Similarity — Multiple Choice Questions

In the adjoining figure, △ABC ~ △QPR.
Then ∠R is

1. 60°

2. 50°

3. 70°

4. 80°

Answer

Given, △ABC ~ △QPR

∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R

∠C = 180° - (70° + 50°) = 180° - 120° = 60°.

∴ ∠R = 60°.

Hence, Option 1 is the correct option.

In the adjoining figure, △ABC ~ △QPR.

The value of x is

1. 2.25 cm

2. 4 cm

3. 4.5 cm

4. 5.25 cm

Answer

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

$\therefore \dfrac{AC}{QR} = \dfrac{BC}{PR} \\[1em] \Rightarrow \dfrac{6}{3} = \dfrac{4.5}{x} \\[1em] \Rightarrow 2 = \dfrac{4.5}{x} \\[1em] \Rightarrow x = \dfrac{4.5}{2} \\[1em] \Rightarrow x = 2.25$

Hence, Option 1 is the correct option.

In the adjoining figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to

1. 50°

2. 30°

3. 60°

4. 100°

Answer

Considering △APB and △CPD,

$\dfrac{AP}{PD} = \dfrac{6}{5} \text{ and } \dfrac{BP}{CP} = \dfrac{3}{2.5} = \dfrac{6}{5}.$ and ∠APB = ∠CPD (Vertically opposite angles are equal)

∴ △APB ~ △CPD

Hence, ∠PAB = ∠PDC = 30°

∠PBA = 180° - (∠PAB + ∠APB) = 180° - (30° + 50°) = 180° - 80° = 100°.

Hence, Option 4 is the correct option.

In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are

1. congruent but not similar

2. similar but not congruent

3. neither congruent nor similar

4. congruent as well as similar

Answer

Given,

∠B = ∠E, ∠F = ∠C and AB = 3DE.

∴ Two angles of the one triangles are equal to corresponding two angles of the other, but sides are not equal.

∴ Triangles are similar but not congruent.

Hence, Option 2 is the correct option.

The adjoining figure, AB || DE. The length of CD is

1. 2.5 cm

2. 2.7 cm

3. $\dfrac{10}{3}$ cm

4. 3.5 cm

Answer

Given AB || DE.

Considering △ABC and △DEC,

∠ ACB = ∠ DCE (Vertically opposite angles are equal)
∠ ABC = ∠ CDE (Alternate angles are equal)

Hence, by AA axiom △ABC ~ △DEC.

Since, both the triangles are similar, hence ratio of their corresponding sides are equal,

$\therefore \dfrac{AB}{DE} = \dfrac{BC}{CD} \\[1em] \Rightarrow \dfrac{5}{3} = \dfrac{4.5}{CD} \\[1em] \Rightarrow CD = \dfrac{3 \times 4.5}{5} \\[1em] \Rightarrow CD = 2.7$

Hence, Option 2 is the correct option.

If △PQR ~ △ABC, PQ = 6 cm, AB = 8 cm and perimeter of △ABC is 36 cm, then perimeter of △PQR is

1. 20.25 cm

2. 27 cm

3. 48 cm

4. 64 cm

Answer

Let perimeter of △PQR be x cm.

Since triangles are similar,

$\therefore \dfrac{\text{PQ}}{\text{AB}} = \dfrac{\text{Perimeter of △PQR}}{\text{Perimeter of △ABC}} \\[1em] \Rightarrow \dfrac{6}{8} = \dfrac{x}{36} \\[1em] \Rightarrow x = \dfrac{6 \times 36}{8} \\[1em] \Rightarrow x = 27.$

∴ Perimeter of △PQR = 27 cm.

Hence, Option 2 is the correct option.

In the adjoining figure, DE || BC and all measurements are in centimetres. The length of AE is

1. 2 cm

2. 2.25 cm

3. 3.5 cm

4. 4 cm

Answer

Given DE || BC.

Considering △ABC and △ADE,

∠ A = ∠ A (Common angles)
∠ ABC = ∠ ADE (Corresponding angles are equal)

Hence, by AA axiom △ABC ~ △ADE.

Let the length of AE be x cm.

Since, both the triangles are similar,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \\[1em] \Rightarrow \dfrac{3}{3 + 4} = \dfrac{x}{x + 3} \\[1em] \Rightarrow \dfrac{3}{7} = \dfrac{x}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 7x \\[1em] \Rightarrow 3x + 9 = 7x \\[1em] \Rightarrow 4x = 9 \\[1em] \Rightarrow x = 2.25$

∴ Length of AE = 2.25 cm.

Hence, Option 2 is the correct option.

In the adjoining figure, PQ || CA and all lengths are given in centimetres. The length of BC is

1. 6.4 cm

2. 7.4 cm

3. 8 cm

4. 9 cm

Answer

Given PQ || CA.

Considering △ABC and △PBQ,

∠ B = ∠ B (Common angles)
∠ CAB = ∠ QPB (Corresponding angles are equal)

Hence, by AA axiom △ABC ~ △PBQ.

Let length of QC be x cm.

Since triangles are similar,

$\therefore \dfrac{BQ}{BC} = \dfrac{BP}{BA} \\[1em] \Rightarrow \dfrac{BQ}{BQ + QC} = \dfrac{BP}{BP + PA} \\[1em] \Rightarrow \dfrac{5}{5 + x} = \dfrac{4}{4 + 2.4} \\[1em] \Rightarrow \dfrac{5 \times 6.4}{4} = x + 5 \\[1em] \Rightarrow 8 = x + 5 \\[1em] \Rightarrow x = 3.$

BC = BQ + QC = 5 + x = 5 + 3 = 8 cm.

Hence, Option 3 is the correct option.

In the adjoining figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is

1. 4 cm

2. 3.6 cm

3. 2 cm

4. 3 cm

Answer

Given MN || QR.

Considering △PMN and △PQR,

∠P = ∠P (Common angles)
∠PMN = ∠PQR (Corresponding angles are equal)

Hence, by AA axiom △PMN ~ △PQR.

Let length of PM be x cm.

Since triangles are similar by basic proportionality theorem,

$\therefore \dfrac{PM}{MQ} = \dfrac{PN}{NR} \\[1em] \Rightarrow \dfrac{PM}{PQ - PM} = \dfrac{PN}{NR} \\[1em] \Rightarrow \dfrac{x}{5 - x} = \dfrac{3.6}{2.4} \\[1em] \Rightarrow \dfrac{x}{5 - x} = \dfrac{36}{24} \\[1em] \Rightarrow 24x = 36(5 - x) \\[1em] \Rightarrow 24x = 180 - 36x \\[1em] \Rightarrow 24x + 36x = 180 \\[1em] \Rightarrow 60x = 180 \\[1em] \Rightarrow x = 3.$

∴ Length of PM = 3 cm.

Hence, Option 4 is the correct option.

It is given that △ABC ~ △PQR with $\dfrac{BC}{QR} = \dfrac{1}{3}$, then $\dfrac{\text{area of △PQR}}{\text{area of △ABC}}$ is equal to

1. 9

2. 3

3. $\dfrac{1}{3}$

4. $\dfrac{1}{9}$

Answer

Given $\dfrac{BC}{QR} = \dfrac{1}{3}$

So, $\dfrac{QR}{BC} = \dfrac{3}{1}$.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △PQR}}{\text{Area of △ABC}} = \dfrac{QR^2}{BC^2} \\[1em] = \dfrac{3^2}{1^2} \\[1em] = \dfrac{9}{1} \\[1em] = 9.$

Hence, Option 1 is the correct option.

If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio

1. 9 : 4

2. 3 : 2

3. 2 : 3

4. 16 : 81

Answer

Given, ratio of the areas of the two similar triangles = 4 : 9

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\text{Ratio of corresponding sides } = \sqrt{\text{Ratio of the areas of similar triangles}} = \sqrt{4} : \sqrt{9}$ = 2 : 3.

Hence, Option 3 is the correct option.

If △ABC ~ △PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of △ABC and △PQR is

1. 8 : 6

2. 3 : 4

3. 9 : 16

4. 16 : 9

Answer

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{BC^2}{QR^2} \\[1em] = \dfrac{8^2}{6^2} \\[1em] = \dfrac{64}{36} \\[1em] = \dfrac{16}{9} \\[1em] = 16 : 9.$

Hence, Option 4 is the correct option.

If △ABC ~ △QRP, $\dfrac{\text{area of △ABC}}{\text{area of △PQR}} = \dfrac{9}{4}$, AB = 18 cm and BC = 15 cm, then the length of PR is equal to

1. 10 cm

2. 12 cm

3. $\dfrac{20}{3}$ cm

4. 8 cm

Answer

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{BC^2}{PR^2}$

Given, $\dfrac{\text{area of △ABC}}{\text{area of △PQR}} = \dfrac{9}{4}$.

So,

$\Rightarrow \dfrac{BC^2}{PR^2} = \dfrac{9}{4} \\[1em] \Rightarrow \dfrac{15^2}{PR^2} = \dfrac{9}{4} \\[1em] \Rightarrow PR^2 = \dfrac{225 \times 4}{9} \\[1em] \Rightarrow PR^2 = 100 \\[1em] \Rightarrow PR = \sqrt{100} \\[1em] \Rightarrow PR = 10.$

∴ PR = 10 cm.

Hence, Option 1 is the correct option.

If △ABC ~ △PQR, area of △ABC = 81 cm², area of △PQR = 144 cm² and QR = 6 cm, then length of BC is

1. 4 cm

2. 4.5 cm

3. 9 cm

4. 12 cm

Answer

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{BC^2}{QR^2} \\[1em] \Rightarrow \dfrac{BC^2}{6^2} = \dfrac{81}{144} \\[1em] \Rightarrow BC^2 = \dfrac{81 \times 36}{144} \\[1em] \Rightarrow BC^2 = \dfrac{2916}{144} \\[1em] \Rightarrow BC^2 = 20.25 \\[1em] \Rightarrow BC = \sqrt{20.25} \\[1em] \Rightarrow BC = 4.5$

∴ BC = 4.5 cm.

Hence, Option 2 is the correct option.

In the adjoining figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of △ABC to area of △BDE is

1. 4 : 1

2. 9 : 1

3. 9 : 4

4. 3 : 2

Answer

Given DE || CA.

Considering △BDE and △BCA,

∠B = ∠B (Common angles)
∠BDE = ∠BCA (Corresponding angles are equal)

Hence, by AA axiom △BDE ~ △BCA.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2}$ .....(i)

Given,

$\Rightarrow \dfrac{BD}{DC} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{BD}{BC - BD} = \dfrac{2}{1} \\[1em] \Rightarrow BD = 2(BC - BD) \\[1em] \Rightarrow BD = 2BC - 2BD \\[1em] \Rightarrow BD + 2BD = 2BC \\[1em] \Rightarrow 3BD = 2BC \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{3}{2}.$

Putting this value in (i) we get,

$\Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2} \\[1em] = \dfrac{3^2}{2^2} \\[1em] = \dfrac{9}{4} \\[1em] = 9 : 4.$

Hence, Option 3 is the correct option.

If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is

1. 2 : 1

2. 1 : 2

3. 1 : 4

4. 4 : 1

Answer

Since triangles ABC and BDE are equilateral triangles so, each angle will be equal to 60°.

Since all angles are equal to 60°.

Hence, by AAA axiom △ABC ~ △BDE.

Since D is the midpoint of BC so,

$\Rightarrow BC = 2BD \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{2}{1}.$

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2} \\[1em] = \dfrac{2^2}{1^2} \\[1em] = \dfrac{4}{1} \\[1em] = 4 : 1.$

Hence, Option 4 is the correct option.

The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is

1. 9 cm

2. 7 cm

3. 6 cm

4. 4.5 cm

Answer

Let the altitude of bigger triangle be x cm.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$\therefore \dfrac{\text{Area of bigger triangle}}{\text{Area of smaller triangle}} = \dfrac{(\text{Bigger triangle altitude})^2}{(\text{Smaller triangle altitude})^2} \\[1em] \Rightarrow \dfrac{81}{49} = \dfrac{x^2}{(3.5)^2} \\[1em] \Rightarrow x^2 = \dfrac{81 \times (3.5)^2}{49} \\[1em] \Rightarrow x^2 = \dfrac{992.25}{49} \\[1em] \Rightarrow x^2 = 20.25 \\[1em] \Rightarrow x = \sqrt{20.25} \\[1em] \Rightarrow x = 4.5$

Hence, altitude of the bigger triangle is 4.5 cm.

Hence, Option 4 is the correct option.

Given △ABC ~ △PQR, area of △ABC = 54 cm² and area of △PQR = 24 cm². If AD and PM are medians of △'s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is

1. $\dfrac{49}{9}$ cm

2. $\dfrac{20}{3}$ cm

3. 15 cm

4. 22.5 cm

Answer

Given, △ABC ~ △PQR.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding medians.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{AD^2}{PM^2} \\[1em] \Rightarrow \dfrac{54}{24} = \dfrac{x^2}{10^2} \\[1em] \Rightarrow x^2 = \dfrac{54 \times 100}{24} \\[1em] \Rightarrow x^2 = \dfrac{5400}{24} \\[1em] \Rightarrow x^2 = 225 \\[1em] \Rightarrow x = 15.$

Hence, length of AD = 15 cm.

Hence, Option 3 is the correct option.

In the given diagram, △ ABC ~ △ PQR and $\dfrac{AD}{PS} = \dfrac{3}{8}$. The value of AB : PQ is :

1. 8 : 3

2. 3 : 5

3. 3 : 8

4. 5 : 8

Answer

Given,

△ ABC ~ △ PQR

⇒ ∠B = ∠Q (Corresponding angles of similar triangle are equal)

In △ ABD and △ PQS,

⇒ ∠B = ∠Q (Proved above)

⇒ ∠D = ∠S (Both equal to 90°)

∴ △ ABD ~ △ PQS (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{PQ} = \dfrac{AD}{PS} = \dfrac{3}{8}$.

∴ AB : PQ = 3 : 8.

Hence, Option 3 is the correct option.

In the given diagram, ∆ABC ∼ ∆PQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively then:

1. ∆ABC ∼ ∆PQS

2. ∆ABD ∼ ∆PQS

3. ∆ABD ∼ ∆PSR

4. ∆ABC ∼ ∆PSR

Answer

Given,

∆ABC ∼ ∆PQR

∴ ∠A = ∠P

⇒ $\dfrac{∠A}{2} = \dfrac{∠P}{2}$

⇒ ∠BAD = ∠QPS

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]

In ∆ABD ∼ ∆PQS,

⇒ ∠BAD = ∠QPS

⇒ ∠B = ∠Q

∴ ∆ABD ∼ ∆PQS (By A.A. axiom)

Hence, Option 2 is the correct option.