Chapter 6 Triangles (Additional Questions)

(B) Similar figures

Explanation:

Two figures are said to be congruent if they have exactly the same shape and the same size.

Two figures are said to be similar if they have the same shape but can have different sizes.

The question asks for figures having the same shape but not necessarily the same size. This is the definition of Similar figures.

Therefore, the correct option is (B).

(A) Similar

Explanation:

Congruent figures have the same shape and the same size.

Similar figures have the same shape but not necessarily the same size.

If two figures are congruent, they have the same shape and the same size. Since having the same shape is a requirement for similarity, and having the same size does not contradict the definition of similarity (which says "not necessarily the same size"), all congruent figures are also similar figures.

The converse is not true; similar figures are not necessarily congruent as they may have different sizes.

Options (C) and (D) are incorrect because congruent figures can be any shape, not just circles or squares.

Therefore, all congruent figures are Similar.

(C) AA/AAA

Explanation:

The similarity criteria for triangles are:

SSS (Side-Side-Side) Similarity: If the corresponding sides of two triangles are proportional, then they are similar.

SAS (Side-Angle-Side) Similarity: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

AA (Angle-Angle) Similarity: If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar. Since the sum of angles in a triangle is $180^\circ$, if two angles are equal, the third angle must also be equal. Hence, this is also known as AAA (Angle-Angle-Angle) Similarity.

RHS (Right angle-Hypotenuse-Side) Congruence: This is a congruence criterion for right-angled triangles, not a general similarity criterion. There is a related similarity criterion for right triangles based on proportional sides, but RHS itself is for congruence.

In the given problem, we are given that $\angle A = \angle P$, $\angle B = \angle Q$, and $\angle C = \angle R$. All three corresponding angles are equal.

This matches the condition for the AA/AAA similarity criterion.

Therefore, the triangles are similar by the AA/AAA criterion.

(C) $\frac{8}{3}$ cm

Given:

In $\triangle ABC$, D is a point on AB and E is a point on AC.

$DE \parallel BC$

AD = 3 cm

DB = 2 cm

AE = 4 cm

To Find:

The length of EC.

Solution:

In $\triangle ABC$, we are given that $DE \parallel BC$.

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

Applying the BPT to $\triangle ABC$ with line segment DE parallel to BC, we have:

Substitute the given values into equation (i):

Now, we solve for EC:

So, the length of EC is $\frac{8}{3}$ cm.

This matches option (C).

(C) SAS

Explanation:

Let's examine the given options and their definitions as similarity criteria:

(A) AA (Angle-Angle) Similarity: This criterion states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. It involves angles, not sides and the included angle.

(B) SSS (Side-Side-Side) Similarity: This criterion states that if the corresponding sides of two triangles are proportional, then the triangles are similar. It involves three sides, not two sides and the included angle.

(C) SAS (Side-Angle-Side) Similarity: This criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion explicitly involves two sides and the included angle.

(D) ASA (Angle-Side-Angle): This is a criterion for congruence of triangles, not similarity. It states that if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.

Based on the definitions, the criterion that involves two sides and the included angle for similarity is SAS.

Therefore, the correct option is (C).

(C) $9/25$

Explanation:

We are given that $\triangle ABC \sim \triangle PQR$.

We are also given the ratio of corresponding sides $AB$ and $PQ$:

According to the theorem on the areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Mathematically, this can be stated as:

Substitute the value from equation (i) into equation (ii):

Calculate the square of the ratio:

So, the ratio of the areas is $\frac{9}{25}$.

This matches option (C).

(C) Pythagoras Theorem

Explanation:

Let's consider the definitions of the theorems given in the options:

(A) Thales Theorem: One version states that if A, B, and C are distinct points on a circle where the line segment AC is a diameter, then the angle $\angle ABC$ is a right angle ($90^\circ$). Another version is related to parallel lines intersecting transversals (same as BPT).

(B) Basic Proportionality Theorem (BPT): States that if a line parallel to one side of a triangle intersects the other two sides at distinct points, then it divides the two sides in the same ratio.

(C) Pythagoras Theorem: States that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (legs). If the legs have lengths $a$ and $b$ and the hypotenuse has length $c$, then the theorem is expressed as $a^2 + b^2 = c^2$.

(D) Midpoint Theorem: States that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

The statement "In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides" is precisely the definition of the Pythagoras Theorem.

Therefore, the correct option is (C).

(A) 10 cm

Given:

In right triangle ABC, right-angled at B.

AB = 6 cm

BC = 8 cm

To Find:

The length of AC.

Solution:

Since $\triangle ABC$ is a right-angled triangle with the right angle at B, AC is the hypotenuse.

We can find the length of the hypotenuse using the Pythagoras Theorem.

The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Applying the theorem to $\triangle ABC$:

Substitute the given values of AB and BC into equation (i):

Calculate the squares of the lengths:

Add the squared values:

To find the length of AC, take the square root of both sides:

The length of AC is 10 cm.

This matches option (A).

Note that option (D) is $\sqrt{100}$ cm, which is equal to 10 cm. Option (A) provides the simplified value.

(B) Q

Explanation:

The given equation is $PQ^2 + QR^2 = PR^2$.

This equation is in the form of the Pythagoras Theorem, which relates the sides of a right-angled triangle.

The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (legs).

The converse of the Pythagoras Theorem states that if in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

In the given equation $PQ^2 + QR^2 = PR^2$, the side $PR$ is the side whose square is equal to the sum of the squares of the other two sides ($PQ$ and $QR$). Therefore, $PR$ must be the hypotenuse of the triangle.

In a right-angled triangle, the hypotenuse is always the side opposite the right angle.

The vertex opposite the side $PR$ is the vertex Q.

Therefore, the right angle must be at vertex Q.

This means $\angle Q = 90^\circ$.

The correct option is (B).

(A) Sides

(B) Altitudes

(C) Medians

(D) Angle bisectors

(E) Perimeters

Explanation:

Let $\triangle ABC$ and $\triangle PQR$ be two similar triangles, i.e., $\triangle ABC \sim \triangle PQR$.

According to the theorem on the areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

This confirms that option (A) is correct.

For similar triangles, the ratio of corresponding linear elements is equal to the ratio of corresponding sides.

Specifically, if the ratio of corresponding sides is $k = \frac{AB}{PQ}$, then:

The ratio of corresponding altitudes is also $k$.

Therefore, the square of the ratio of corresponding altitudes is $k^2$, which is equal to the ratio of the areas.

This confirms that option (B) is correct.

The ratio of corresponding medians is also $k$.

Therefore, the square of the ratio of corresponding medians is $k^2$, which is equal to the ratio of the areas.

This confirms that option (C) is correct.

The ratio of corresponding angle bisectors is also $k.

Therefore, the square of the ratio of corresponding angle bisectors is $k^2$, which is equal to the ratio of the areas.

This confirms that option (D) is correct.

The ratio of the perimeters of two similar triangles is also equal to the ratio of their corresponding sides.

Since $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$, we have $AB = k \cdot PQ$, $BC = k \cdot QR$, $AC = k \cdot PR$.

Therefore, the square of the ratio of perimeters is $k^2$, which is equal to the ratio of the areas.

This confirms that option (E) is correct.

Thus, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, altitudes, medians, angle bisectors, and perimeters.

(C) 16:81

Explanation:

Let the two similar triangles be $\triangle T_1$ and $\triangle T_2$.

We are given that the ratio of their perimeters is 4:9.

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

From (i), the ratio of corresponding sides is:

The theorem on the areas 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.

Substitute the ratio of sides from equation (ii) into equation (iii):

Calculate the square:

So, the ratio of the areas is 16:81.

This matches option (C).

(B) (5, 12, 13)

Explanation:

A Pythagorean triplet is a set of three positive integers $a$, $b$, and $c$, such that $a^2 + b^2 = c^2$. In a right-angled triangle, these three integers represent the lengths of the sides, with $c$ being the length of the hypotenuse (the longest side).

We need to check which of the given options satisfies this condition. For each triplet $(a, b, c)$, we check if the sum of the squares of the two smaller numbers equals the square of the largest number.

(A) (2, 3, 4):

Here, $a=2$, $b=3$, $c=4$. We check if $a^2 + b^2 = c^2$ or $2^2 + 3^2 = 4^2$.

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$2^2 + 3^2 = 4 + 9 = 13$

Since $13 \neq 16$, (2, 3, 4) is not a Pythagorean triplet.

(B) (5, 12, 13):

Here, $a=5$, $b=12$, $c=13$. We check if $a^2 + b^2 = c^2$ or $5^2 + 12^2 = 13^2$.

$5^2 = 25$

$12^2 = 144$

$13^2 = 169$

$5^2 + 12^2 = 25 + 144 = 169$

Since $169 = 169$, (5, 12, 13) is a Pythagorean triplet.

(C) (6, 8, 9):

Here, $a=6$, $b=8$, $c=9$. We check if $a^2 + b^2 = c^2$ or $6^2 + 8^2 = 9^2$.

$6^2 = 36$

$8^2 = 64$

$9^2 = 81$

$6^2 + 8^2 = 36 + 64 = 100$

Since $100 \neq 81$, (6, 8, 9) is not a Pythagorean triplet.

(D) (7, 24, 26):

Here, $a=7$, $b=24$, $c=26$. We check if $a^2 + b^2 = c^2$ or $7^2 + 24^2 = 26^2$.

$7^2 = 49$

$24^2 = 576$

$26^2 = 676$

$7^2 + 24^2 = 49 + 576 = 625$

Since $625 \neq 676$, (7, 24, 26) is not a Pythagorean triplet.

Only option (B) satisfies the condition of a Pythagorean triplet.

(C) Basic Proportionality Theorem

Explanation:

Let's define the relevant theorems:

The Basic Proportionality Theorem (BPT), also known as Thales's Theorem, states: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

That is, if in $\triangle ABC$, $DE \parallel BC$ (where D is on AB and E is on AC), then $\frac{AD}{DB} = \frac{AE}{EC}$.

The Converse of the Basic Proportionality Theorem states: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

That is, if in $\triangle ABC$, a line segment DE intersects sides AB and AC at D and E respectively such that $\frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

The statement given in the question: "In $\triangle ABC$, if a line segment DE intersects sides AB and AC at D and E respectively such that $\frac{AD}{DB} = \frac{AE}{EC}$, then DE is parallel to BC" matches the definition of the Converse of the Basic Proportionality Theorem.

The other options are incorrect:

(A) Pythagoras Theorem relates the sides of a right-angled triangle ($a^2 + b^2 = c^2$).

(B) Area Similarity Theorem relates the ratio of areas of similar triangles to the square of the ratio of their corresponding sides.

(D) Midpoint Theorem is a special case derived from the BPT, dealing specifically with a line segment joining the midpoints of two sides.

Therefore, the correct option is (C).

(C) Similar, similar

Explanation:

Consider a right triangle $\triangle ABC$, with the right angle at vertex B ($\angle ABC = 90^\circ$). Let BD be the altitude drawn from B to the hypotenuse AC, such that D lies on AC.

The altitude BD divides the original triangle $\triangle ABC$ into two smaller triangles: $\triangle ADB$ and $\triangle BDC$.

Let's examine the angles in these triangles:

In $\triangle ABC$: $\angle ABC = 90^\circ$. Let $\angle BAC = \angle A$ and $\angle BCA = \angle C$. We know that $\angle A + \angle C = 90^\circ$ (sum of angles in a triangle).

In $\triangle ADB$:

Since the sum of angles in $\triangle ADB$ is $180^\circ$, $\angle ABD = 180^\circ - 90^\circ - \angle A = 90^\circ - \angle A$.

Since $\angle C = 90^\circ - \angle A$, we have $\angle ABD = \angle C$.

Comparing $\triangle ABC$ and $\triangle ADB$:

By AA similarity criterion, $\triangle ABC \sim \triangle ADB$.

In $\triangle BDC$:

Since the sum of angles in $\triangle BDC$ is $180^\circ$, $\angle DBC = 180^\circ - 90^\circ - \angle C = 90^\circ - \angle C$.

Since $\angle A = 90^\circ - \angle C$, we have $\angle DBC = \angle A$.

Comparing $\triangle ABC$ and $\triangle BDC$:

By AA similarity criterion, $\triangle ABC \sim \triangle BDC$.

Now, let's compare $\triangle ADB$ and $\triangle BDC$:

Since $\angle A + \angle C = 90^\circ$, the remaining angles are equal ($\angle ABD = \angle C$ and $\angle DAB = \angle DBC$).

By AA similarity criterion, $\triangle ADB \sim \triangle BDC$.

Therefore, the altitude drawn from the vertex of the right angle to the hypotenuse divides the triangle into two triangles which are similar to the original triangle and also similar to each other.

This matches option (C).

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

Explanation:

Let's examine the Assertion (A) and the Reason (R).

Assertion (A): All equilateral triangles are similar.

An equilateral triangle has all three sides equal in length and all three interior angles equal.

The sum of angles in a triangle is $180^\circ$. Since all three angles in an equilateral triangle are equal, each angle must be $\frac{180^\circ}{3} = 60^\circ$.

Two triangles are similar if their corresponding angles are equal (AA or AAA similarity criterion).

Consider any two equilateral triangles. Both triangles have all their angles equal to $60^\circ$. Thus, the corresponding angles of any two equilateral triangles are equal.

Therefore, by the AAA similarity criterion, all equilateral triangles are similar.

The Assertion (A) is True.

Reason (R): All equilateral triangles have each angle equal to $60^\circ$.

As explained above, in an equilateral triangle, all three angles are equal, and their sum is $180^\circ$. Thus, each angle is indeed $60^\circ$.

The Reason (R) is True.

Now let's check if R is the correct explanation for A.

The similarity of equilateral triangles (Assertion A) is based on the fact that their corresponding angles are equal. Reason (R) provides exactly this fact – that each angle in an equilateral triangle is $60^\circ$. This directly leads to the conclusion that the corresponding angles of any two equilateral triangles are equal, which is the condition for similarity by the AAA criterion.

Therefore, Reason (R) is the correct explanation for Assertion (A).

Both A and R are true, and R correctly explains A.

This matches option (A).

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

Explanation:

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Consider a triangle with sides $a, b, c$. The assertion says that if $c^2 = a^2 + b^2$, then the angle opposite side $c$ is $90^\circ$. This is a fundamental theorem in geometry.

The Assertion (A) is True.

Reason (R): This is the converse of Pythagoras Theorem.

The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. That is, if an angle is $90^\circ$, then $c^2 = a^2 + b^2$ (where $c$ is the hypotenuse and $a, b$ are the other sides).

The Converse of Pythagoras Theorem states that if in a triangle, the square of one side is equal to the sum of the squares of the other two sides ($c^2 = a^2 + b^2$), then the angle opposite the first side (side $c$) is a right angle ($90^\circ$).

The statement in Assertion (A) is exactly the statement of the Converse of Pythagoras Theorem.

The Reason (R) is True.

Furthermore, Reason (R) provides the correct identification and explanation for why Assertion (A) is true. Assertion (A) is true *because* it is the Converse of Pythagoras Theorem, which is a proven theorem.

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

(A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d)

Explanation:

Let's match the similarity criteria with the information they use:

(i) AA (Angle-Angle) Similarity: This criterion states that if two angles of one triangle are equal to two corresponding angles of another triangle, then the triangles are similar. It uses the information about Two angles.

So, (i) matches with (c).

(ii) SSS (Side-Side-Side) Similarity: This criterion states that if the corresponding sides of two triangles are proportional, then the triangles are similar. It uses the information about Proportional sides.

So, (ii) matches with (b).

(iii) SAS (Side-Angle-Side) Similarity: This criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. It uses the information about Proportional sides and included angle.

So, (iii) matches with (a).

(iv) RHS (Right angle-Hypotenuse-Side): This is a criterion used to prove the congruence of two right-angled triangles. It is not a general similarity criterion.

So, (iv) matches with (d).

Combining the matches, we get:

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

This set of matches corresponds to option (A).

(A) 30 m

Given:

Horizontal distance from the camera to the stick = 2 m

Height of the stick = 3 m

Horizontal distance from the camera to the building = 20 m

Assume the camera/eye is at the origin (0,0) and the ground is the x-axis.

To Find:

The height of the building.

Solution:

According to the specific coordinate system assumption provided in the problem, the camera (eye) is located at the origin $(0,0)$ and the ground corresponds to the x-axis.

The stick is held vertically at a horizontal distance of 2 m from the camera. Since its base is on the ground (x-axis) and its height is 3 m, the coordinates of the top of the stick are $(2, 3)$.

The building is located at a horizontal distance of 20 m from the camera. Its base is on the ground (x-axis). Let the height of the building be $H$ meters. The coordinates of the top of the building are $(20, H)$.

Since the top of the stick aligns with the top of the building when viewed through the camera lens (at the origin), the three points - the camera $(0,0)$, the top of the stick $(2,3)$, and the top of the building $(20, H)$ - are collinear.

For three points to be collinear, the slope between any two pairs of points must be equal.

The slope of the line segment joining the camera $(0,0)$ and the top of the stick $(2,3)$ is:

The slope of the line segment joining the camera $(0,0)$ and the top of the building $(20, H)$ is:

Since the points are collinear, the slopes must be equal:

Equating the expressions for the slopes from (i) and (ii):

To solve for $H$, multiply both sides by 20:

Thus, the height of the building is 30 meters.

This result corresponds to option (A).

Note: The initial information about the camera being 1.5 m from the ground seems to be superseded by the explicit instruction to place the camera at the origin with the ground as the x-axis for setting up the coordinate system and similar triangles.

(C) 4:9

Explanation:

Let the two similar triangles be $\triangle T_1$ and $\triangle T_2$.

Let the ratio of their corresponding altitudes be $h_1 : h_2$. We are given:

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding altitudes.

Let Area($\triangle T_1$) and Area($\triangle T_2$) be the areas of the two triangles.

Substitute the given ratio of altitudes from equation (i):

Calculate the square:

So, the ratio of the areas is 4:9.

This matches option (C).

(A) True

Explanation:

We are given a right-angled triangle $\triangle ABC$ with $\angle A = 90^\circ$. AD is the altitude from the vertex A to the hypotenuse BC, which means $AD \perp BC$ and D is a point on BC. Thus, $\angle ADB = 90^\circ$ and $\angle ADC = 90^\circ$.

Consider the triangles $\triangle ABD$ and $\triangle CAD$.

In $\triangle ABC$, since $\angle A = 90^\circ$, we have $\angle B + \angle C = 180^\circ - 90^\circ = 90^\circ$.

In $\triangle ABD$, $\angle ADB = 90^\circ$. Therefore, $\angle B + \angle BAD = 180^\circ - 90^\circ = 90^\circ$.

From the angles in $\triangle ABC$, we have $\angle C = 90^\circ - \angle B$.

Comparing equations (i) and (ii), we get $\angle BAD = \angle C$.

Now let's consider $\triangle CAD$. $\angle ADC = 90^\circ$. Therefore, $\angle CAD + \angle C = 180^\circ - 90^\circ = 90^\circ$.

From the angles in $\triangle ABC$, we have $\angle B = 90^\circ - \angle C$.

Comparing equations (iii) and (iv), we get $\angle CAD = \angle B$.

Now we compare $\triangle ABD$ and $\triangle CAD$ based on their angles:

1. $\angle ADB = \angle ADC$ ($90^\circ$ each)

2. $\angle BAD = \angle C$ (Proved above)

3. $\angle B = \angle CAD$ (Proved above)

Since all three corresponding angles are equal, by the AAA similarity criterion, $\triangle ABD \sim \triangle CAD$.

Let's check the order of vertices in the similarity statement $\triangle ABD \sim \triangle CAD$. The corresponding angles are $\angle A_{\triangle ABD} (\angle BAD) \leftrightarrow \angle C_{\triangle CAD}$, $\angle B_{\triangle ABD} (\angle B) \leftrightarrow \angle A_{\triangle CAD} (\angle CAD)$, $\angle D_{\triangle ABD} (\angle ADB) \leftrightarrow \angle D_{\triangle CAD} (\angle ADC)$.

The order in the statement $\triangle ABD \sim \triangle CAD$ means A corresponds to C, B corresponds to A, and D corresponds to D. This matches our angle correspondences:

A (in $\triangle ABD$) $\leftrightarrow$ C (in $\triangle CAD$)

B (in $\triangle ABD$) $\leftrightarrow$ A (in $\triangle CAD$)

D (in $\triangle ABD$) $\leftrightarrow$ D (in $\triangle CAD$)

So, the statement $\triangle ABD \sim \triangle CAD$ is correct in terms of similarity and vertex order.

Therefore, the statement is True.

(A) 12 cm

Given:

$\triangle ABC \sim \triangle DEF$

Area$(\triangle ABC) = 48 \text{ cm}^2$

Area$(\triangle DEF) = 108 \text{ cm}^2$

BC = 8 cm

To Find:

The length of EF.

Solution:

We are given that $\triangle ABC \sim \triangle DEF$.

According to the theorem on the areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

The side BC in $\triangle ABC$ corresponds to the side EF in $\triangle DEF$.

Therefore, we can write the ratio of areas as:

Substitute the given values into equation (i):

Simplify the fraction on the left side. Both 48 and 108 are divisible by common factors like 2, 4, 6, 12. The greatest common divisor is 12.

So the equation becomes:

Take the square root of both sides:

Now, solve for EF by cross-multiplication:

Divide by 2:

The length of EF is 12 cm.

This matches option (A).

(C) All isosceles triangles are similar.

Explanation:

Let's analyze each statement:

(A) All squares are similar.

A square is a quadrilateral with four equal sides and four right angles ($90^\circ$). For any two squares, their corresponding angles are always equal (all are $90^\circ$). The ratio of corresponding sides will always be constant for any two squares (the ratio of their side lengths). Since corresponding angles are equal and corresponding sides are proportional, all squares are similar. This statement is TRUE.

(B) All circles are similar.

A circle is defined by its center and radius. All circles have the same shape. Any circle can be transformed into any other circle by a combination of translation and uniform scaling (changing the radius proportionally). Therefore, all circles are similar. This statement is TRUE.

(C) All isosceles triangles are similar.

An isosceles triangle has at least two sides of equal length, and the angles opposite these sides are equal. However, the angle measures can vary between different isosceles triangles. For example, an isosceles triangle with angles $50^\circ, 50^\circ, 80^\circ$ is isosceles. Another isosceles triangle with angles $70^\circ, 70^\circ, 40^\circ$ is also isosceles. These two triangles have different angle measures, so their corresponding angles are not equal. Therefore, they are not similar. This statement is FALSE.

(D) All congruent figures are similar.

Congruent figures have the same shape and the same size. Similar figures have the same shape but not necessarily the same size. If two figures are congruent, they have the same shape, and the ratio of their corresponding sides is 1:1. Since they have the same shape and corresponding sides are proportional (ratio 1), they satisfy the conditions for similarity. Thus, all congruent figures are similar (with a similarity ratio of 1). This statement is TRUE.

The statement that is FALSE is (C).

(B) Area$(\triangle ADE) = \frac{1}{4}$ Area$(\triangle ABC)$

Explanation:

We are given that in $\triangle ABC$, D and E are the midpoints of sides AB and AC respectively.

According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

Thus, we have $DE \parallel BC$ and $DE = \frac{1}{2}BC$.

Now, let's consider $\triangle ADE$ and $\triangle ABC$.

Since $DE \parallel BC$, we can consider AB as a transversal line intersecting parallel lines DE and BC. This gives:

Similarly, considering AC as a transversal line intersecting parallel lines DE and BC:

Also, the angle at vertex A is common to both triangles:

Since two corresponding angles of $\triangle ADE$ and $\triangle ABC$ are equal ($\angle ADE = \angle ABC$ and $\angle DAE = \angle BAC$), by the AA similarity criterion, we have:

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

The corresponding sides are AD and AB, AE and AC, and DE and BC.

Since D is the midpoint of AB, $AD = \frac{1}{2}AB$. So, $\frac{AD}{AB} = \frac{1}{2}$.

Since E is the midpoint of AC, $AE = \frac{1}{2}AC$. So, $\frac{AE}{AC} = \frac{1}{2}$.

From the Midpoint Theorem, $DE = \frac{1}{2}BC$. So, $\frac{DE}{BC} = \frac{1}{2}$.

The ratio of corresponding sides is $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = \frac{1}{2}$.

Now, applying the area similarity theorem:

Substitute the ratio of sides:

Calculate the square:

So, we have:

Multiplying both sides by Area$(\triangle ABC)$, we get:

Thus, the area of $\triangle ADE$ is $\frac{1}{4}$ the area of $\triangle ABC$.

This matches option (B).

(A) 12 m

Given:

Length of the ladder (Hypotenuse) = 15 m

Distance of the foot of the ladder from the wall (Base) = 9 m

To Find:

Height the ladder reaches on the wall (Perpendicular height).

Solution:

The ladder leaning against the wall forms a right-angled triangle with the wall and the ground. The ladder is the hypotenuse, the distance from the wall to the foot of the ladder is one leg, and the height the ladder reaches on the wall is the other leg.

Let $H$ be the height the ladder reaches on the wall.

According to the Pythagoras Theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Applying the theorem, we have:

Calculate the squares:

$9^2 = 9 \times 9 = 81$

$15^2 = 15 \times 15 = 225$

Substitute these values back into equation (i):

Subtract 81 from both sides to find $H^2$:

Take the square root of both sides to find H:

The height the ladder reaches on the wall is 12 m.

This matches option (A).

(A) 3, 4, 5

(B) 6, 8, 10

(D) 5, 12, 13

(E) 1, 1, $\sqrt{2}$

Explanation:

According to the Converse of Pythagoras Theorem, a triangle with side lengths $a$, $b$, and $c$ is a right triangle if and only if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides. That is, if $c$ is the longest side, then $a^2 + b^2 = c^2$.

We will check this condition for each option:

(A) (3, 4, 5):

The longest side is 5.

$9 + 16 = 25$

$25 = 25$

The condition is satisfied. So, (3, 4, 5) can be the side lengths of a right triangle. This is a common Pythagorean triplet.

(B) (6, 8, 10):

The longest side is 10.

$36 + 64 = 100$

$100 = 100$

The condition is satisfied. So, (6, 8, 10) can be the side lengths of a right triangle. This is a multiple of the (3, 4, 5) triplet ($2 \times 3, 2 \times 4, 2 \times 5$).

(C) (7, 9, 11):

The longest side is 11.

$49 + 81 = 121$

$130 = 121$

The condition is not satisfied ($130 \neq 121$). So, (7, 9, 11) cannot be the side lengths of a right triangle.

(D) (5, 12, 13):

The longest side is 13.

$25 + 144 = 169$

$169 = 169$

The condition is satisfied. So, (5, 12, 13) can be the side lengths of a right triangle. This is a common Pythagorean triplet.

(E) (1, 1, $\sqrt{2}$):

The longest side is $\sqrt{2}$.

$1 + 1 = 2$

$2 = 2$

The condition is satisfied. So, (1, 1, $\sqrt{2}$) can be the side lengths of a right triangle (specifically, an isosceles right triangle). This is a valid Pythagorean triplet where one side is not an integer.

The sets of side lengths that form a right triangle are (A), (B), (D), and (E).

(A) 8 cm

Given:

Two similar triangles, let's call them $\triangle T_L$ (larger) and $\triangle T_S$ (smaller).

Ratio of their areas: $\frac{\text{Area}(\triangle T_L)}{\text{Area}(\triangle T_S)} = \frac{100}{64}$

Median of the larger triangle ($m_L$) = 10 cm

To Find:

The corresponding median of the smaller triangle ($m_S$).

Solution:

We are given that the two triangles are similar.

According to the theorem relating areas and corresponding linear elements of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Mathematically, this can be written as:

Substitute the given values into equation (i):

Simplify the fraction on the left side:

So, the equation becomes:

Take the square root of both sides:

Now, solve for $m_S$ by cross-multiplication:

Divide both sides by 5:

The corresponding median of the smaller triangle is 8 cm.

This matches option (A).

(B) Yes, by converse of BPT.

Given:

In $\triangle PQR$, S is on PQ and T is on PR.

PS = 3

SQ = 4

PT = 4.5

TR = 6

To Determine:

If ST is parallel to QR.

Solution:

According to the Converse of the Basic Proportionality Theorem (BPT), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

To check if ST is parallel to QR, we need to determine if the line segment ST divides the sides PQ and PR in the same ratio. That is, we need to check if $\frac{PS}{SQ} = \frac{PT}{TR}$.

Let's calculate the ratio $\frac{PS}{SQ}$:

Now let's calculate the ratio $\frac{PT}{TR}$:

To compare the ratios, let's simplify the fraction $\frac{4.5}{6}$. We can multiply the numerator and denominator by 10 to remove the decimal, and then simplify:

Divide the numerator and denominator by their greatest common divisor, which is 15:

So, the ratio $\frac{PT}{TR} = \frac{3}{4}$.

Comparing the ratios from (i) and (ii):

Since $\frac{PS}{SQ} = \frac{PT}{TR}$, the line segment ST divides sides PQ and PR in the same ratio.

By the Converse of Basic Proportionality Theorem, if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore, ST is parallel to QR.

The correct option is (B).

(B) Equal, proportional

Explanation:

By the definition of similar triangles, two triangles are similar if and only if:

1. Their corresponding angles are equal.

2. Their corresponding sides are proportional.

Let $\triangle ABC \sim \triangle PQR$. This similarity implies:

AND

Based on this definition, the corresponding angles are equal and the corresponding sides are proportional.

Option (B) correctly states this relationship.

Option (A) is incorrect because angles are equal, not proportional, and sides are proportional, not equal (unless the triangles are congruent).

Option (C) is incorrect because sides are proportional, not necessarily equal (equal sides imply congruence).

Option (D) is incorrect because angles are equal, not proportional (unless the angles are all $0^\circ$, which is not possible for a triangle).

Therefore, the correct option is (B).

(C) 0.04 m

Given:

Actual height of the building = 50 m

Height of the model = 1 m

Actual width of a window = 2 m

To Find:

The width of the corresponding window on the scale model.

Solution:

The scale model and the actual building are similar figures. In similar figures, the ratio of corresponding lengths is constant. This ratio is called the scale factor.

We can find the scale factor by comparing the height of the model to the actual height of the building:

This means that any linear dimension on the model is $\frac{1}{50}$ times the corresponding linear dimension on the actual building.

The width of the window on the model will be the actual width of the window multiplied by the scale factor.

Substitute the given actual width and the calculated scale factor:

Calculate the result:

Simplify the fraction:

So, the width of the model window is $\frac{1}{25}$ m.

To express this as a decimal, we perform the division $1 \div 25$:

This matches option (C).

(D) 0.4 m$^2$

Given:

Actual floor area ($A_a$) = $1000 \text{ m}^2$

Actual height of the building = 50 m

Height of the model = 1 m

To Find:

The floor area of the scale model ($A_m$).

Solution:

The actual building and its scale model are similar figures.

The ratio of corresponding linear dimensions is called the scale factor.

The ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding linear dimensions (the scale factor).

Substitute the given actual area ($A_a$) and the calculated scale factor ($k$) into equation (ii):

Calculate the square of the scale factor:

So, the equation becomes:

To solve for $A_m$, multiply both sides of the equation by 1000:

Simplify the fraction:

So, $A_m = \frac{2}{5} \text{ m}^2$.

To express this as a decimal:

The floor area of the scale model is $0.4 \text{ m}^2$.

This matches option (D).

(B) $DG \cdot GF$

Explanation:

Consider the right-angled triangle $\triangle DEF$, where $\angle E = 90^\circ$. EG is the altitude drawn from the right angle vertex E to the hypotenuse DF, such that G is a point on DF.

The altitude to the hypotenuse in a right triangle creates two smaller triangles ($\triangle DEG$ and $\triangle EFG$) that are similar to the original triangle and also similar to each other.

Let's consider the two smaller triangles, $\triangle DEG$ and $\triangle EFG$.

In $\triangle DEG$, $\angle DGE = 90^\circ$ (since EG is the altitude).

In $\triangle EFG$, $\angle EGF = 90^\circ$ (since EG is the altitude).

In the original triangle $\triangle DEF$, $\angle D + \angle F = 90^\circ$ (sum of acute angles in a right triangle).

In $\triangle DEG$, $\angle DEG = 180^\circ - 90^\circ - \angle D = 90^\circ - \angle D$. Since $\angle F = 90^\circ - \angle D$, we have $\angle DEG = \angle F$.

In $\triangle EFG$, $\angle GEF = 180^\circ - 90^\circ - \angle F = 90^\circ - \angle F$. Since $\angle D = 90^\circ - \angle F$, we have $\angle GEF = \angle D$.

Now, let's compare $\triangle DEG$ and $\triangle EFG$ based on their angles:

1. $\angle DGE = \angle EGF = 90^\circ$

2. $\angle EDG = \angle GEF$ (since $\angle EDG = \angle D$ and $\angle GEF = \angle D$)

3. $\angle DEG = \angle EFG$ (since $\angle DEG = \angle F$ and $\angle EFG = \angle F$)

By AAA (or AA) similarity criterion, $\triangle DEG \sim \triangle EFG$. We need to be careful with the order of vertices to match corresponding angles.

Correct correspondence: $\angle D \leftrightarrow \angle GEF$, $\angle G \leftrightarrow \angle G$, $\angle E \leftrightarrow \angle F$.

So, $\triangle DGE \sim \triangle EGF$.

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

Using the sides opposite the corresponding angles we identified earlier: $\angle D \leftrightarrow \angle GEF$, $\angle DEG \leftrightarrow \angle F$, $\angle DGE \leftrightarrow \angle EGF$.

Sides opposite: $(EG, DE, DG)$ in $\triangle DEG$ correspond to $(GF, EF, EG)$ in $\triangle EFG$.

Ratio of sides opposite $\angle D$ and $\angle GEF$: $\frac{EG}{GF}$

Ratio of sides opposite $\angle DEG$ and $\angle F$: $\frac{DG}{EG}$

Ratio of sides opposite $\angle DGE$ and $\angle EGF$: $\frac{DE}{EF}$

Thus, the proportion is $\frac{EG}{GF} = \frac{DG}{EG} = \frac{DE}{EF}$.

Taking the first two ratios:

Cross-multiplying, we get:

This identity states that the square of the altitude to the hypotenuse in a right triangle is equal to the product of the segments into which the hypotenuse is divided.

This matches option (B).

(C) ASA

Explanation:

Let's review the criteria for similarity and congruence of triangles.

Similarity Criteria for Triangles:

1. AA (Angle-Angle) or AAA (Angle-Angle-Angle): If two (or three) corresponding angles of two triangles are equal, then the triangles are similar.

2. SSS (Side-Side-Side) Similarity: If the ratios of the lengths of corresponding sides of two triangles are equal (i.e., corresponding sides are proportional), then the triangles are similar.

3. SAS (Side-Angle-Side) Similarity: If in two triangles, one pair of corresponding angles are equal and the sides including these angles are proportional, then the triangles are similar.

Common Congruence Criteria for Triangles:

1. SSS (Side-Side-Side) Congruence: If three sides of one triangle are equal in length to the corresponding three sides of another triangle, then the triangles are congruent.

2. SAS (Side-Angle-Side) Congruence: If two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.

3. ASA (Angle-Side-Angle) Congruence: If two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, then the triangles are congruent.

4. AAS (Angle-Angle-Side) Congruence: If two angles and a non-included side of one triangle are equal to the corresponding two angles and the corresponding non-included side of another triangle, then the triangles are congruent.

5. RHS (Right angle-Hypotenuse-Side) Congruence: If in two right-angled triangles, the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle, then the triangles are congruent.

Comparing the given options with the list of similarity criteria:

(A) AA is a similarity criterion.

(B) SSS is a similarity criterion (proportional sides).

(C) ASA is a congruence criterion, not a similarity criterion.

(D) SAS is a similarity criterion (proportional sides and equal included angle).

The criterion that is NOT a criterion for similarity of triangles is ASA.

(C) Right-angled

Explanation:

Given:

The lengths of the sides of a triangle are $p^2+q^2$, $p^2-q^2$, and $2pq$.

For these lengths to form a valid triangle and be positive, we assume $p > q > 0$.

To Determine:

The type of triangle.

Solution:

Let the side lengths of the triangle be $a = p^2-q^2$, $b = 2pq$, and $c = p^2+q^2$.

We need to determine if these side lengths satisfy the condition for a right-angled triangle using the Converse of the Pythagoras Theorem.

The Converse of the Pythagoras Theorem states that if in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Assuming $p > q > 0$, the longest side among $p^2+q^2$, $p^2-q^2$, and $2pq$ is $p^2+q^2$. Let's call the longest side $c = p^2+q^2$, and the other two sides $a = p^2-q^2$ and $b = 2pq$.

We check if $a^2 + b^2 = c^2$.

Calculate the sum of the squares of the two shorter sides ($a^2 + b^2$):

Expand the terms $(p^2-q^2)^2$ and $(2pq)^2$:

Now, calculate the square of the longest side ($c^2$):

Expand the term $(p^2+q^2)^2$:

Comparing the results from equation (ii) and equation (iii):

The square of one side ($p^2+q^2$) is equal to the sum of the squares of the other two sides ($p^2-q^2$ and $2pq$).

Therefore, by the Converse of the Pythagoras Theorem, the angle opposite the side with length $p^2+q^2$ is a right angle ($90^\circ$).

Hence, the triangle is a Right-angled triangle.

This matches option (C).

These side lengths $(p^2-q^2, 2pq, p^2+q^2)$ are a general form for generating Pythagorean triplets (when p and q are integers). Depending on the specific values of $p$ and $q$, the triangle might also be scalene (if $p^2-q^2 \neq 2pq$) or isosceles (if $p^2-q^2 = 2pq$), but it is always a right-angled triangle for valid $p, q$. The primary classification based on this relationship is "Right-angled".

(D) 4/49

Given:

In $\triangle ABC$, D is a point on AB and E is a point on AC.

$DE \parallel BC$

$\frac{AD}{DB} = \frac{2}{5}$

To Find:

The ratio $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)}$.

Solution:

We are given that $DE \parallel BC$ in $\triangle ABC$.

Consider $\triangle ADE$ and $\triangle ABC$.

Since $DE \parallel BC$, AB is a transversal intersecting DE and BC.

By the AA similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Therefore, $\triangle ADE \sim \triangle ABC$.

The theorem on the areas 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.

The ratio of areas is $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\frac{AD}{AB}\right)^2$.

We are given the ratio $\frac{AD}{DB} = \frac{2}{5}$. This means that for some constant $k$, $AD = 2k$ and $DB = 5k$.

The length of side AB is the sum of AD and DB:

Now we can find the ratio of the corresponding side AD to AB:

Apply the area similarity theorem using this ratio:

Calculate the square:

So, the ratio of the areas is $\frac{4}{49}$.

This matches option (D).

(B) $144 \text{ cm}^2$

Given:

Two similar triangles, let's call them $\triangle T_S$ (smaller) and $\triangle T_L$ (larger).

Ratio of corresponding sides: $\frac{\text{Side of } \triangle T_S}{\text{Side of } \triangle T_L} = \frac{1}{3}$

Area of the smaller triangle (Area($\triangle T_S$)) = $16 \text{ cm}^2$

To Find:

The area of the larger triangle (Area($\triangle T_L$)).

Solution:

We are given that the two triangles are similar.

According to the theorem on the areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Substitute the given ratio of sides and the area of the smaller triangle into equation (i):

Calculate the square of the ratio:

So, the equation becomes:

Now, solve for Area$(\triangle T_L)$ by cross-multiplication:

The area of the larger triangle is $144 \text{ cm}^2$.

This matches option (B).

(B) AD = DB.

Given:

In $\triangle ABC$, D is a point on side AB and E is a point on side AC.

Area$(\triangle ADE) =$ Area$(\triangle BDE)$.

To Determine:

What the equality of areas implies about the points D and E.

Solution:

Consider the two triangles $\triangle ADE$ and $\triangle BDE$.

Both triangles share the common vertex E.

Their bases, AD and DB, lie on the same straight line segment AB.

The area of a triangle is given by the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

When considering AD as the base of $\triangle ADE$ and DB as the base of $\triangle BDE$, the height of both triangles with respect to these bases is the perpendicular distance from the common vertex E to the line containing AB (which is the line AB itself).

Let $h$ be the perpendicular distance from point E to the line AB.

The area of $\triangle ADE$ with base AD is:

The area of $\triangle BDE$ with base DB is:

We are given that Area$(\triangle ADE) =$ Area$(\triangle BDE)$.

Since E is on AC and AC is a side of the triangle (assuming a non-degenerate triangle), E is not on the line AB, so the height $h > 0$. Also, $\frac{1}{2} \neq 0$. We can divide both sides of the equation by $\frac{1}{2} h$.

This equality implies that the point D is the midpoint of the side AB.

Let's evaluate the given options:

(A) DE is parallel to BC. This would be true if D and E were *both* midpoints of AB and AC respectively (by the Midpoint Theorem, which is a consequence of BPT). However, the given condition Area$(\triangle ADE) =$ Area$(\triangle BDE)$ only implies that D is the midpoint of AB. E can be any point on AC, and the areas would still be equal as long as D is the midpoint of AB. Thus, DE is not necessarily parallel to BC.

(B) AD = DB. As derived above, the equality of areas directly implies that AD = DB.

(C) AE = EC. This means E is the midpoint of AC. The given condition Area$(\triangle ADE) =$ Area$(\triangle BDE)$ does not provide any information about the position of E on AC. E could be any point on AC, and the area equality would still hold if AD=DB. Thus, AE = EC is not implied.

(D) Both B and C. Since option (C) is false, this option is also false.

Therefore, the only implication of Area$(\triangle ADE) =$ Area$(\triangle BDE)$ is that AD = DB.

(A) Yes, by AA similarity.

Explanation:

We are given two isosceles triangles, and their corresponding angles are equal.

Let the two isosceles triangles be $\triangle ABC$ and $\triangle PQR$.

If their corresponding angles are equal, it means:

One of the criteria for similarity of triangles is the AA (Angle-Angle) similarity criterion. This criterion states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since we are given that all three corresponding angles are equal ($\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$), the condition for AA similarity is satisfied (if two angles are equal, the third must also be equal, so AAA is also applicable).

Therefore, if two triangles (whether isosceles or not) have their corresponding angles equal, they are necessarily similar.

The fact that the triangles are isosceles is additional information, but the condition of equal corresponding angles is sufficient to conclude similarity.

Option (A) correctly identifies that they are necessarily similar and points to the relevant similarity criterion (AA similarity, which is equivalent to AAA similarity in this context).

Option (B) is incorrect because equality of corresponding angles is a sufficient condition for similarity; proportionality of sides is a consequence, not a necessary additional condition when angles are already equal.

Option (C) is incorrect; similar triangles are not necessarily congruent.

Option (D) is incorrect; they are always similar if their corresponding angles are equal.

Thus, if two isosceles triangles have their corresponding angles equal, they are necessarily similar by AA similarity.

(A) 10 cm

Given:

Lengths of the diagonals of a rhombus are $d_1 = 12$ cm and $d_2 = 16$ cm.

To Find:

The length of each side of the rhombus.

Solution:

Let the rhombus be ABCD, and let the diagonals AC and BD intersect at point O.

We know that the diagonals of a rhombus bisect each other at right angles.

This means that $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ$.

Also, the diagonals bisect each other, so the point O is the midpoint of both diagonals.

Length of half of diagonal AC is $AO = OC = \frac{1}{2} \times AC = \frac{1}{2} \times 16 \text{ cm} = 8 \text{ cm}$.

Length of half of diagonal BD is $BO = OD = \frac{1}{2} \times BD = \frac{1}{2} \times 12 \text{ cm} = 6 \text{ cm}$.

Consider the triangle $\triangle AOB$. This is a right-angled triangle at O.

The sides of $\triangle AOB$ are AO, BO, and AB. AB is one of the sides of the rhombus and is the hypotenuse of $\triangle AOB$.

We can use the Pythagoras Theorem in $\triangle AOB$ to find the length of the side AB.

Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Applying the theorem to $\triangle AOB$:

Substitute the lengths of AO and BO into equation (i):

Calculate the squares:

Add the squared values:

To find the length of AB, take the square root of both sides:

Since all sides of a rhombus are equal in length, the length of each side of the rhombus is 10 cm.

This matches option (A).

Note that option (D) is $\sqrt{12^2+16^2}$ cm, which calculates the length of the diagonal squared, not the side length. It should be $\sqrt{(12/2)^2 + (16/2)^2} = \sqrt{6^2 + 8^2}$.

(C) $\triangle QMR \sim \triangle PQR$

Given:

In $\triangle PQR$, $\angle Q = 90^\circ$.

QM is perpendicular to PR (QM $\perp$ PR).

To Determine:

Which of the given similarity statements is TRUE.

Solution:

We have a right-angled triangle $\triangle PQR$, with the right angle at Q. QM is the altitude drawn from the vertex of the right angle Q to the hypotenuse PR. Point M lies on PR, and $\angle QMP = 90^\circ$, $\angle QMR = 90^\circ$.

The altitude drawn from the vertex of the right angle to the hypotenuse divides the triangle into two smaller triangles which are similar to the original triangle and also similar to each other.

The three triangles are $\triangle PQR$ (the original triangle), $\triangle PQM$, and $\triangle RQM$.

Let's analyze the angles in each triangle:

In $\triangle PQR$: $\angle PQR = 90^\circ$. The sum of the acute angles is $\angle P + \angle R = 180^\circ - 90^\circ = 90^\circ$.

In $\triangle PQM$: $\angle QMP = 90^\circ$. The sum of the acute angles is $\angle P + \angle PQM = 180^\circ - 90^\circ = 90^\circ$.

From $\triangle PQR$, we know $\angle R = 90^\circ - \angle P$. Comparing with (i), we get $\angle PQM = \angle R$.

In $\triangle RQM$: $\angle QMR = 90^\circ$. The sum of the acute angles is $\angle R + \angle RQM = 180^\circ - 90^\circ = 90^\circ$.

From $\triangle PQR$, we know $\angle P = 90^\circ - \angle R$. Comparing with (ii), we get $\angle RQM = \angle P$.

Now let's check the similarity statements by comparing corresponding angles based on the vertex order given in each option.

(A) $\triangle PQM \sim \triangle RQM$

This statement implies the following angle correspondences:

$\angle P \leftrightarrow \angle R$

$\angle Q_{\triangle PQM} (\angle PQM) \leftrightarrow \angle Q_{\triangle RQM} (\angle RQM)$

$\angle M_{\triangle PQM} (\angle QMP) \leftrightarrow \angle M_{\triangle RQM} (\angle QMR)$

We know that $\angle QMP = 90^\circ$ and $\angle QMR = 90^\circ$, so the correspondence $\angle QMP \leftrightarrow \angle QMR$ is True.

However, $\angle P$ is not necessarily equal to $\angle R$ (unless $\triangle PQR$ is an isosceles right triangle).

Also, $\angle PQM = \angle R$ and $\angle RQM = \angle P$. So, $\angle PQM$ is not necessarily equal to $\angle RQM$ unless $\angle P = \angle R$.

Thus, this similarity statement is FALSE for a general right triangle due to the incorrect vertex order.

(B) $\triangle PQM \sim \triangle RQP$

This statement implies the following angle correspondences:

$\angle P \leftrightarrow \angle R$

$\angle Q_{\triangle PQM} (\angle PQM) \leftrightarrow \angle Q_{\triangle RQP} (\angle PQR)$

$\angle M_{\triangle PQM} (\angle QMP) \leftrightarrow \angle P_{\triangle RQP} (\angle QPR)$

$\angle QMP = 90^\circ$, $\angle PQR = 90^\circ$, $\angle QPR = \angle P$. The correspondence $\angle PQM \leftrightarrow \angle PQR$ is $\angle PQM \leftrightarrow 90^\circ$, which is generally false. The correspondence $\angle QMP \leftrightarrow \angle QPR$ is $90^\circ \leftrightarrow \angle P$, which is generally false.

Thus, this similarity statement is FALSE.

(C) $\triangle QMR \sim \triangle PQR$

This statement implies the following angle correspondences:

$\angle Q_{\triangle QMR} (\angle RQM) \leftrightarrow \angle P_{\triangle PQR}$

$\angle M_{\triangle QMR} (\angle QMR) \leftrightarrow \angle Q_{\triangle PQR} (\angle PQR)$

$\angle R_{\triangle QMR} (\angle MRQ) \leftrightarrow \angle R_{\triangle PQR} (\angle PRQ)$

We check if these angle equalities hold:

$\angle RQM = \angle P$ (Proved above). $\checkmark$

So, $\angle QMR = \angle PQR$. $\checkmark$

$\angle MRQ$ in $\triangle QMR$ is the same angle as $\angle PRQ$ (or $\angle R$) in $\triangle PQR$. This is a common angle.

So, $\angle R_{\triangle QMR} = \angle R_{\triangle PQR}$. $\checkmark$

Since all three corresponding angles are equal, the similarity statement $\triangle QMR \sim \triangle PQR$ is TRUE.

(D) $\triangle MPQ \sim \triangle MRP$

$\triangle MPQ$ is the same as $\triangle PQM$. $\triangle MRP$ is the same as $\triangle RMP$. The statement is $\triangle PQM \sim \triangle RMP$.

This statement implies the following angle correspondences:

$\angle M_{\triangle PQM} (\angle QMP) \leftrightarrow \angle M_{\triangle RMP} (\angle RMP)$

$\angle P_{\triangle PQM} \leftrightarrow \angle R_{\triangle RMP}$

$\angle Q_{\triangle PQM} (\angle PQM) \leftrightarrow \angle P_{\triangle RMP} (\angle MRP)$

$\angle QMP = 90^\circ$ and $\angle RMP = 90^\circ$. So $\angle QMP \leftrightarrow \angle RMP$ is True.

$\angle P$ is not necessarily equal to $\angle R$.

$\angle PQM = \angle R$ and $\angle MRP = \angle P$. So $\angle PQM \leftrightarrow \angle MRP$ is $\angle R \leftrightarrow \angle P$, which is generally false.

Thus, this similarity statement is FALSE due to incorrect vertex order.

Based on the analysis, only option (C) provides a correct similarity statement with the correct vertex order.

(B) Yes, by SAS

Given:

In $\triangle ABC$: $AB=10$ m, $BC=12$ m, $\angle B = 80^\circ$.

In $\triangle XYZ$: $XY=15$ m, $YZ=18$ m, $\angle Y = 80^\circ$.

To Determine:

If $\triangle ABC$ is similar to $\triangle XYZ$ and by which criterion.

Solution:

We are given the lengths of two sides and the measure of the included angle for both triangles.

Let's check if the SAS (Side-Angle-Side) similarity criterion can be applied.

The SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

We compare the given information for $\triangle ABC$ and $\triangle XYZ$:

1. Check the included angles: The angle included between sides AB and BC in $\triangle ABC$ is $\angle B$. The angle included between sides XY and YZ in $\triangle XYZ$ is $\angle Y$.

Thus, $\angle B = \angle Y$. The included angles are equal.

2. Check the proportionality of the sides including the angles: The sides including $\angle B$ are AB and BC. The sides including $\angle Y$ are XY and YZ.

Let's check the ratio of corresponding sides AB and XY:

Now, let's check the ratio of corresponding sides BC and YZ:

Since both ratios are equal, the corresponding sides including the equal angles are proportional:

Since $\angle B = \angle Y$ and the sides including these angles are proportional ($\frac{AB}{XY} = \frac{BC}{YZ}$), by the SAS similarity criterion, $\triangle ABC \sim \triangle XYZ$. Note that the vertex order matches the correspondence: B corresponds to Y, AB corresponds to XY, and BC corresponds to YZ.

Therefore, the two triangles are similar by the SAS criterion.

This matches option (B).

Definition of Similar Figures:

Two figures are said to be similar if they have the same shape but not necessarily the same size.

Specifically, for polygons, two polygons are similar if:

(i) Their corresponding angles are equal.

(ii) The lengths of their corresponding sides are in the same ratio (proportional).

Difference between Similar and Congruent Figures:

While similar figures have the same shape, congruent figures have both the same shape and the same size.

Think of it this way:

• Similar figures are like a photograph and an enlargement of that photograph. They look the same, but one is bigger or smaller than the other. The ratio of corresponding lengths is constant, but the lengths themselves are generally different.
• Congruent figures are like two identical copies of the same photograph. They are exactly the same in both shape and size. If you place one on top of the other, they would perfectly overlap. The ratio of corresponding lengths is 1:1, meaning the corresponding lengths are equal.

In summary:

• Similar Figures: Same shape, different or same size. Corresponding angles are equal, corresponding sides are proportional.
• Congruent Figures: Same shape and same size. Corresponding angles are equal, corresponding sides are equal (which is a special case of proportionality with ratio 1).

Therefore, all congruent figures are similar, but not all similar figures are congruent.

Let's analyze the relationship between similar and congruent figures.

Are all congruent figures similar?

The answer is Yes.

Justification:

Recall the definitions:

• Similar figures have the same shape and their corresponding angles are equal, and their corresponding sides are in proportion.
• Congruent figures have the same shape and the same size. This means their corresponding angles are equal, and their corresponding sides are equal in length.

If two figures are congruent, their corresponding angles are equal (satisfying condition (i) for similarity). Also, since their corresponding sides are equal in length, the ratio of the lengths of any pair of corresponding sides is always $1:1$. This constant ratio $1$ is a proportionality constant, satisfying condition (ii) for similarity.

Therefore, since congruent figures satisfy all the conditions required for similarity, all congruent figures are indeed similar.

Example:

Consider two squares, Square A and Square B, both with side length $5$ cm.

Square A and Square B are congruent because they have the same shape (square) and the same size (side length $5$ cm).

Let's check if they are similar:

• Corresponding angles are all $90^\circ$ for both squares. They are equal.
• The ratio of corresponding sides is $\frac{5 \text{ cm}}{5 \text{ cm}} = 1$. This ratio is constant for all pairs of corresponding sides.

Since both conditions for similarity are met, Square A and Square B are also similar.

Are all similar figures congruent?

The answer is No.

Justification:

Similar figures are only required to have the same shape and proportional corresponding sides. They do not necessarily have the same size.

For figures to be congruent, they must have the same size as well as the same shape. If two similar figures have different sizes, they cannot be congruent.

Example:

Consider two squares, Square C with side length $2$ cm and Square D with side length $4$ cm.

Let's check if they are similar:

• Corresponding angles are all $90^\circ$ for both squares. They are equal.
• The ratio of corresponding sides is $\frac{2 \text{ cm}}{4 \text{ cm}} = \frac{1}{2}$. This ratio is constant for all pairs of corresponding sides.

Since both conditions for similarity are met, Square C and Square D are similar.

Now, let's check if they are congruent:

Square C has side length $2$ cm, and Square D has side length $4$ cm. Their sizes are different.

Since their sizes are different, Square C and Square D are not congruent.

This example shows that two figures can be similar without being congruent.

Conclusion:

All congruent figures are similar, but not all similar figures are congruent.

For two polygons with the same number of sides to be similar, the following two conditions must be satisfied:

Condition 1: Corresponding Angles

The corresponding angles of the two polygons must be equal.

For example, if we have two pentagons ABCDE and PQRST such that polygon ABCDE is similar to polygon PQRST, then:

$\angle A = \angle P$

$\angle B = \angle Q$

$\angle C = \angle R$

$\angle D = \angle S$

$\angle E = \angle T$

Condition 2: Corresponding Sides

The lengths of the corresponding sides of the two polygons must be in the same ratio (i.e., they must be proportional).

Using the same pentagons ABCDE and PQRST, if they are similar, then:

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CD}{RS} = \frac{DE}{ST} = \frac{EA}{TP} = k$ (where $k$ is a positive constant known as the scale factor or ratio of similarity).

Both conditions must be met for the polygons to be similar. If only one condition is met, the polygons are not necessarily similar.

Basic Proportionality Theorem (BPT) or Thales Theorem:

The theorem states:

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.

Let's illustrate this with a triangle ABC.

Suppose DE is a line segment drawn inside triangle ABC such that it is parallel to side BC and intersects side AB at point D and side AC at point E.

According to the BPT, if $DE \parallel BC$, then the ratio of the segments on sides AB and AC are equal.

This is the mathematical statement of the Basic Proportionality Theorem.

Given:

In $\triangle ABC$, $DE \parallel BC$.

$AD = 3$ cm

$DB = 4$ cm

$AE = 4.5$ cm

To Find:

The length of $EC$.

Solution:

In $\triangle ABC$, we are given that the line segment $DE$ is parallel to the side $BC$, and it intersects the other two sides $AB$ and $AC$ at distinct points $D$ and $E$ respectively.

According to the Basic Proportionality Theorem (BPT), also known as Thales Theorem, 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.

Therefore, by BPT, we have:

Now, substitute the given values into the equation:

To find the value of $EC$, we can cross-multiply:

Calculate the product on the right side:

So, the equation becomes:

Now, divide both sides by $3$ to solve for $EC$:

$\text{EC} = 6$

Thus, the length of $EC$ is $6$ cm.

Answer:

The length of $EC$ is $6$ cm.

Given:

In $\triangle PQR$, points $S$ and $T$ are on sides $PQ$ and $PR$ respectively.

$\frac{PS}{SQ} = \frac{PT}{TR}$

$\angle PST = \angle PQR$

To Prove:

$\triangle PQR$ is an isosceles triangle.

Proof:

We are given that points $S$ and $T$ are on sides $PQ$ and $PR$ of $\triangle PQR$ such that the line segment $ST$ divides these sides in the same ratio:

$\frac{PS}{SQ} = \frac{PT}{TR}$

By the Converse of the Basic Proportionality Theorem (BPT), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore, $ST \parallel QR$.

When a transversal line intersects two parallel lines, the corresponding angles are equal.

Considering transversal $PQ$ intersecting parallel lines $ST$ and $QR$, the corresponding angles are $\angle PST$ and $\angle PQR$. Thus, $\angle PST = \angle PQR$.

Considering transversal $PR$ intersecting parallel lines $ST$ and $QR$, the corresponding angles are $\angle PTS$ and $\angle PRQ$. Thus, $\angle PTS = \angle PRQ$.

We are also given that $\angle PST = \angle PQR$.

This given condition matches the first equality from the corresponding angles.

We want to prove that $\triangle PQR$ is an isosceles triangle. A triangle is isosceles if two of its sides are equal or if two of its angles are equal. Let's aim to prove that $\angle PQR = \angle PRQ$.

We have the following relationships:

$\angle PQR = \angle PST$ (Given)

$\angle PRQ = \angle PTS$ (From $ST \parallel QR$)

If we can show that $\angle PST = \angle PTS$, then it will follow that $\angle PQR = \angle PTS$, and since $\angle PTS = \angle PRQ$, we will have $\angle PQR = \angle PRQ$.

Let's look at $\triangle PST$. The sum of angles in $\triangle PST$ is $180^\circ$:

$\angle P + \angle PST + \angle PTS = 180^\circ$

Now consider $\triangle PQR$. The sum of angles in $\triangle PQR$ is $180^\circ$:

$\angle P + \angle PQR + \angle PRQ = 180^\circ$

We are given $\angle PST = \angle PQR$. Let's substitute this into the angle sum equation for $\triangle PST$:

$\angle P + \angle PQR + \angle PTS = 180^\circ$

Compare this equation with the angle sum equation for $\triangle PQR$:

$\angle P + \angle PQR + \angle PTS = \angle P + \angle PQR + \angle PRQ$

Subtract $\angle P + \angle PQR$ from both sides of the equation:

$\angle PTS = \angle PRQ$

This confirms the second corresponding angle equality which we derived from $ST \parallel QR$.

Let's use the given $\angle PST = \angle PQR$ and the derived $\angle PTS = \angle PRQ$ to show $\angle PQR = \angle PRQ$.

Substitute $\angle PQR = \angle PST$ into the equation $\angle PQR = \angle PRQ$ (which we want to prove):

$\angle PST = \angle PRQ$

Now, substitute $\angle PRQ = \angle PTS$ into this equation:

$\angle PST = \angle PTS$

So, the given conditions imply that in $\triangle PST$, the angles $\angle PST$ and $\angle PTS$ are equal.

In $\triangle PST$, since $\angle PST = \angle PTS$, the sides opposite these angles are equal. The side opposite $\angle PTS$ is $PS$, and the side opposite $\angle PST$ is $PT$.

Therefore, $PS = PT$.

Now, substitute $PS = PT$ into the given ratio $\frac{PS}{SQ} = \frac{PT}{TR}$:

$\frac{PS}{SQ} = \frac{PS}{TR}$

Since $S$ is a point on side $PQ$, $PS$ is a length and must be non-zero (otherwise $S$ would coincide with $P$, and $ST$ would not be a line segment intersecting $PQ$ and $PR$). Since $PS \neq 0$, we can cancel $PS$ from both sides:

$\frac{1}{SQ} = \frac{1}{TR}$

This implies $SQ = TR$.

Now consider the lengths of sides $PQ$ and $PR$:

$PQ = PS + SQ$

$PR = PT + TR$

Substitute $PS = PT$ and $SQ = TR$:

$PQ = PS + SQ$

$PR = PS + SQ$

Therefore, $PQ = PR$.

Since two sides of $\triangle PQR$ ($PQ$ and $PR$) are equal in length, $\triangle PQR$ is an isosceles triangle.

Hence, proved.

The Angle-Angle (AA) or Angle-Angle-Angle (AAA) similarity criterion for triangles states:

AA (Angle-Angle) Similarity Criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

That is, if in $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$ and $\angle B = \angle E$, then $\triangle ABC \sim \triangle DEF$.

AAA (Angle-Angle-Angle) Similarity Criterion: If the three angles of one triangle are respectively equal to the three angles of another triangle, then the two triangles are similar.

That is, if in $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$, then $\triangle ABC \sim \triangle DEF$.

Note that the AAA criterion is equivalent to the AA criterion because if two pairs of corresponding angles are equal, the third pair must also be equal (since the sum of angles in a triangle is $180^\circ$).

The Side-Side-Side (SSS) Similarity Criterion states:

If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and hence the two triangles are similar.

That is, if for two triangles $\triangle ABC$ and $\triangle PQR$,

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$

then $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$, and $\triangle ABC \sim \triangle PQR$.

The Side-Angle-Side (SAS) Similarity Criterion states:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

That is, if for two triangles $\triangle ABC$ and $\triangle PQR$,

$\angle A = \angle P$

and $\frac{AB}{PQ} = \frac{AC}{PR}$,

then $\triangle ABC \sim \triangle PQR$.

Given:

$\triangle ABE \sim \triangle ACD$

To Prove:

$\frac{AB}{AC} = \frac{AE}{AD}$

Proof:

We are given that $\triangle ABE$ is similar to $\triangle ACD$.

When two triangles are similar, their corresponding sides are proportional.

For similar triangles $\triangle ABE$ and $\triangle ACD$, the correspondence between vertices is $A \leftrightarrow A$, $B \leftrightarrow C$, and $E \leftrightarrow D$.

Therefore, the ratio of corresponding sides is equal:

$\frac{AB}{AC} = \frac{BE}{CD} = \frac{AE}{AD}$

From this proportionality, we can extract the required equality.

Considering the first and the third ratios, we get:

$\frac{AB}{AC} = \frac{AE}{AD}$

Hence, the required result is proved.

The theorem about the ratio of the areas of two similar triangles states:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Mathematically, if $\triangle ABC \sim \triangle PQR$, then

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$.

Given that the ratio of the corresponding sides of two similar triangles is $2:3$.

Let the two similar triangles be $\triangle ABC$ and $\triangle PQR$.

Let the ratio of their corresponding sides be $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{2}{3}$.

According to the theorem on the ratio of the areas of two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Ratio of areas = $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2$

Substitute the given ratio of sides:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{2}{3}\right)^2$

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{2^2}{3^2} = \frac{4}{9}$

Thus, the ratio of their areas is $4:9$.

The ratio of the areas of the two similar triangles is $4:9$.

Given:

Area of the first similar triangle = $64$ sq cm

Area of the second similar triangle = $121$ sq cm

To Find:

The ratio of their corresponding sides.

Solution:

Let $\text{Area}_1$ and $\text{Area}_2$ be the areas of the two similar triangles, and let $s_1$ and $s_2$ be the lengths of their corresponding sides.

According to the theorem on the ratio of areas of two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

So, we have:

$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{s_1}{s_2}\right)^2$

Substitute the given areas:

$\frac{64}{121} = \left(\frac{s_1}{s_2}\right)^2$

To find the ratio of the sides $\frac{s_1}{s_2}$, we take the square root of both sides of the equation:

$\frac{s_1}{s_2} = \sqrt{\frac{64}{121}}$

$\frac{s_1}{s_2} = \frac{\sqrt{64}}{\sqrt{121}}$

$\frac{s_1}{s_2} = \frac{8}{11}$

Thus, the ratio of their corresponding sides is $8:11$.

The ratio of the corresponding sides of the two similar triangles is $8:11$.

The Pythagoras Theorem states:

In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (legs).

If $a$ and $b$ are the lengths of the legs of a right-angled triangle and $c$ is the length of the hypotenuse, then the theorem can be expressed as:

$a^2 + b^2 = c^2$

Given:

A right-angled triangle.

Length of the hypotenuse, $c = 10$ cm.

Length of one side (leg), $a = 6$ cm.

To Find:

The length of the other side (leg), $b$.

Solution:

In a right-angled triangle, according to the Pythagoras Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

The theorem is given by:

$a^2 + b^2 = c^2$

where $a$ and $b$ are the lengths of the legs, and $c$ is the length of the hypotenuse.

Substitute the given values into the formula:

$(6 \text{ cm})^2 + b^2 = (10 \text{ cm})^2$

Calculate the squares:

$36 \text{ cm}^2 + b^2 = 100 \text{ cm}^2$

To find $b^2$, subtract $36 \text{ cm}^2$ from both sides:

$b^2 = 100 \text{ cm}^2 - 36 \text{ cm}^2$

$b^2 = 64 \text{ cm}^2$

To find $b$, take the square root of both sides:

$b = \sqrt{64 \text{ cm}^2}$

$b = 8 \text{ cm}$

The length of the other side is $8$ cm.

The length of the other side is $8$ cm.

The Converse of the Pythagoras Theorem states:

If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

In other words, if for a triangle with sides $a$, $b$, and $c$, the relation $a^2 + b^2 = c^2$ holds, then the angle opposite the side with length $c$ is a $90^\circ$ angle, and thus the triangle is a right-angled triangle.

Given:

Lengths of the sides of a triangle are $8$ cm, $15$ cm, and $17$ cm.

To Check:

If these sides form a right-angled triangle.

Justification using the converse of the Pythagoras Theorem:

According to the converse of the Pythagoras Theorem, if the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle, and the triangle is a right-angled triangle.

Let the sides be $a = 8$ cm, $b = 15$ cm, and $c = 17$ cm.

The longest side is $17$ cm. Let's check if the square of the longest side is equal to the sum of the squares of the other two sides.

Square of the longest side:

$c^2 = (17 \text{ cm})^2 = 17 \times 17 \text{ cm}^2 = 289 \text{ cm}^2$

Sum of the squares of the other two sides:

$a^2 + b^2 = (8 \text{ cm})^2 + (15 \text{ cm})^2$

$a^2 + b^2 = (8 \times 8) \text{ cm}^2 + (15 \times 15) \text{ cm}^2$

$a^2 + b^2 = 64 \text{ cm}^2 + 225 \text{ cm}^2$

$a^2 + b^2 = 289 \text{ cm}^2$

Compare the results:

We have $c^2 = 289 \text{ cm}^2$ and $a^2 + b^2 = 289 \text{ cm}^2$.

Thus, $c^2 = a^2 + b^2$.

Since the square of the longest side is equal to the sum of the squares of the other two sides, by the converse of the Pythagoras Theorem, the triangle with sides 8 cm, 15 cm, and 17 cm is a right-angled triangle. The right angle is opposite the side of length 17 cm.

Yes, the sides 8 cm, 15 cm, and 17 cm do form a right-angled triangle.

Given:

Speed of the first aeroplane flying North = $1000$ km/hr

Speed of the second aeroplane flying West = $1200$ km/hr

Time elapsed = $1\frac{1}{2}$ hours = $1.5$ hours

To Find:

The distance between the two aeroplanes after $1.5$ hours.

Solution:

Let the airport be denoted by point $O$.

The first aeroplane flies due North from $O$.

The distance covered by the first aeroplane in $1.5$ hours is:

Distance = Speed $\times$ Time

Distance North = $1000 \text{ km/hr} \times 1.5 \text{ hours} = 1500 \text{ km}$.

Let this distance be represented by $OA$, where $A$ is the position of the first aeroplane after $1.5$ hours. So, $OA = 1500$ km.

The second aeroplane flies due West from $O$.

The distance covered by the second aeroplane in $1.5$ hours is:

Distance West = $1200 \text{ km/hr} \times 1.5 \text{ hours} = 1800 \text{ km}$.

Let this distance be represented by $OB$, where $B$ is the position of the second aeroplane after $1.5$ hours. So, $OB = 1800$ km.

Since North and West directions are perpendicular to each other, the paths of the two aeroplanes ($OA$ and $OB$) form the two legs of a right-angled triangle $AOB$ at $O$.

The distance between the two planes after $1.5$ hours is the length of the hypotenuse $AB$ of the right-angled triangle $AOB$.

According to the Pythagoras Theorem in right-angled $\triangle AOB$:

$AB^2 = OA^2 + OB^2$

Substitute the values of $OA$ and $OB$:

$AB^2 = (1500 \text{ km})^2 + (1800 \text{ km})^2$

$AB^2 = (15 \times 100)^2 + (18 \times 100)^2$

$AB^2 = 15^2 \times 100^2 + 18^2 \times 100^2$

$AB^2 = 225 \times 10000 + 324 \times 10000$

$AB^2 = (225 + 324) \times 10000$

$AB^2 = 549 \times 10000$

To find $AB$, take the square root of both sides:

$AB = \sqrt{549 \times 10000}$

$AB = \sqrt{549} \times \sqrt{10000}$

$AB = \sqrt{549} \times 100$

Now, we need to simplify $\sqrt{549}$. Let's find the prime factors of 549.

$\begin{array}{c|cc} 3 & 549 \\ \hline 3 & 183 \\ \hline 61 & 61 \\ \hline & 1 \end{array}$

$549 = 3 \times 3 \times 61 = 3^2 \times 61$.

So, $\sqrt{549} = \sqrt{3^2 \times 61} = 3\sqrt{61}$.

Substitute this back into the expression for $AB$:

$AB = 3\sqrt{61} \times 100$

$AB = 300\sqrt{61}$ km.

The two planes will be $300\sqrt{61}$ km apart after $1\frac{1}{2}$ hours.

The distance between the two planes after $1\frac{1}{2}$ hours will be $300\sqrt{61}$ km.

Given:

Length of the ladder (hypotenuse), $c = 10$ m.

Height the ladder reaches on the wall (one leg), $a = 8$ m.

To Find:

The distance of the foot of the ladder from the base of the wall (the other leg), $b$.

Solution:

The ladder, the wall, and the ground form a right-angled triangle. The ladder is the hypotenuse, the wall is one leg, and the distance from the foot of the ladder to the wall is the other leg.

According to the Pythagoras Theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$a^2 + b^2 = c^2$

Substitute the given values:

$(8 \text{ m})^2 + b^2 = (10 \text{ m})^2$

$64 \text{ m}^2 + b^2 = 100 \text{ m}^2$

Subtract $64 \text{ m}^2$ from both sides to find $b^2$:

$b^2 = 100 \text{ m}^2 - 64 \text{ m}^2$

$b^2 = 36 \text{ m}^2$

Take the square root of both sides to find $b$:

$b = \sqrt{36 \text{ m}^2}$

$b = 6 \text{ m}$

The distance of the foot of the ladder from the base of the wall is $6$ m.

The distance of the foot of the ladder from the base of the wall is $6$ m.

Given:

In $\triangle ABC$, points $D$ is on side $AB$ and $E$ is on side $AC$.

$\angle ADE = \angle ABC$

To Show:

$\triangle ADE \sim \triangle ABC$

Proof:

Consider triangles $\triangle ADE$ and $\triangle ABC$.

We are given that $\angle ADE = \angle ABC$.

Both triangles share the angle $\angle A$.

Now, in $\triangle ADE$ and $\triangle ABC$, we have two pairs of corresponding angles that are equal:

$\angle ADE = \angle ABC$

$\angle DAE = \angle BAC$

According to the Angle-Angle (AA) Similarity Criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Therefore, by AA similarity criterion:

$\triangle ADE \sim \triangle ABC$

Hence, it is shown that $\triangle ADE \sim \triangle ABC$.

This theorem is known as the Converse of the Basic Proportionality Theorem (BPT) or the Converse of Thales's Theorem.

It states:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Explanation:

Consider a triangle $\triangle ABC$.

Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$.

Draw a line segment $DE$ connecting the points $D$ and $E$.

The theorem says that if the line segment $DE$ divides the sides $AB$ and $AC$ such that the ratio of the segments on $AB$ is equal to the ratio of the segments on $AC$, then the line $DE$ must be parallel to the third side $BC$.

In mathematical terms, if in $\triangle ABC$, we have points $D$ on $AB$ and $E$ on $AC$ such that:

$\frac{AD}{DB} = \frac{AE}{EC}$

Then the theorem concludes that:

$DE \parallel BC$

This theorem is the reverse of the Basic Proportionality Theorem, which states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the two sides in the same ratio.

Given:

In $\triangle ABC$, $\angle B = 90^\circ$.

$D$ is any point on side $AB$.

$E$ is any point on side $BC$.

To Prove:

$AE^2 + CD^2 = AC^2 + DE^2$

Proof:

In the given figure, consider the right-angled triangle $\triangle ABE$, which is right-angled at $B$.

By applying the Pythagoras Theorem to $\triangle ABE$, we get:

$AE^2 = AB^2 + BE^2$

Now, consider the right-angled triangle $\triangle CDB$, which is right-angled at $B$.

By applying the Pythagoras Theorem to $\triangle CDB$, we get:

$CD^2 = DB^2 + BC^2$

Add the equations for $AE^2$ and $CD^2$:

$AE^2 + CD^2 = (AB^2 + BE^2) + (DB^2 + BC^2)$

$AE^2 + CD^2 = AB^2 + BE^2 + DB^2 + BC^2$ ... (1)

Now, consider the right-angled triangle $\triangle ABC$, which is right-angled at $B$.

By applying the Pythagoras Theorem to $\triangle ABC$, we get:

$AC^2 = AB^2 + BC^2$

Consider the right-angled triangle $\triangle DBE$, which is right-angled at $B$.

By applying the Pythagoras Theorem to $\triangle DBE$, we get:

$DE^2 = DB^2 + BE^2$

Add the equations for $AC^2$ and $DE^2$:

$AC^2 + DE^2 = (AB^2 + BC^2) + (DB^2 + BE^2)$

$AC^2 + DE^2 = AB^2 + BC^2 + DB^2 + BE^2$ ... (2)

Comparing equation (1) and equation (2), we see that the Right Hand Sides are equal:

$AB^2 + BE^2 + DB^2 + BC^2 = AB^2 + BC^2 + DB^2 + BE^2$

Therefore, the Left Hand Sides must also be equal:

$AE^2 + CD^2 = AC^2 + DE^2$

Hence, the required result is proved.

Given:

Length of the vertical pole = $6$ m

Length of the pole's shadow = $4$ m

Length of the tower's shadow = $28$ m

To Find:

The height of the tower.

Solution:

Let the height of the pole be $h_p$ and the length of its shadow be $s_p$.

Let the height of the tower be $h_t$ and the length of its shadow be $s_t$.

At the same time of day, the sun's rays make the same angle with the ground.

The vertical pole and its shadow form a right-angled triangle with the ground.

The vertical tower and its shadow also form a right-angled triangle with the ground.

Since the angle of elevation of the sun is the same for both at the same time, the two triangles formed are similar by the Angle-Angle (AA) Similarity Criterion.

In similar triangles, the ratio of corresponding sides is equal.

We can set up a proportion between the height and shadow length for the pole and the tower:

$\frac{\text{Height of Pole}}{\text{Length of Pole's Shadow}} = \frac{\text{Height of Tower}}{\text{Length of Tower's Shadow}}$

Substitute the given values into the proportion:

$\frac{h_p}{s_p} = \frac{h_t}{s_t}$

$\frac{6 \text{ m}}{4 \text{ m}} = \frac{h_t}{28 \text{ m}}$

Now, we solve for $h_t$:

$h_t = \frac{6}{4} \times 28 \text{ m}$

$h_t = \frac{3}{2} \times 28 \text{ m}$

$h_t = 3 \times \frac{28}{2} \text{ m}$

$h_t = 3 \times 14 \text{ m}$

$h_t = 42 \text{ m}$

The height of the tower is $42$ m.

The height of the tower is $42$ m.

Given:

In $\triangle ABC$ and $\triangle DEF$, the ratio of corresponding sides is:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{2}{3}$

Area$(\triangle ABC) = 32$ sq cm

To Find:

Area$(\triangle DEF)$

Solution:

Since the ratio of corresponding sides of $\triangle ABC$ and $\triangle DEF$ is equal, the two triangles are similar by the Side-Side-Side (SSS) Similarity Criterion.

Thus, $\triangle ABC \sim \triangle DEF$.

According to the theorem on the ratio of the areas of two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2$

We are given that $\frac{AB}{DE} = \frac{2}{3}$ and Area$(\triangle ABC) = 32$ sq cm.

Substitute these values into the formula:

$\frac{32 \text{ cm}^2}{\text{Area}(\triangle DEF)} = \left(\frac{2}{3}\right)^2$

$\frac{32 \text{ cm}^2}{\text{Area}(\triangle DEF)} = \frac{2^2}{3^2}$

$\frac{32 \text{ cm}^2}{\text{Area}(\triangle DEF)} = \frac{4}{9}$

Now, we need to solve for Area$(\triangle DEF)$. We can cross-multiply:

$4 \times \text{Area}(\triangle DEF) = 32 \text{ cm}^2 \times 9$

$\text{Area}(\triangle DEF) = \frac{32 \text{ cm}^2 \times 9}{4}$

Simplify the expression:

$\text{Area}(\triangle DEF) = \cancel{\frac{32}{4}}^{8} \times 9 \text{ cm}^2$

$\text{Area}(\triangle DEF) = 8 \times 9 \text{ cm}^2$

$\text{Area}(\triangle DEF) = 72 \text{ cm}^2$

The area of triangle $DEF$ is 72 sq cm.

The area of $\triangle DEF$ is $72$ sq cm.

Basic Proportionality Theorem (BPT) or Thales Theorem states:

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.

Given:

A triangle $\triangle ABC$.

A line $DE$ is drawn parallel to side $BC$, intersecting sides $AB$ and $AC$ at distinct points $D$ and $E$ respectively.

So, $DE \parallel BC$.

To Prove:

The line $DE$ divides the sides $AB$ and $AC$ in the same ratio. That is,

$\frac{AD}{DB} = \frac{AE}{EC}$

Construction:

Join $BE$ and $CD$.

Draw $EM \perp AB$ and $DN \perp AC$.

Proof:

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

Consider $\triangle ADE$. Using base $AD$ and height $EM$:

$\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EM$

Consider $\triangle BDE$. Using base $DB$ and height $EM$ (since $EM$ is the perpendicular distance from $E$ to line $AB$):

$\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EM$

Now, find the ratio of the areas of $\triangle ADE$ and $\triangle BDE$:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EM}{\frac{1}{2} \times DB \times EM}$

Cancel out the common terms $\frac{1}{2}$ and $EM$:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{AD}{DB}$ ... (1)

Now, consider $\triangle ADE$ again. Using base $AE$ and height $DN$:

$\text{Area}(\triangle ADE) = \frac{1}{2} \times AE \times DN$

Consider $\triangle CDE$. Using base $EC$ and height $DN$ (since $DN$ is the perpendicular distance from $D$ to line $AC$):

$\text{Area}(\triangle CDE) = \frac{1}{2} \times EC \times DN$

Now, find the ratio of the areas of $\triangle ADE$ and $\triangle CDE$:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DN}{\frac{1}{2} \times EC \times DN}$

Cancel out the common terms $\frac{1}{2}$ and $DN$:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{AE}{EC}$ ... (2)

Triangles $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$.

Triangles on the same base and between the same parallels have equal areas.

Therefore,

$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$ ... (3)

From equations (1), (2), and (3), we can conclude that:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)}$

Substitute the ratios from (1) and (2):

$\frac{AD}{DB} = \frac{AE}{EC}$

Thus, the line $DE$ divides the sides $AB$ and $AC$ in the same ratio.

Hence, the Basic Proportionality Theorem is proved.

The Converse of the Basic Proportionality Theorem (BPT) states:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given:

A triangle $\triangle ABC$.

A line $DE$ intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively, such that

$\frac{AD}{DB} = \frac{AE}{EC}$

To Prove:

$DE \parallel BC$

Proof:

We will prove this theorem by contradiction.

Assume that the line $DE$ is not parallel to $BC$.

Then, there must be some other line through $D$ that is parallel to $BC$.

Let us draw a line $DE'$ through point $D$, parallel to $BC$, where $E'$ is a point on $AC$.

So, $DE' \parallel BC$.

By the Basic Proportionality Theorem (which states that if a line is parallel to one side of a triangle and intersects the other two sides in distinct points, then the other two sides are divided in the same ratio), since $DE' \parallel BC$, we have:

$\frac{AD}{DB} = \frac{AE'}{E'C}$ ... (1)

However, we are given that the line $DE$ divides sides $AB$ and $AC$ in the same ratio:

$\frac{AD}{DB} = \frac{AE}{EC}$ ... (Given)

From equation (1) and the given condition, we can equate the ratios:

$\frac{AE'}{E'C} = \frac{AE}{EC}$

Now, let's add 1 to both sides of the equation:

$\frac{AE'}{E'C} + 1 = \frac{AE}{EC} + 1$

Find a common denominator for both sides:

$\frac{AE' + E'C}{E'C} = \frac{AE + EC}{EC}$

From the figure, $AE' + E'C = AC$ and $AE + EC = AC$.

Substitute these into the equation:

$\frac{AC}{E'C} = \frac{AC}{EC}$

Since $AC \neq 0$ (as it is a side of a triangle), we can cancel $AC$ from both sides:

$\frac{1}{E'C} = \frac{1}{EC}$

This implies:

$E'C = EC$

Since $E'$ and $E$ are both points on the side $AC$, and their distances from $C$ are equal, they must be the same point.

Thus, the point $E'$ must coincide with the point $E$.

Our initial assumption that $DE$ is not parallel to $BC$ led us to draw a line $DE'$ parallel to $BC$, and we found that this line $DE'$ is actually the same line as $DE$.

Therefore, the line $DE$ must be parallel to $BC$.

$DE \parallel BC$

Hence, the converse of the Basic Proportionality Theorem is proved.

Angle-Angle-Angle (AAA) Similarity Criterion states:

If in two triangles, the corresponding angles are equal, then their corresponding sides are in the same ratio (are proportional) and hence the two triangles are similar.

Given:

Two triangles, $\triangle ABC$ and $\triangle DEF$.

Their corresponding angles are equal:

$\angle A = \angle D$

$\angle B = \angle E$

$\angle C = \angle F$

To Prove:

The corresponding sides are in the same ratio:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$

And hence, the triangles are similar:

$\triangle ABC \sim \triangle DEF$

Construction:

Assume $AB < DE$ and $AC < DF$.

Cut off $DP$ on $DE$ equal to $AB$ ($DP = AB$).

Cut off $DQ$ on $DF$ equal to $AC$ ($DQ = AC$).

Join $PQ$.

Proof:

Consider $\triangle ABC$ and $\triangle DPQ$.

Therefore, by the Side-Angle-Side (SAS) congruence criterion, $\triangle ABC \cong \triangle DPQ$.

Since corresponding parts of congruent triangles are equal (CPCTC):

$\angle B = \angle DPQ$

and $\angle C = \angle DQP$.

We are given that $\angle B = \angle E$.

So, $\angle DPQ = \angle E$.

These are corresponding angles formed by the transversal $DE$ intersecting lines $PQ$ and $EF$. Since they are equal, $PQ$ must be parallel to $EF$.

$PQ \parallel EF$

Now, consider $\triangle DEF$ with line $PQ$ parallel to $EF$.

By the Basic Proportionality Theorem (BPT), if a line is parallel to one side of a triangle and intersects the other two sides, it divides the two sides in the same ratio.

So, in $\triangle DEF$, since $PQ \parallel EF$, we have:

$\frac{DP}{PE} = \frac{DQ}{QF}$

Adding 1 to both sides:

$\frac{DP}{PE} + 1 = \frac{DQ}{QF} + 1$

$\frac{DP + PE}{PE} = \frac{DQ + QF}{QF}$

$\frac{DE}{PE} = \frac{DF}{QF}$

Taking the reciprocal of both sides:

$\frac{PE}{DE} = \frac{QF}{DF}$

We know that $PE = DE - DP$ and $QF = DF - DQ$. Substitute these:

$\frac{DE - DP}{DE} = \frac{DF - DQ}{DF}$

$1 - \frac{DP}{DE} = 1 - \frac{DQ}{DF}$

$\frac{DP}{DE} = \frac{DQ}{DF}$

Substitute the constructions $DP = AB$ and $DQ = AC$:

$\frac{AB}{DE} = \frac{AC}{DF}$ ... (4)

By performing a similar construction (e.g., cutting segments on $ED$ and $EF$ corresponding to $AB$ and $BC$) or by symmetry of the theorem, we can similarly prove that:

$\frac{AB}{DE} = \frac{BC}{EF}$ ... (5)

Combining equations (4) and (5), we get the required proportionality of sides:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$

Since the corresponding angles are equal and the corresponding sides are proportional, the two triangles are similar by the definition of similar triangles.

Thus, $\triangle ABC \sim \triangle DEF$.

The case where $AB = DE$ and $AC = DF$ leads to congruence by ASA ($\angle A = \angle D, AB = DE, \angle B = \angle E$), which is a special case of similarity with ratio 1:1. The case where $AB > DE$ or $AC > DF$ can be handled by considering the ratio of sides of $\triangle DEF$ to $\triangle ABC$.

Hence, the AAA similarity criterion is proved.

Side-Side-Side (SSS) Similarity Criterion states:

If the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Given:

Two triangles, $\triangle ABC$ and $\triangle DEF$, such that the ratio of their corresponding sides is equal:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$

To Prove:

Their corresponding angles are equal: $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$.

And hence, the triangles are similar: $\triangle ABC \sim \triangle DEF$.

Construction:

Assume, without loss of generality, that $AB \leq DE$. If $AB=DE$, then by the given proportionality, $AB=DE$, $BC=EF$, $AC=DF$, which means $\triangle ABC \cong \triangle DEF$ by SSS congruence. Congruent triangles are similar with a ratio of 1. So, we consider the case where $AB < DE$.

On side $DE$, cut off a line segment $DP$ such that $DP = AB$.

On side $DF$, cut off a line segment $DQ$ such that $DQ = AC$.

Join $PQ$.

Proof:

We are given that:

$\frac{AB}{DE} = \frac{AC}{DF}$

By construction, $AB = DP$ and $AC = DQ$. Substituting these into the equation:

$\frac{DP}{DE} = \frac{DQ}{DF}$

Now consider $\triangle DEF$. We have points $P$ on $DE$ and $Q$ on $DF$ such that $\frac{DP}{DE} = \frac{DQ}{DF}$.

This can be rewritten as:

$\frac{DE}{DP} = \frac{DF}{DQ}$

Subtract 1 from both sides:

$\frac{DE}{DP} - 1 = \frac{DF}{DQ} - 1$

$\frac{DE - DP}{DP} = \frac{DF - DQ}{DQ}$

$\frac{PE}{DP} = \frac{QF}{DQ}$

Taking reciprocals:

$\frac{DP}{PE} = \frac{DQ}{QF}$

By the Converse of the Basic Proportionality Theorem, since the line segment $PQ$ divides the sides $DE$ and $DF$ in the same ratio, $PQ$ must be parallel to $EF$.

$PQ \parallel EF$

Since $PQ \parallel EF$ and $DE$ is a transversal, the corresponding angles are equal:

$\angle DPQ = \angle E$

Since $PQ \parallel EF$ and $DF$ is a transversal, the corresponding angles are equal:

$\angle DQP = \angle F$

Now, consider $\triangle DPQ$ and $\triangle DEF$. We have $\angle D$ common to both triangles, $\angle DPQ = \angle E$, and $\angle DQP = \angle F$.

Thus, $\triangle DPQ \sim \triangle DEF$ by the Angle-Angle-Angle (AAA) Similarity Criterion.

Since $\triangle DPQ \sim \triangle DEF$, the ratio of their corresponding sides is equal:

$\frac{DP}{DE} = \frac{DQ}{DF} = \frac{PQ}{EF}$

We are given $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$ and by construction $DP = AB$ and $DQ = AC$.

Substitute $DP=AB$ and $DQ=AC$ into the similarity ratio for $\triangle DPQ$ and $\triangle DEF$:

$\frac{AB}{DE} = \frac{AC}{DF} = \frac{PQ}{EF}$

Compare this with the given ratio $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$.

From $\frac{AB}{DE} = \frac{PQ}{EF}$ and $\frac{AB}{DE} = \frac{BC}{EF}$, we get $\frac{PQ}{EF} = \frac{BC}{EF}$.

Since $EF \neq 0$, we can multiply both sides by $EF$ to get $PQ = BC$.

Now, consider $\triangle ABC$ and $\triangle DPQ$.

Therefore, $\triangle ABC \cong \triangle DPQ$ by the Side-Side-Side (SSS) congruence criterion.

Since congruent triangles have equal corresponding angles:

$\angle A = \angle D$

$\angle B = \angle DPQ$

$\angle C = \angle DQP$

We previously showed that $\angle DPQ = \angle E$ and $\angle DQP = \angle F$ (because $PQ \parallel EF$).

Combining these results, we have:

$\angle A = \angle D$

$\angle B = \angle E$

$\angle C = \angle F$

Since the corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal, the two triangles are similar.

$\triangle ABC \sim \triangle DEF$

Hence, the SSS similarity criterion is proved.

Side-Angle-Side (SAS) Similarity Criterion states:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given:

Two triangles, $\triangle ABC$ and $\triangle DEF$.

$\angle A = \angle D$

The sides including these angles are proportional:

$\frac{AB}{DE} = \frac{AC}{DF}$

To Prove:

$\triangle ABC \sim \triangle DEF$

Construction:

Assume, without loss of generality, that $AB \leq DE$. If $AB=DE$, then by the given proportionality, $AB=DE$, $AC=DF$, and $\angle A = \angle D$. By SAS congruence, $\triangle ABC \cong \triangle DEF$, which implies similarity. So, we consider the case where $AB < DE$.

On side $DE$, cut off a line segment $DP$ such that $DP = AB$.

On side $DF$, cut off a line segment $DQ$ such that $DQ = AC$.

Join $PQ$.

Proof:

Consider $\triangle ABC$ and $\triangle DPQ$.

Therefore, by the Side-Angle-Side (SAS) congruence criterion, $\triangle ABC \cong \triangle DPQ$.

Since corresponding parts of congruent triangles are equal (CPCTC), we have:

$\angle B = \angle DPQ$

$\angle C = \angle DQP$

Also, $\frac{AB}{DE} = \frac{AC}{DF}$ (Given).

Substituting $AB=DP$ and $AC=DQ$ (By construction):

$\frac{DP}{DE} = \frac{DQ}{DF}$

Now consider $\triangle DEF$ with line segment $PQ$ dividing sides $DE$ and $DF$. The ratio of the segments from vertex $D$ is equal: $\frac{DP}{DE} = \frac{DQ}{DF}$.

By the Converse of the Basic Proportionality Theorem, if a line divides two sides of a triangle proportionally, then the line is parallel to the third side.

So, $PQ \parallel EF$.

Now consider $\triangle DPQ$ and $\triangle DEF$.

Since $PQ \parallel EF$, $DE$ is a transversal, so corresponding angles are equal:

$\angle DPQ = \angle E$

Since $PQ \parallel EF$, $DF$ is a transversal, so corresponding angles are equal:

$\angle DQP = \angle F$

Thus, in $\triangle DPQ$ and $\triangle DEF$, all corresponding angles are equal: $\angle D = \angle D$, $\angle DPQ = \angle E$, and $\angle DQP = \angle F$.

Therefore, by the Angle-Angle-Angle (AAA) Similarity Criterion, $\triangle DPQ \sim \triangle DEF$.

From the congruence $\triangle ABC \cong \triangle DPQ$, we know that $\triangle ABC$ has the same angles as $\triangle DPQ$.

Since $\triangle DPQ$ is similar to $\triangle DEF$, $\triangle ABC$ also has the same angles as $\triangle DEF$.

$\angle A = \angle D$ (Given)

$\angle B = \angle DPQ$ (From $\triangle ABC \cong \triangle DPQ$) and $\angle DPQ = \angle E$ (From $PQ \parallel EF$). Thus $\angle B = \angle E$.

$\angle C = \angle DQP$ (From $\triangle ABC \cong \triangle DPQ$) and $\angle DQP = \angle F$ (From $PQ \parallel EF$). Thus $\angle C = \angle F$.

Since the corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal, by the definition of similar triangles (or AAA similarity), the triangles are similar.

$\triangle ABC \sim \triangle DEF$

Hence, the SAS similarity criterion is proved.

Theorem on the Areas of Similar Triangles states:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given:

Two triangles $\triangle ABC$ and $\triangle PQR$ such that $\triangle ABC \sim \triangle PQR$.

This means that their corresponding angles are equal ($\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$) and their corresponding sides are proportional ($\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$).

To Prove:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$

Construction:

Draw altitude $AM \perp BC$ in $\triangle ABC$, where $M$ is on $BC$.

Draw altitude $PN \perp QR$ in $\triangle PQR$, where $N$ is on $QR$.

Proof:

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

$\text{Area}(\triangle ABC) = \frac{1}{2} \times BC \times AM$

$\text{Area}(\triangle PQR) = \frac{1}{2} \times QR \times PN$

The ratio of their areas is:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times QR \times PN} = \frac{BC}{QR} \times \frac{AM}{PN}$ ... (1)

Now, consider $\triangle ABM$ and $\triangle PQN$.

Since $\triangle ABC \sim \triangle PQR$, their corresponding angles are equal.

Also, by construction, the altitudes are perpendicular to the bases:

So, $\angle AMB = \angle PNQ$.

In $\triangle ABM$ and $\triangle PQN$, we have two pairs of corresponding angles equal ($\angle B = \angle Q$ and $\angle AMB = \angle PNQ$).

Therefore, by the Angle-Angle (AA) Similarity Criterion, $\triangle ABM \sim \triangle PQN$.

Since $\triangle ABM \sim \triangle PQN$, the ratio of their corresponding sides is equal:

$\frac{AB}{PQ} = \frac{AM}{PN}$ ... (2)

We are given that $\triangle ABC \sim \triangle PQR$, which implies the ratio of corresponding sides is equal:

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$ ... (3)

From equations (2) and (3), we can see that the ratio of altitudes $\frac{AM}{PN}$ is equal to the ratio of the corresponding sides:

$\frac{AM}{PN} = \frac{BC}{QR}$

Now, substitute this relationship back into equation (1):

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{BC}{QR} \times \frac{AM}{PN} = \frac{BC}{QR} \times \frac{BC}{QR}$

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{BC}{QR}\right)^2$

Since the ratio of all corresponding sides is equal (from equation 3), we can also write:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$

Thus, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Hence, the theorem is proved.

Pythagoras Theorem states:

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given:

A right-angled triangle $\triangle ABC$, right-angled at $B$.

Let the sides opposite to vertices $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. So, $AC$ is the hypotenuse with length $b$, $BC$ has length $a$, and $AB$ has length $c$.

$\angle B = 90^\circ$.

To Prove:

$AC^2 = AB^2 + BC^2$

(or $b^2 = c^2 + a^2$)

Construction:

Draw a perpendicular $BD$ from the vertex $B$ to the hypotenuse $AC$.

Let $D$ be a point on $AC$. So, $BD \perp AC$.

Proof:

In $\triangle ADB$ and $\triangle ABC$:

Therefore, by the Angle-Angle (AA) Similarity Criterion, $\triangle ADB \sim \triangle ABC$.

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

$\frac{AD}{AB} = \frac{AB}{AC} = \frac{BD}{BC}$

From the first two ratios, we get:

$\frac{AD}{AB} = \frac{AB}{AC}$

$AD \times AC = AB^2$ ... (1)

Now, consider $\triangle BDC$ and $\triangle ABC$:

Therefore, by the Angle-Angle (AA) Similarity Criterion, $\triangle BDC \sim \triangle ABC$.

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

$\frac{BD}{AB} = \frac{DC}{BC} = \frac{BC}{AC}$

From the last two ratios, we get:

$\frac{DC}{BC} = \frac{BC}{AC}$

$DC \times AC = BC^2$ ... (2)

Add equation (1) and equation (2):

$AD \times AC + DC \times AC = AB^2 + BC^2$

Factor out $AC$ from the left side:

$AC \times (AD + DC) = AB^2 + BC^2$

From the figure, the point $D$ lies on the line segment $AC$, so $AD + DC = AC$.

Substitute this into the equation:

$AC \times (AC) = AB^2 + BC^2$

$AC^2 = AB^2 + BC^2$

Using the side lengths $a$, $b$, and $c$:

$b^2 = c^2 + a^2$

Thus, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Hence, the Pythagoras Theorem is proved.

The Converse of the Pythagoras Theorem states:

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Given:

A triangle $\triangle ABC$ in which the square of one side, say $AC$, is equal to the sum of the squares of the other two sides, $AB$ and $BC$.

So, $AC^2 = AB^2 + BC^2$.

To Prove:

The angle opposite to side $AC$ is a right angle, i.e., $\angle B = 90^\circ$.

Construction:

Construct a triangle $\triangle PQR$, which is right-angled at $Q$, such that $PQ = AB$ and $QR = BC$.

Proof:

In the right-angled triangle $\triangle PQR$, by the Pythagoras Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$PR^2 = PQ^2 + QR^2$

By construction, we have $PQ = AB$ and $QR = BC$. Substitute these values into the equation:

$PR^2 = AB^2 + BC^2$ ... (1)

We are given that in $\triangle ABC$:

$AC^2 = AB^2 + BC^2$ ... (Given)

Comparing equation (1) and the given condition for $\triangle ABC$, we see that their Right Hand Sides are equal:

$AB^2 + BC^2 = AC^2$ and $AB^2 + BC^2 = PR^2$.

Therefore, their Left Hand Sides must be equal:

$AC^2 = PR^2$

Taking the square root of both sides (since lengths are positive):

$AC = PR$

Now, consider $\triangle ABC$ and $\triangle PQR$.

Since all three corresponding sides are equal, $\triangle ABC$ is congruent to $\triangle PQR$ by the Side-Side-Side (SSS) congruence criterion.

$\triangle ABC \cong \triangle PQR$

Since corresponding angles of congruent triangles are equal (CPCTC):

$\angle B = \angle Q$

By construction, $\triangle PQR$ is right-angled at $Q$, so $\angle Q = 90^\circ$.

Therefore, $\angle B = 90^\circ$.

Thus, if the square of one side of a triangle is equal to the sum of the squares of the other two sides, the angle opposite the first side is a right angle.

Hence, the converse of the Pythagoras Theorem is proved.

Given:

$\triangle ABC$ is an equilateral triangle.

$D$ is a point on side $BC$ such that $BD = \frac{1}{3} BC$.

To Prove:

$9 AD^2 = 7 AB^2$

Construction:

Draw $AE \perp BC$ from vertex $A$ to side $BC$.

Proof:

In an equilateral triangle, the altitude from a vertex to the opposite side bisects the base and the angle at the vertex.

Since $\triangle ABC$ is equilateral and $AE \perp BC$, $E$ is the midpoint of $BC$.

Let $AB = BC = AC = s$.

Then $BE = EC = \frac{1}{2} BC = \frac{s}{2}$.

We are given that $BD = \frac{1}{3} BC$.

Substitute $BC = s$: $BD = \frac{1}{3} s$.

Now, consider the segment $DE$. Since $E$ is the midpoint of $BC$, $BE = s/2$. Since $D$ is on $BC$ and $BD = s/3$, $D$ must lie between $B$ and $E$ because $s/3 < s/2$.

The length of $DE$ is the difference between $BE$ and $BD$:

$DE = BE - BD = \frac{s}{2} - \frac{s}{3}$

Find a common denominator:

$DE = \frac{3s}{6} - \frac{2s}{6} = \frac{3s - 2s}{6} = \frac{s}{6}$

Now, consider the right-angled triangle $\triangle AEB$ (since $AE \perp BC$).

By the Pythagoras Theorem in $\triangle AEB$:

$AB^2 = AE^2 + BE^2$

Substitute $AB=s$ and $BE=s/2$:

$s^2 = AE^2 + \left(\frac{s}{2}\right)^2$

$s^2 = AE^2 + \frac{s^2}{4}$

Solve for $AE^2$:

$AE^2 = s^2 - \frac{s^2}{4} = \frac{4s^2 - s^2}{4} = \frac{3s^2}{4}$

Now, consider the right-angled triangle $\triangle AED$ (since $AE \perp BC$, $\angle AED = 90^\circ$).

By the Pythagoras Theorem in $\triangle AED$:

$AD^2 = AE^2 + DE^2$

Substitute the values we found for $AE^2$ and $DE$:

$AD^2 = \frac{3s^2}{4} + \left(\frac{s}{6}\right)^2$

$AD^2 = \frac{3s^2}{4} + \frac{s^2}{36}$

Find a common denominator (36):

$AD^2 = \frac{3s^2 \times 9}{4 \times 9} + \frac{s^2}{36}$

$AD^2 = \frac{27s^2}{36} + \frac{s^2}{36}$

$AD^2 = \frac{27s^2 + s^2}{36} = \frac{28s^2}{36}$

Simplify the fraction $\frac{28}{36}$:

$\frac{28}{36} = \frac{\cancel{4} \times 7}{\cancel{4} \times 9} = \frac{7}{9}$

So, $AD^2 = \frac{7s^2}{9}$.

Multiply both sides by 9:

$9 AD^2 = 7 s^2$

Since $AB = s$, we have $AB^2 = s^2$. Substitute this into the equation:

$9 AD^2 = 7 AB^2$

This is the required result.

Hence, proved.

Given:

In $\triangle ABC$, $\angle C = 90^\circ$.

$P$ is a point on side $CA$.

$Q$ is a point on side $CB$.

To Prove:

$AQ^2 + BP^2 = AB^2 + PQ^2$

Proof:

Consider the right-angled triangle $\triangle ACQ$, which is right-angled at $C$.

By the Pythagoras Theorem in $\triangle ACQ$:

$AQ^2 = AC^2 + CQ^2$ ... (1)

Consider the right-angled triangle $\triangle BCP$, which is right-angled at $C$.

By the Pythagoras Theorem in $\triangle BCP$:

$BP^2 = BC^2 + CP^2$ ... (2)

Add equation (1) and equation (2):

$AQ^2 + BP^2 = (AC^2 + CQ^2) + (BC^2 + CP^2)$

$AQ^2 + BP^2 = AC^2 + CQ^2 + BC^2 + CP^2$ ... (3)

Now, consider the right-angled triangle $\triangle ABC$, which is right-angled at $C$.

By the Pythagoras Theorem in $\triangle ABC$:

$AB^2 = AC^2 + BC^2$ ... (4)

Consider the right-angled triangle $\triangle PCQ$, which is right-angled at $C$.

By the Pythagoras Theorem in $\triangle PCQ$:

$PQ^2 = CP^2 + CQ^2$ ... (5)

Add equation (4) and equation (5):

$AB^2 + PQ^2 = (AC^2 + BC^2) + (CP^2 + CQ^2)$

$AB^2 + PQ^2 = AC^2 + BC^2 + CP^2 + CQ^2$ ... (6)

Comparing equation (3) and equation (6), we see that the Right Hand Sides are identical:

$AC^2 + CQ^2 + BC^2 + CP^2 = AC^2 + BC^2 + CP^2 + CQ^2$

Therefore, the Left Hand Sides must also be equal:

$AQ^2 + BP^2 = AB^2 + PQ^2$

Hence, the required result is proved.

Theorem Statement:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given:

Two triangles $\triangle ABC$ and $\triangle PQR$ such that $\triangle ABC \sim \triangle PQR$.

$AD$ is a median of $\triangle ABC$, so $D$ is the midpoint of $BC$.

$PS$ is a median of $\triangle PQR$, so $S$ is the midpoint of $QR$.

To Prove:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AD}{PS}\right)^2$

Proof:

Since $\triangle ABC \sim \triangle PQR$, the ratio of their corresponding sides is equal:

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ ... (1)

Also, the corresponding angles are equal:

$\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$

Since $AD$ is the median to $BC$, $D$ is the midpoint of $BC$. Thus, $BC = 2BD$.

Since $PS$ is the median to $QR$, $S$ is the midpoint of $QR$. Thus, $QR = 2QS$.

Substitute $BC = 2BD$ and $QR = 2QS$ into the ratio of sides (1):

$\frac{AB}{PQ} = \frac{2BD}{2QS} = \frac{AC}{PR}$

$\frac{AB}{PQ} = \frac{BD}{QS} = \frac{AC}{PR}$ ... (2)

Now, consider $\triangle ABD$ and $\triangle PQS$.

From (2), we have $\frac{AB}{PQ} = \frac{BD}{QS}$.

Also, we know that $\angle B = \angle Q$ (from the similarity of $\triangle ABC$ and $\triangle PQR$).

So, in $\triangle ABD$ and $\triangle PQS$, we have:

$\frac{AB}{PQ} = \frac{BD}{QS}$

and the included angle $\angle B = \angle Q$.

Therefore, by the Side-Angle-Side (SAS) Similarity Criterion, $\triangle ABD \sim \triangle PQS$.

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

$\frac{AB}{PQ} = \frac{BD}{QS} = \frac{AD}{PS}$ ... (3)

Now, we use the theorem on the areas of similar triangles, which states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

For $\triangle ABC \sim \triangle PQR$:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2$

From equation (3), we have $\frac{AB}{PQ} = \frac{AD}{PS}$ (the ratio of corresponding medians).

Substitute this into the area ratio equation:

$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AD}{PS}\right)^2$

Thus, the ratio of the areas of the two similar triangles is equal to the square of the ratio of their corresponding medians.

Hence, proved.

Theorem Statement:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on either side of the perpendicular are similar to the whole triangle and also to each other.

Given:

A right-angled triangle $\triangle ABC$, right-angled at $B$ ($\angle ABC = 90^\circ$).

$BD$ is drawn perpendicular to the hypotenuse $AC$, where $D$ is a point on $AC$.

So, $\angle BDA = 90^\circ$ and $\angle BDC = 90^\circ$.

To Prove:

1. $\triangle ADB \sim \triangle ABC$

2. $\triangle BDC \sim \triangle ABC$

3. $\triangle ADB \sim \triangle BDC$

Proof:

Part 1: Proving $\triangle ADB \sim \triangle ABC$

Consider $\triangle ADB$ and $\triangle ABC$.

Since two corresponding angles are equal, by the Angle-Angle (AA) Similarity Criterion:

$\triangle ADB \sim \triangle ABC$

This proves the first part.

Part 2: Proving $\triangle BDC \sim \triangle ABC$

Consider $\triangle BDC$ and $\triangle ABC$.

Since two corresponding angles are equal, by the Angle-Angle (AA) Similarity Criterion:

$\triangle BDC \sim \triangle ABC$

This proves the second part.

Part 3: Proving $\triangle ADB \sim \triangle BDC$

From Part 1, we have $\triangle ADB \sim \triangle ABC$.

From Part 2, we have $\triangle BDC \sim \triangle ABC$.

Since both $\triangle ADB$ and $\triangle BDC$ are similar to the same triangle $\triangle ABC$, they are also similar to each other by the transitive property of similarity.

Therefore, $\triangle ADB \sim \triangle BDC$.

Alternatively, we can prove this directly using angles:

In $\triangle ABC$, $\angle A + \angle ABC + \angle C = 180^\circ$.

Since $\angle ABC = 90^\circ$, $\angle A + 90^\circ + \angle C = 180^\circ$, which gives $\angle A + \angle C = 90^\circ$.

In right-angled $\triangle ADB$, $\angle A + \angle ABD + \angle ADB = 180^\circ$.

Since $\angle ADB = 90^\circ$, $\angle A + \angle ABD + 90^\circ = 180^\circ$, which gives $\angle A + \angle ABD = 90^\circ$.

Comparing $\angle A + \angle C = 90^\circ$ and $\angle A + \angle ABD = 90^\circ$, we get $\angle C = \angle ABD$.

Now, consider $\triangle ADB$ and $\triangle BDC$.

Since two corresponding angles are equal, by the Angle-Angle (AA) Similarity Criterion:

$\triangle ADB \sim \triangle BDC$

This proves the third part.

Combining all three parts, the triangles on either side of the perpendicular are similar to the whole triangle and also to each other.

Hence, proved.