# CRE 2 - Similarity of Triangles | Geometry - Triangles

**CRE 2 - Similarity of Triangles | Geometry - Triangles**

A triangle has sides a, b and c and the measure of the angles opposite to these sides are X, Y and Z respectively. Find the measures of the angles for the triangle with sides 2a, 2b and 2c.

- (a)
X/2, Y/2 and Z/2

- (b)
2X, 2Y and 2Z

- (c)
X, Y and Z

- (d)
4X, 4Y and 4Z

Answer: Option C

**Explanation** :

The sides of the two triangles are a, b, c and 2a, 2b and 2c respectively.

∴ The corresponding sides for the two triangles are in the ratio 1 : 2.

∴ The two triangles are similar.

∴ The corresponding angles of both the triangles are equal.

∴ The measures of the angles in the second triangle are X, Y and Z respectively.

Hence, option (c).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

A line DE is drawn by joining the midpoints of the sides AB and AC of ∆ABC with area 12 sq. units. What is the area of ∆ADE (in sq. units)?

- (a)
6

- (b)
5

- (c)
4

- (d)
3

Answer: Option D

**Explanation** :

∵ DE joins the midpoints of AB and AC

∴ According to the basic proportionality theorem, side DE of ∆ADE is parallel to side BC of ∆ABC and

BC = 2 × DE

In ∆ADE and ∆ABC,

∠ADE = ∠ABC (Corresponding angles) and ∠A is common to both the triangles.

∴ ∆ADE ∼ ∆ABC ...(By A-A test of similarity)

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

A(ΔADE) : A(ΔABC) = (DE)^{2} : (BC)^{2} = 1/4

∴ A(ΔADE)/12 = 1/4

∴ A(ΔADE) = 12/4 = 3 sq. units

Hence, option (d).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In ∆ABC, AB = 15, AC = 10 and BC = 6. D and E are points on AB and AC respectively such that DE || BC. If DB = 10, DE =?

Answer: 2

**Explanation** :

In ∆ABC and ∆ ADE

∠ABC = ∠ADE (Corresponding angles) and ∠A is common to both the triangles.

∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)

⇒ AD/AB=DE/BC

⇒ (15 - 10)/15 = DE/6

⇒ DE = 2

Hence, 2.

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In the figure below, AB ∥ CD. Find the value of x.

- (a)
2

- (b)
3

- (c)
1/2

- (d)
None of these

Answer: Option B

**Explanation** :

Given, AB || CD.

Consider ∆AOB and ∆COD

∠AOB = ∠COD (Vertically opposite angles) and

∠ABD = ∠CDB (alternate interior angles)

∴ ∆AOB ~ ∆COD

⇒ AO/CO = BO/DO

⇒ 4/(4x - 4) = (2x - 1)/(2x + 4)

⇒ 1/(x - 1) = (2x - 1)/(2x + 4)

⇒ 2x + 4 = 2x^{2} - 3x + 1

⇒ 2x^{2} – 5x – 3 = 0

⇒ (2x + 1)(x - 3) = 0

⇒ x = -1/2 or 3

Since x has to be positive, x = 3

Hence, option (b).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

ΔABC and ΔDEF are similar and their areas be respectively 81 cm^{2} and 144 cm^{2}. If EF = 9 cm, BC is?

- (a)
10.5

- (b)
10.8

- (c)
12

- (d)
6.75

- (e)
None of these

Answer: Option D

**Explanation** :

Given, ΔABC ~ ΔDEF

∴ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective sides.

In the two similar triangles side BC in ∆ABC is corresponding to side EF in ∆DEF.

⇒ $\frac{Area\left(\u2206ABC\right)}{Area\left(\u2206DEF\right)}={\left(\frac{BC}{EF}\right)}^{2}$

⇒ $\frac{81}{144}={\left(\frac{BC}{9}\right)}^{2}$

⇒ $\frac{9}{12}=\frac{BC}{9}$

⇒ BC = 6.75

Hence, option (d).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

Two isosceles triangles have equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding heights.

- (a)
3 : 4

- (b)
4 : 3

- (c)
9 : 16

- (d)
Can't be determined

Answer: Option A

**Explanation** :

Let the triangles be ∆ABC and ∆DEF.

Given vertical angles are equal, ∠A = ∠D = x (say)

In Isosceles ∆ABC, ∠B = ∠C = (180 - x)/2

Similarly, in Isosceles ∆DEF, ∠E = ∠F = (180 - x)/2

In ΔABC ~ ΔDEF

∠A = ∠D (vertically opposite angles)

∠B = ∠E = (180 - x)/2

∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)

⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective heights.

⇒ ${\left(\frac{{h}_{1}}{{h}_{2}}\right)}^{2}=\frac{Area\left(\u2206ABC\right)}{Area\left(\u2206DEF\right)}$

⇒ ${\left(\frac{{h}_{1}}{{h}_{2}}\right)}^{2}=\frac{9}{16}$

⇒ h_{1} : h_{2} = 3 : 4

Hence, option (a).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In ∆ABC, DE is a line segment intersecting AB at D and AC at E such that DE ∥ BC. DE divides ∆ ABC in two parts equal in area. Find BD/AB?

- (a)
1 : 1

- (b)
1 : 2

- (c)
(√2 - 1)/√2

- (d)
None of these

Answer: Option C

**Explanation** :

Given, Area(∆ADE) = Area(∎BDEC)

∴ Area(∆ABC) = 2 × Area(∆ADE)

Now, let’s consider ∆ABC and ∆ ADE

∠ABC = ∠ADE (Corresponding angles) and ∠A is common to both the triangles.

∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)

⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective heights.

⇒ ${\left(\frac{AB}{AD}\right)}^{2}=\frac{Area\left(\u2206ABC\right)}{Area\left(\u2206DEF\right)}=\frac{2}{1}$

⇒ AB = √2 × AD.

⇒ BD = AB – AD = (√2 – 1) × AD.

∴ $\frac{BD}{AB}=\frac{\left(\sqrt{2}\u2013\mathrm{}1\right)\times \mathrm{}\mathrm{AD}}{\sqrt{2}\times \mathrm{}\mathrm{AD}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

Hence, option (c).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In the figure below, PS = 32 cm, PQ = 18 cm, SR = 25 cm and ∠QRP = ∠QSR. Find the ratio of the perimeters of the triangle PQR and PRS.

- (a)
5 : 4

- (b)
4 : 3

- (c)
2 : 3

- (d)
13 : 11

Answer: Option B

**Explanation** :

In ∆RPS and ∆QPR

∠RPS = ∠QPR (∵ P is common angle)

∠PSR = ∠PRQ (Given)

∴ ∆RPS ~∆QPR

⇒ $\frac{PS}{PR}=\frac{PR}{PQ}=\frac{SR}{RQ}$

⇒ $\frac{32}{PR}=\frac{PR}{18}=\frac{25}{RQ}$

Solving first two ratios we get, PR = 24.

Solving last two ratios we get RQ = 75/4 = 18.75

∴ Perimeter of ∆RPS = 24 + 32 + 25 = 81

∴ Perimeter of ∆QPR = 18 + 24 + 18.75 = 60.75

⇒ Ratio of perimeter of ∆RPS and ∆QPR = 81 : 60.75 = 4 : 3.

Hence, option (b).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

P is the mid-point of side BC of ∆ABC. If Q is mid-point of AP and BQ when produced meets AC at L. Is LA = 1/3(CA)?

Type 1: if your answer is yes.

Type 2: if your answer is no.

Type 3: if answer cannot be determined.

Answer: 1

**Explanation** :

Draw a line from P parallel to BL which meets AC at M.

Now, let’s consider ∆APM and ∆AQL

∠QAL is common in both triangles, and

∠AQL = ∠APM (Corresponding angles ∵ QL || PM)

∴ ∆APM ∼ ∆AQL ...(By A-A test of similarity)

⇒ AQ/AP = AL/AM

⇒ 1/2 = AL/AM

⇒ AL = LM …(1)

Similarly, ∆CPM ~ ∆CBL, and

⇒ CM = LM …(2)

From (1) and (2), we get

CM = ML = LA = 1/3 × AC

Hence, 1.

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In the figure below, PB and QA are perpendiculars to segment AB. If PO = 5 cm, QO = 7 cm and area ∆POB = 150 sq.cm., find area of ∆QOA.

Answer: 294

**Explanation** :

In ∆PBO and ∆QAO

∠B = ∠A = 90°

∠POB = ∠QOA (vertically opposite angles)

∴ ∆PBO ∼ ∆QAO ...(By A-A test of similarity)

⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective sides.

⇒ ${\left(\frac{PO}{QO}\right)}^{2}=\frac{Area\left(\u2206PBO\right)}{Area\left(\u2206QAO\right)}$

⇒ ${\left(\frac{5}{7}\right)}^{2}=\frac{150}{Area\left(\u2206QAO\right)}$

⇒ Area(∆QAO) = 294 sq.cm.

Hence, 294.

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

Area of triangle ABC is 144 cm^{2}. Another triangle is formed by joining the mid-points of sides of ABC. Yet another triangle is formed by joining the mid-points of the new triangle to obtain a third triangle. Find the area of the smallest triangle.

- (a)
72

- (b)
36

- (c)
12

- (d)
9

Answer: Option D

**Explanation** :

When mid-points of a triangle is formed, we get 4 smaller triangles and area of each of these smaller triangles is one-fourth the area of the larger triangle.

Thus, if area of triangle ABC is 144, the area of triangle formed by joining the midpoints will be 144/4 = 36.

The area of the third triangle will be 36/4 = 9cm^{2}.

Hence, option (d).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In a right triangle ABC right angled at B, perpendicular BD is drawn from B to AC. AB = 10 and AD = 4, find CD?

- (a)
14

- (b)
21

- (c)
25

- (d)
Cannot be determined

Answer: Option B

**Explanation** :

In a right triangle, perpendicular drawn from right vertex to hypotenuse divides the triangle in three similar triangles.

⇒ AB^{2} = AD × AC

⇒ 10^{2} = 4 × AC

⇒ AC = 25

∴ CD = AC - AD = 25 - 4 = 21

Hence, option (b).

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**CRE 2 - Similarity of Triangles | Geometry - Triangles**

In the figure given below, DE || BC = 1/3 of BC. If area of triangle ADE = 20 cm^{2}, then what is the area (in cm^{2}) of triangle DEC?

- (a)
40

- (b)
60

- (c)
80

- (d)
120

Answer: Option A

**Explanation** :

Let us draw AF which is perpendicular to both DE and BC

∆ADE ≈ ∆ABC (since DE || BC)

⇒ AF : AG = DE : BC = 1 : 3

⇒ AF : FG = 1 : 2

For ∆ADE and ∆DEC

Base is same for both i.e., DE while height of ∆DEC which is FG is twice that of height of ∆ADE which is AF

∴ Area of ∆DEC will be twice that of ∆ADE

⇒ Area of ∆DEC = 2 × 20 = 40 cm^{2}.

Hence, option (a).

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