, we drew $B_3$C' parallel to $B_2$C.

Consider the triangle $\Delta BB_3$C'. Since $B_2$C $\parallel$ $B_3$C' and C lies on BC', by the Basic Proportionality Theorem (BPT) in $\Delta BB_3$C', we have:

By construction, the points $B_1, B_2, B_3$ are marked at equal distances on BY. Let the equal distance be $k$.

Then, $BB_2 = 2k$ and $BB_3 = 3k$.

So, the ratio $\frac{\text{BB}_2}{\text{BB}_3} = \frac{2k}{3k} = \frac{2}{3}$.

Therefore,

This implies $\text{BC'} = \frac{3}{2}\text{BC}$.

2. By construction, we drew A'C' parallel to AC.

Consider the triangle $\Delta$ABC and $\Delta$A'BC'.

$\angle$ABC = $\angle$A'BC' (Common angle)

$\angle$BCA = $\angle$BC'A' (Corresponding angles since A'C' $\parallel$ AC)

Therefore, by AA similarity criterion, $\Delta$A'BC' is similar to $\Delta$ABC.

3. For similar triangles, the ratio of corresponding sides is equal:

From step 1, we know $\frac{\text{BC'}}{\text{BC}} = \frac{3}{2}$.

Therefore,

This shows that the sides of the constructed triangle $\Delta$A'BC' are $\frac{3}{2}$ times the corresponding sides of the original triangle $\Delta$ABC, which is the required scale factor.

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Are the two triangles congruent?

No, the two triangles $\Delta$ABC and $\Delta$A'BC' are not congruent.

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Reasons:

Two triangles are congruent if they are exactly the same size and shape. This means all corresponding sides are equal in length, and all corresponding angles are equal in measure.

We have constructed $\Delta$A'BC' to be similar to $\Delta$ABC with a scale factor of $\frac{3}{2}$.

Since the scale factor is $\frac{3}{2} \neq 1$, the corresponding sides of the two triangles are not equal in length.

For example, A'B = $\frac{3}{2}$ AB = $\frac{3}{2} \times 4$ cm = 6 cm, while AB = 4 cm. Clearly, A'B $\neq$ AB.

While similar triangles do have equal corresponding angles (all three angles are equal), having equal corresponding angles is a condition for similarity (AA or AAA criterion), not congruence.

The note mentioning "all the three angles and two sides of the two triangles are equal" is misleading in the context of the scale factor $\frac{3}{2}$. For similar triangles with a scale factor not equal to 1, only the angles are equal; the sides are proportional with the ratio being the scale factor. If two triangles had all three angles equal AND two pairs of corresponding sides equal, then the third pair of sides would also have to be equal (due to similarity ratios), making them congruent (by SSS or ASA/SAS if angle was included). However, here the scale factor ensures sides are *not* equal.

Since the corresponding sides are not equal (as the scale factor is not 1), the triangles are not congruent.