Similarity of Triangles — Multiple Choice Questions

If ΔABC and ΔDEF are similar triangles in which ∠A = 47° and ∠E = 83°, then ∠C equals :

1. 50°

2. 60°

3. 70°

4. 80°

Answer

Given,

ΔABC ∼ ΔDEF.

Thus, corresponding angles are equal.

∠B = ∠E = 83°

We know that, sum of all angles in triangle is equal to 180°.

∠A + ∠B + ∠C = 180°

47° + 83° + ∠C = 180°

130° + ∠C = 180°

∠C = 180° - 130°

∠C = 50°.

Hence, option 1 is the correct option.

If ΔABC ∼ ΔQRP, then the corresponding proportional sides are :

1. $\dfrac{AB}{PQ} = \dfrac{BC}{RP}$

2. $\dfrac{AB}{QR} = \dfrac{BC}{QP}$

3. $\dfrac{AC}{QR} = \dfrac{BC}{RP}$

4. $\dfrac{AB}{QR} = \dfrac{BC}{RP}$

Answer

Given,

ΔABC ∼ ΔQRP.

$\therefore \dfrac{AB}{QR} = \dfrac{BC}{RP}$

Hence, option 4 is the correct option.

If in ΔABC and ΔPQR, we have $\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$, then:

1. ΔPQR ∼ ΔCAB

2. ΔPQR ∼ ΔABC

3. ΔBCA ∼ ΔPQR

4. ΔCBA ∼ ΔPQR

Answer

Given,

The two triangles are ΔABC and ΔPQR.

$\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$

By SSS similarity criterion, if the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and the triangles are similar.

To find the correspondence of vertices:

The side AB corresponds to QR. The vertex opposite to AB is C, and opposite to QR is P.

C = P

The side BC corresponds to PR. The vertex opposite to BC is A, and opposite to PR is Q.

A = Q

The side CA corresponds to PQ. The vertex opposite to CA is B, and opposite to PQ is R.

B = R

∴ΔPQR ∼ ΔCAB

Hence, option 1 is the correct option.

In the given figure, AB = 24 cm, AC = 18 cm, DE = 12 cm, DF = 9 cm and ∠BAC = ∠EDF. Then ΔABC ∼ ΔEDF by the condition :

1. AAA

2. SAS

3. SSS

4. AAS

Answer

$\dfrac{AB}{DE} = \dfrac{24}{12} = 2$

$\dfrac{AC}{DF} = \dfrac{18}{9} = 2$

Thus,

$\dfrac{AB}{DE} = \dfrac{AC}{DF}$

∠BAC = ∠EDF [Given ]

∴ ΔABC ∼ ΔEDF (By S.A.S. axiom)

Hence, option 2 is the correct option.

In the given diagram, △ ABC and △ DEF (not drawn to scale) are such that ∠C = ∠F and $\dfrac{AB}{DE} = \dfrac{BC}{EF}$, then

1. △ ABC ~ △ DEF

2. △ BCA ~ △ DEF

3. △ CAB ~ △ DEF

4. the similarity of given triangles cannot be determined.

Answer

Given,

⇒ ∠C = ∠F

Since, the sides containing the angle may or may not be proportional,

i.e., we don't know if $\dfrac{AC}{DF}$ and $\dfrac{BC}{EF}$ are equal or not equal.

∴ Similarity of given triangles cannot be determined.

Hence, Option 4 is the correct option.

If ΔABC and ΔDEF are so related that $\dfrac{AB}{FD} = \dfrac{BC}{DE} = \dfrac{CA}{EF}$, then which of the following is true?

1. ∠A = ∠E and ∠B = ∠D

2. ∠B = ∠F and ∠C = ∠D

3. ∠A = ∠F and ∠B = ∠D

4. ∠C = ∠F and ∠A = ∠D

Answer

Given,

$\dfrac{AB}{FD} = \dfrac{BC}{DE} = \dfrac{CA}{EF}$

∴ ΔABC ∼ ΔDEF BY SSS theorem.

Side AB corresponds to side FD. Therefore, vertex A corresponds to vertex F and vertex B corresponds to vertex D.

Side BC corresponds to side DE. Therefore, vertex B corresponding to D. It also implies vertex C corresponds to vertex E. 

Since the triangles are similar, the corresponding angles are also similar,

∠A = ∠F and ∠B = ∠D

Hence, option 3 is the correct option.

In the given figure, PQ is parallel to TR, then by using condition of similarity:

1. $\dfrac{PQ}{RT} = \dfrac{OP}{OT} = \dfrac{OQ}{OR}$

2. $\dfrac{PQ}{RT} = \dfrac{OP}{OR} = \dfrac{OQ}{OT}$

3. $\dfrac{PQ}{RT} = \dfrac{OR}{OP} = \dfrac{OQ}{OT}$

4. $\dfrac{PQ}{RT} = \dfrac{OP}{OR} = \dfrac{OT}{OQ}$

Answer

Given,

ΔPOQ and ΔROT

∠POQ = ∠TOR [vertically opposite angles are equal]

∠OPQ = ∠ORT [Alternate interior angles are equal]

∴ ΔPOQ ∼ ΔROT (By A.A. axiom)

We know that,

The ratio of corresponding sides in similar triangles are proportional.

$\dfrac{PQ}{RT} = \dfrac{OP}{OR} = \dfrac{OQ}{OT}$

Hence, option 2 is the correct option.

D and E are points on the sides AB and AC respectively of ΔABC. In which of the following cases DE ∥ BC?

1. AD = 3 cm, BD = 8 cm, AC = 8 cm, AE = 3 cm

2. AD = 5 cm, BD = 6 cm, AE = 6 cm, CE = 5 cm

3. AB = 18 cm, AD = 8 cm, AE = 12 cm, EC = 15 cm

4. none of these

Answer

Checking option 3,

Given,

In ΔABC, DE is parallel to BC if and only if it divides the sides AB and AC in the same ratio

From figure,

DB = AB - AD = 18 - 8 = 10 cm

$\dfrac{AD}{DB} = \dfrac{8}{10} = \dfrac{4}{5}$

$\dfrac{AE}{EC} = \dfrac{12}{15} = \dfrac{4}{5}$

$\therefore \dfrac{AD}{DB} = \dfrac{AE}{EC}$

AB = 18 cm, AD = 8 cm, AE = 12 cm, EC = 15 cm DE is parallel to BC holds true.

Hence, option 3 is the correct option.

In ΔABC, DE ∥ BC such that $\dfrac{AD}{DB} = \dfrac{3}{5}$. If AC = 5.6 cm, then AE = ?

1. 2.1 cm

2. 2.8 cm

3. 3.1 cm

4. 4.2 cm

Answer

By basic proportionality theorem,

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the same ratio.

Since DE ∥ BC in ΔABC,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{AE}{AC - AE} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{AE}{5.6 - AE} \\[1em] \Rightarrow 3 \times (5.6 - AE) = 5AE \\[1em] \Rightarrow 16.8 - 3AE = 5AE \\[1em] \Rightarrow 5AE + 3AE = 16.8 \\[1em] \Rightarrow 8AE = 16.8 \\[1em] \Rightarrow AE = \dfrac{16.8}{8} \\[1em] \Rightarrow AE = 2.1 \text{ cm.}$

Hence, option 1 is the correct option.

In ΔABC, DE ∥ BC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB = (3x + 4) cm and EC = (3x) cm.
Then x equals:

1. 2.5

2. 3

3. 4

4. 5

Answer

By basic proportionality theorem,

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the same ratio.

Since DE ∥ BC in ΔABC,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{(7x - 4)}{(3x + 4)} = \dfrac{(5x - 2)}{3x} \\[1em] \Rightarrow 3x \times (7x - 4) = (5x - 2) \times (3x + 4) \\[1em] \Rightarrow 21x^2 - 12x = 15x^2 + 20x -6x -8 \\[1em] \Rightarrow 21x^2 - 12x - 15x^2 - 20x + 6x + 8 = 0\\[1em] \Rightarrow 21x^2 - 15x^2 - 12x - 20x + 6x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 26x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 24x - 2x + 8 = 0 \\[1em] \Rightarrow 6x(x - 4) - 2(x - 4) = 0 \\[1em] \Rightarrow (6x - 2)(x - 4) = 0 \\[1em] \Rightarrow (6x - 2) = 0 \text{ or } (x - 4) = 0 \text{[Using Zero-product rule]}\\[1em] \Rightarrow 6x= 2 \text{ or } x = 4 \\[1em] \Rightarrow x= \dfrac{2}{6} \text{ or } x = 4 \\[1em] \Rightarrow x= \dfrac{1}{3} \text{ or } x = 4 \\[1em]$

x cannot be $\dfrac{1}{3}$ as it yields negative AD and AE values. Therefore x = 4.

Hence, option 3 is the correct option.

It is given that ΔABC ∼ ΔDFE. If ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm, then which of the following is true?

1. DE = 12 cm, ∠F = 50°

2. DE = 12 cm, ∠F = 100°

3. EF = 12 cm, ∠D = 100°

4. EF = 12 cm, ∠D = 30°

Answer

Given,

∠A = 30°

∠C = 50°

In triangle ABC,

∠A + ∠B + ∠C = 180°

∠B = 180° - (∠A + ∠C)

∠B = 180° - (30° + 50°)

∠B = 180° - 80°

∠B = 100°.

As the triangles are similar, the corresponding angles will be equal.

So, ∠F = ∠B = 100°

Since the triangles are similar, the ratio of corresponding sides is proportional,

$\Rightarrow \dfrac{AB}{DF} = \dfrac{AC}{DE} = \dfrac{BC}{FE} \\[1em] \text{Let's consider} \\[1em] \Rightarrow \dfrac{AB}{DF} = \dfrac{AC}{DE} \\[1em] \Rightarrow \dfrac{5}{7.5} = \dfrac{8}{DE} \\[1em] \Rightarrow DE = \dfrac{8 \times 7.5}{5} \\[1em] \Rightarrow DE = \dfrac{60}{5} \\[1em] \Rightarrow DE = 12 \text{ cm.}$

DE = 12 cm, ∠F = 100°

Hence, option 2 is the correct option.

In ΔDEF and ΔPQR, it is given that ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?

1. $\dfrac{DE}{PQ} = \dfrac{EF}{RP}$

2. $\dfrac{DE}{QR} = \dfrac{DF}{PQ}$

3. $\dfrac{EF}{PR} = \dfrac{DF}{PQ}$

4. $\dfrac{EF}{RP} = \dfrac{DE}{QR}$

Answer

Given, in ΔDEF and ΔPQR:

∠D = ∠Q

∠E = ∠R

∴ ΔDEF ∼ ΔQRP by AA similarity.

The ratio of corresponding sides is:

$\dfrac{DE}{QR} = \dfrac{DF}{QP}$

$\dfrac{DE}{PQ} = \dfrac{EF}{RP}$ do not hold true.

Hence, option 1 is the correct option.

In ΔABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then DE is equal to:

1. 4 cm

2. 5 cm

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

4. 15 cm

Answer

From figure,

AC = AE + EC = 2 + 3 = 5 cm.

In ΔABC and ΔADE,

∠A = ∠A [Common angle]

∠ABC = ∠ADE [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE (By A.A. axiom).

In similar triangles, ratios of corresponding sides are equal.

$\therefore \dfrac{AE}{AC} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{DE}{10} \\[1em] \Rightarrow DE = \dfrac{2 \times 10}{5} \\[1em] \Rightarrow DE = \dfrac{20}{5} \\[1em] \Rightarrow DE = 4 \text{ cm.}$

Hence, option 1 is the correct option.

ΔABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. ΔDEF ∼ ΔABC. If EF = 4 cm, then the perimeter of ΔDEF is :

1. 7.5 cm

2. 15 cm

3. 22.5 cm

4. 30 cm

Answer

Given,

ΔDEF ∼ ΔABC

Perimeter of ΔABC = AB + BC + CA = 3 + 2 + 2.5 = 7.5 cm

Since the two triangles are similar, we can write:

$\Rightarrow \dfrac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}} = \dfrac{BC}{EF} \\[1em] \Rightarrow \dfrac{7.5}{\text{Perimeter of ΔDEF}} = \dfrac{2}{4} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{7.5 \times 4}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{30}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = 15 \text{ cm.}$

Hence, option 2 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.

In the given figure, ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is equal to:

1. 5 : 3

2. 3 : 5

3. 3 : 2

4. 2 : 3

Answer

Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB ∥ CD.

By basic proportionality theorm,

$\Rightarrow \dfrac{PD}{DB} = \dfrac{PC}{CA} \\[1em] \Rightarrow \dfrac{PD}{DB} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{PD}{DB} = \dfrac{3}{2} \\[1em] \Rightarrow PD : DB = 3 : 2.$

Hence, option 3 is the correct option.

In the adjoining figure, DE ∥ BC. If AD : DB = 3 : 1 and EA = 3.3 cm, then AC equals:

1. 1.1 cm

2. 4 cm

3. 4.4 cm

4. 5.5 cm

Answer

Since, DE // BC,

By basic proportionality theorem,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{3}{1} = \dfrac{3.3}{EC} \\[1em] \Rightarrow EC = \dfrac{3.3}{3} \\[1em] \Rightarrow EC = 1.1 \text{ cm.}$

From figure,

AC = AE + EC

AC = 3.3 + 1.1 = 4.4 cm.

Hence, option 3 is the correct option.

In the adjoining figure, ∠ADE = ∠ABC, AE = 8 cm, EB = 7 cm, BC = 9 cm, AD = 10 cm and DC = 2 cm. Then the length of DE is:

1. 6 cm

2. 6.75 cm

3. 7.8 cm

4. 13.5 cm

Answer

From figure,

AB = AE + EB = 8 + 7 = 15 cm

AC = AD + DC = 10 + 2 = 12 cm

In ΔABC and ΔADE

∠BAC = ∠DAE [Common angle]

∠ABC = ∠ADE [Given]

∴ ΔABC ∼ ΔADE (By A.A. axiom).

We know that,

In similar triangles the ratio of corresponding sides are equal.

$\Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC} \\[1em] \text{Let's Consider} \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{10}{15} = \dfrac{DE}{9} \\[1em] \Rightarrow DE = \dfrac{10 \times 9}{15} \\[1em] \Rightarrow DE = \dfrac{90}{15} \\[1em] \Rightarrow DE = 6 \text{ cm.}$

Hence, option 1 is the correct option.

In ΔABC, it is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm. Also ΔDEF is given such that EF = 8 cm and ΔDEF ∼ ΔABC. Then the perimeter of ΔDEF is:

1. 22.5 cm

2. 25 cm

3. 27 cm

4. 30 cm

Answer

Given,

ΔDEF ∼ ΔABC

Perimeter of ΔABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm.

Since the triangles are similar,

$\therefore \dfrac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}} = \dfrac{BC}{EF} \\[1em] \Rightarrow \dfrac{22.5}{\text{Perimeter of ΔDEF}} = \dfrac{6}{8} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{22.5 \times 8}{6} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{180}{6} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = 30 \text{ cm.}$

Hence, option 4 is the correct option.

In a ΔABC, AB = 10 cm, AC = 14 cm and BC = 6 cm. If AD is the internal bisector of ∠A, then CD is equal to:

1. 3.5 cm

2. 4.8 cm

3. 7 cm

4. 10.5 cm

Answer

Construction: Draw a line through C parallel to AD, meeting BA produced at E.

Since AD ∥ EC and BE is the transversal:

∠BAD = ∠AEC [Corresponding angles are equal]

Since AD ∥ EC and AC is the transversal:

∠DAC = ∠ACE [Alternate interior angles]

Given AD is the bisector of ∠A:

∠BAD = ∠DAC

∴ ∠AEC = ∠ACE

In ΔACE, sides opposite to equal angles are equal.

∴ AE = AC = 14 cm

Let DC be x,

Now, in ΔBCE, we have AD ∥ EC. By Basic Proportionality Theorem,

$\Rightarrow \dfrac{BD}{DC} = \dfrac{AB}{AE} \\[1em] \Rightarrow \dfrac{BD}{DC} = \dfrac{10}{14} \\[1em] \Rightarrow \dfrac{BD}{DC} = \dfrac{10}{14} \\[1em] \Rightarrow \dfrac{BD}{DC} = \dfrac{5}{7} \\[1em]$

Let CD = x.

Then BD = BC - CD = 6 - x.

Substituting these values into the ratio:

$\Rightarrow \dfrac{6 - x}{x} = \dfrac{5}{7} \\[1em] \Rightarrow 7(6 - x) = 5x \\[1em] \Rightarrow 42 - 7x = 5x \\[1em] \Rightarrow 42 = 5x + 7x \\[1em] \Rightarrow 12x = 42 \\[1em] \Rightarrow x = \dfrac{42}{12} \\[1em] \Rightarrow x = 3.5 \text{ cm.}$

Thus, CD = 3.5 cm.

Hence, option 1 is the correct option.

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

1. 30°

2. 50°

3. 60°

4. 100°

Answer

Considering ΔAPB and ΔCPD,

$\dfrac{AP}{PD} = \dfrac{6}{5} \text{ and } \dfrac{BP}{CP} = \dfrac{3}{2.5} = \dfrac{6}{5}$

∠APB = ∠CPD [Vertically opposite angles are equal]

∴ ΔAPB ∼ ΔDPC (By SAS similarity)

Hence, ∠PAB = ∠PDC = 30°

⇒ ∠PBA = 180° - (∠PAB + ∠APB)

⇒ ∠PBA = 180° - (30° + 50°)

⇒ ∠PBA = 180° - 80°

⇒ ∠PBA = 100°.

Hence, option 4 is the correct option.

ABCD is a trapezium with AB parallel to DC. Then the triangle similar to ΔAOB is:

1. ΔACB

2. ΔADB

3. ΔCOB

4. ΔCOD

Answer

Given,

Consider ΔAOB and ΔCOD

∠AOB = ∠COD [Vertically opposite angles are equal]

∠OBA = ∠ODC [Alternate angles are equal]

Therefore, by AA rule of similarity ΔAOB ~ ΔCOD.

Hence, option 4 is the correct option.

The ratio of the corresponding sides of two similar triangles is 1 : 3. The ratio of their corresponding heights is :

1. 1 : 3

2. 1 : 9

3. 3 : 1

4. 9 : 1

Answer

Given,

The ratio of corresponding sides be 1 : 3.

Heights are linear measures corresponding to sides, so the ratio of the heights will also be 1 : 3.

Hence, option 1 is the correct option.

If in triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B is equal to :

1. 35°

2. 65°

3. 75°

4. 85°

Answer

Given,

ΔABC and ΔDEF

∠A = ∠E = 40° [Equal angles]

AB : ED = AC : EF [Corresponding sides are equal]

∴ ΔABC ∼ ΔEDF by SAS similarity. Therefore,

⇒ ∠A = ∠E = 40°

⇒ ∠B = ∠D = x

⇒ ∠C = ∠F = 65°

The sum of all the angles in a triangle is 180°:

⇒ ∠A + ∠B + ∠C = 180°

⇒ 40° + x + 65° = 180°

⇒ x = 180° - 40° - 65°

⇒ x = 75°

⇒ ∠B = ∠D = 75°.

Hence, option 3 is the correct option.

In the given figure, AB ∥ CD and OA = (2x + 4) cm, OB = (9x − 21) cm, OC = (2x − 1) cm and OD = 3 cm. Then x equals:

1. 2.1

2. 3

3. 4

4. 6

Answer

Given,

OA = (2x + 4) cm

OB = (9x − 21) cm

OC = (2x − 1) cm

OD = 3 cm

In ΔOAB and ΔOCD

∠COD = ∠AOB [Vertically Opposite Angles are equal]

∠OAB = ∠OCD [alternate interior angles are equal]

∴ ΔOAB ∼ ΔOCD by AA similarity. Then ratios of corresponding sides are equal :

$\Rightarrow \dfrac{OA}{OC} = \dfrac{OB}{OD} \\[1em] \Rightarrow \dfrac{(2x + 4)}{(2x - 1)} = \dfrac{(9x - 21)}{3} \\[1em] \Rightarrow 3 \times (2x + 4) = (2x - 1) \times (9x - 21) \\[1em] \Rightarrow 6x + 12 = 18x^2 - 42x -9x +21 \\[1em] \Rightarrow 18x^2 - 42x -9x +21 - 6x - 12 = 0\\[1em] \Rightarrow 18x^2 - 42x - 9x - 6x +21 - 12 = 0\\[1em] \Rightarrow 18x^2 - 57x + 9 = 0\\[1em] \Rightarrow 18x^2 - 54x - 3x + 9 = 0\\[1em] \Rightarrow 18x(x - 3) - 3(x - 3) = 0\\[1em] \Rightarrow (18x - 3)(x - 3) = 0\\[1em] \Rightarrow (18x - 3) = 0 \text{ or }(x - 3) = 0 \\[1em] \Rightarrow 18x = 3 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{3}{18} \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{1}{6} \text{ or } x = 3.$

x cannot be $\dfrac{1}{6}$ as it yields negative OB and OC values. Therefore x = 3.

Hence, option 2 is the correct option.

The shadow of a 5 m long stick is 2 m long. At the same time the length of the shadow of a 12.5 m high tree is:

1. 3 m

2. 3.5 m

3. 4.5 m

4. 5 m

Answer

Given,

Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.

Suppose EA and CA are the shadows of DE and BC respectively.

Now, In ΔABC and ΔADE,

∠ACB = ∠AED = 90°

∠A = ∠A [Common]

∴ ΔABC ∼ ΔADE by AA similarity.Then ratios of corresponding sides are equal:

$\Rightarrow \dfrac{CA}{AE} = \dfrac{BC}{DE} \\[1em] \Rightarrow \dfrac{CA}{2} = \dfrac{12.5}{5} \\[1em] \Rightarrow CA = \dfrac{12.5 \times 2}{5} \\[1em] \Rightarrow CA = \dfrac{25}{5} \\[1em] \Rightarrow CA = 5 \text{ cm.}$

Hence, option 4 is the correct option.

In a right-angled ΔABC, right-angled at A, if AD ⟂ BC such that AD = p, BC = a, CA = b and AB = c, then:

1. $\dfrac{p}{a} = \dfrac{p}{b}$

2. p² = b² + c²

3. p² = b² c²

4. $\dfrac{1}{p^{2}} = \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}}$

Answer

Given,

Area of a right angled triangle = $\dfrac{1}{2} \times \text{ base } \times \text{ height }$

Area of triangle ABC = $\dfrac{1}{2} \times AB \times AC$

Area of triangle ABC = $\dfrac{1}{2} \times bc$ .......(1)

Also,

Area of triangle ABC = $\dfrac{1}{2} \times BC \times AD$

Area of triangle ABC = $\dfrac{1}{2} \times ap$ .....(2)

From equation (1) and (2), we get :

$\dfrac{1}{2}$ bc = $\dfrac{1}{2}$ ap

bc = ap

a = $\dfrac{bc}{p}$

Applying Pythogoras theorem to right-angled triangle ABC:

$\Rightarrow BC^2 = AC^2 + AB^2 \\[1em] \Rightarrow a^2 = b^2 + c^2 \\[1em] \Rightarrow \Big(\dfrac{bc}{p}\Big)^2 = b^2 + c^2 \\[1em] \Rightarrow \dfrac{b^2c^2}{p^2} = b^2 + c^2 \\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{b^2 + c^2}{b^2c^2} \\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{b^2}{b^2c^2} + \dfrac{c^2}{b^2c^2}\\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{1}{c^2} + \dfrac{1}{b^2} .$

Hence, option 4 is the correct option.

In a right-angled ΔABC in which ∠A = 90°, if AD ⟂ BC, then which of the following statements is correct?

1. AB = BD × AD

2. AB² = BC × BD

3. AB² = BD × DC

4. AB² = BC × DC

Answer

Given,

In ΔABC and ΔDBA,

∠ADC = ∠ADB = 90°

∠B = ∠B [Common]

ΔABC ∼ ΔDBA [By AA similarity]

From the similarity ΔABC ∼ ΔDBA, the ratio of corresponding sides is equal:

$\therefore \dfrac{AB}{DB} = \dfrac{BC}{BA} = \dfrac{AC}{DA} \\[1em] \text{ Considering first two terms } \\[1em] \Rightarrow \dfrac{AB}{DB} = \dfrac{BC}{BA} \\[1em] \Rightarrow AB^2 = BC \times BD .$

Hence, option 2 is the correct option.

In the adjoining figure, XY is parallel to BC. If XY divides the triangle into two equal parts, then $\dfrac{AX}{AB}$ equals:

1. $\dfrac{1}{\sqrt{2}}$

2. $\dfrac{1}{2}$

3. $\dfrac{\sqrt{2} + 1}{\sqrt{2}}$

4. $\dfrac{\sqrt{2} - 1}{\sqrt{2}}$

Answer

Given,

In ΔABC and ΔAXY

∠BAC = ∠XAY [Common angle]

∠AXY = ∠ABC [Corresponding angles are equal]

∴ ΔABC ∼ ΔAXY (By A.A. axiom)

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

$\dfrac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \Big(\dfrac{AX}{AB}\Big)^2$ .....(1)

X and Y divides triangle ABC into two equal parts. Therefore,

Area of ΔAXY = $\dfrac{1}{2}$ Area of ΔABC

$\dfrac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \dfrac{1}{2}$ .....(2)

Equating Eqn(1) and Eqn(2) :

$\Rightarrow \dfrac{1}{2} = \Big(\dfrac{AX}{AB}\Big)^2 \\[1em] \Rightarrow \dfrac{AX}{AB} = \dfrac{1}{\sqrt{2}}.$

Hence, option 1 is the correct option.

Which of the following is true in the given figure, where AD is the altitude to the hypotenuse of a right-angled ΔABC?

(I) ΔABD ∼ ΔCAD

(II) ΔADB ≅ ΔCDA

(III) ΔADB ∼ ΔCAB

1. I and II

2. II and III

3. I and III

4. I, II and III

Answer

In ΔABD,

∠B + ∠BAD = 90°

∠BAD = 90° - ∠B ........(1)

In ΔABC,

Since, BC is the hypotenuse, thus angle A = 90°

From figure,

∠DAC = ∠BAC - ∠BAD

∠DAC = 90° - ∠BAD

Substituting value of ∠BAD from equation (1) in above equation, we get :

∠DAC = 90° - (90° - ∠B) = ∠B.

In ΔABD and ΔCAD,

∠DAC = ∠B (Proved above)

∠ADB = ∠ADC (Both equal to 90°)

∴ ΔABD ∼ ΔCAD by AA similarity.

Thus, (I) is true.

or we can say that ΔADB ∼ ΔCDA.

Thus, (II) is true.

In ΔADB and ΔCAB,

∠ADB = ∠CAB = 90°

∠ABD = ∠CBA [Common angle]

∴ ΔADB ∼ ΔCAB by AA similarity.

Thus, (III) is true.

Hence, option 4 is the correct option.

The areas of two similar triangles are 49 cm² and 64 cm² respectively. The ratio of their corresponding sides is:

1. 7 : 8

2. 49 : 64

3. 8 : 7

4. 64 : 49

Answer

Given,

Let, A₁ and A₂ be the areas of two triangle. s₁ and s₂ be the length of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$\Rightarrow \dfrac{A_1}{A_2} = \Big(\dfrac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \dfrac{49}{64} = \Big(\dfrac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \dfrac{s_1}{s_2} = \sqrt{\dfrac{49}{64}} \\[1em] \Rightarrow \dfrac{s_1}{s_2} = \dfrac{7}{8}.$

Ratio between sides = 7 : 8.

Hence, option 1 is the correct option.

The areas of two similar triangles are 12 cm² and 48 cm² respectively. If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is:

1. 0.525 cm

2. 4.2 cm

3. 4.41 cm

4. 8.4 cm

Answer

Given,

Let, A₁ and A₂ be the areas of two triangle. h₁ and h₂ be the height of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding heights.

$\Rightarrow \dfrac{A_1}{A_2} = \Big(\dfrac{h_1}{h_2}\Big)^2 \\[1em] \Rightarrow \dfrac{12}{48} = \Big(\dfrac{2.1}{h_2}\Big)^2 \\[1em] \Rightarrow \dfrac{1}{4} = \Big(\dfrac{2.1}{h_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{1}{4}} = \dfrac{2.1}{h_2} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{2.1}{h_2} \\[1em] \Rightarrow h_2 = 2.1 \times 2 \\[1em] \Rightarrow h_2 = 4.2 \text{ cm.}$

Hence, option 2 is the correct option.

The areas of two similar triangles are 81 cm² and 144 cm². If the largest side of the smaller triangle is 27 cm, then largest side of the larger triangle is:

1. 24 cm

2. 36 cm

3. 48 cm

4. none of these

Answer

Given,

Let, A₁ and A₂ be the areas of two triangle. s₁ and s₂ be the length of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$\Rightarrow \dfrac{A_1}{A_2} = \Big(\dfrac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \dfrac{81}{144} = \Big(\dfrac{27}{s_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{81}{144}} = \dfrac{27}{s_2} \\[1em] \Rightarrow \dfrac{9}{12} = \dfrac{27}{s_2} \\[1em] \Rightarrow s_2 = \dfrac{27 \times 12}{9} \\[1em] \Rightarrow s_2 = \dfrac{324}{9} \\[1em] \Rightarrow s_2 = 36 \text{ cm.}$

Hence, option 2 is the correct option.

ΔABC and ΔDEF are similar to each other. If the ratio of side AB to side DE is $(\sqrt{2} + 1) : \sqrt{3}$, then the ratio of area of ΔABC to that of ΔDEF is:

1. $(3 + 2\sqrt{2}) : 3$

2. $1 : (9 - 6\sqrt{2})$

3. $(9 - 6\sqrt{2}) : 2$

4. $(3 - 2\sqrt{2}) : 3$

Answer

Given,

ΔABC and ΔDEF are similar to each other.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$\Rightarrow \dfrac{\text{Area of triangle ABC}}{\text{Area of triangle DEF}} = \Big(\dfrac{(\sqrt{2} + 1)}{\sqrt{3}}\Big)^2 \\[1em] = \dfrac{(\sqrt{2} + 1)^2}{(\sqrt{3})^2} \\[1em] = \dfrac{(\sqrt{2})^2 + (1)^2 + 2 \times \sqrt{2} \times 1}{3} \\[1em] = \dfrac{2 + 1 + 2\sqrt{2}}{3} \\[1em] = \dfrac{3 + 2\sqrt{2}}{3}.$

Area of triangle ABC : Area of triangle DEF = $(3 + 2\sqrt{2}) : 3$.

Hence, option 1 is the correct option.

If ΔABC ∼ ΔQRP, $\dfrac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \dfrac{9}{4}$, AB = 18 cm and BC = 15 cm, then PR = ?

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

2. 8 cm

3. 10 cm

4. 12 cm

Answer

Given,

AB = 18 cm

BC = 15 cm

Since the triangles are similar, the ratios of Areas of triangles is equal to the ratio of the squares of corresponding sides.

$\Rightarrow \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\dfrac{BC}{PR}\Big)^2 \\[1em] \Rightarrow \dfrac{9}{4} = \Big(\dfrac{15}{PR}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{9}{4}} = \dfrac{15}{PR} \\[1em] \Rightarrow \dfrac{3}{2} = \dfrac{15}{PR} \\[1em] \Rightarrow PR = \dfrac{15 \times 2}{3} \\[1em] \Rightarrow PR = \dfrac{30}{3} \\[1em] \Rightarrow PR = 10 \text{ cm.}$

Hence, option 3 is the correct option.

In the given figure, ∠CAB = 90° and AD ⟂ BC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, then AD equals:

1. 50 cm

2. 60 cm

3. 65 cm

4. 70 cm

Answer

Given,

∠CAB = 90°

AC = 75 cm = 0.75 m

AB = 1 m

BC = 1.25 m

In ΔCAB and ΔADB,

∠CAB = ∠ADB = 90° [AD ⟂ BC]

∠CBA = ∠ABD [Common angle]

∴ ΔCAB ∼ ΔADB by AA similarity. Then ratios of corresponding sides are equal:

$\Rightarrow \dfrac{AC}{AD} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{AC}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{0.75}{1.25} = \dfrac{AD}{1} \\[1em] \Rightarrow AD = \dfrac{0.75}{1.25} \\[1em] \Rightarrow AD = 0.6 \text{ m.}$

AD = 0.6 m = 60 cm.

Hence, option 2 is the correct option.

A street lamp is fixed on a lamp-post at a height of 3.3 m from the ground. A boy 110 cm tall walks away from the base of this lamp post at a speed of 0.8 m/s. The length of the shadow of the boy after 4 seconds is:

1. 1.1 m

2. 1.6 m

3. 2.1 m

4. 2.6 m

Answer

Let AB = 3.3 m be the lamp post and QP = 1.1 m be the position of boy after 4 seconds.

AP is the distance moved in 4s at 0.8m/s = 4(0.8) = 3.2 m

PM is the length of the shadow of boy.

In triangle AMB and PMQ,

∠MAB = ∠MPQ = 90°

∠AMB = ∠PMQ [Common angle]

∴ ΔAMB ∼ ΔPMQ [By AA similarity].

Then ratios of corresponding sides of similar triangles are equal.

$\Rightarrow \dfrac{AM}{PM} = \dfrac{AB}{PQ} \\[1em] \Rightarrow \dfrac{AP + PM}{PM} = \dfrac{AB}{PQ} \\[1em] \Rightarrow \dfrac{3.2 + x}{x} = \dfrac{3.3}{1.1} \\[1em] \Rightarrow 3.2 + x = 3x \\[1em] \Rightarrow 3.2 = 2x \\[1em] \Rightarrow x = \dfrac{3.2}{2} \\[1em] \Rightarrow x = 1.6 \text{ m}.$

Hence, option 2 is the correct option.

In the given diagram (not draw to scale), railway stations A, B, C, P and Q are connected by straight tracks. Track PQ is parallel to BC. The time taken by a train travelling at 90 km/hr to reach B from A by the shortest route is :

1. 8 minutes

2. 12 minutes

3. 16.8 minutes

4. 20 minutes

Answer

In △ APQ and △ ABC,

⇒ ∠PAQ = ∠BAC (Common angle)

⇒ ∠APQ = ∠ABC (Corresponding angles are equal)

∴ △ APQ ~ △ ABC (By A.A. axiom)

From figure,

Let AP = x km.

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{AP}{AB} = \dfrac{PQ}{BC} \\[1em] \Rightarrow \dfrac{x}{AP + PB} = \dfrac{20}{50} \\[1em] \Rightarrow \dfrac{x}{x + 18} = \dfrac{2}{5} \\[1em] \Rightarrow 5x = 2(x + 18) \\[1em] \Rightarrow 5x = 2x + 36 \\[1em] \Rightarrow 5x - 2x = 36 \\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \dfrac{36}{3} \\[1em] \Rightarrow x = 12.$

AB = AP + BP = 12 + 18 = 30 km.

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{AB}{90} = \dfrac{30}{90} = \dfrac{1}{3} \text{ hr} = \dfrac{1}{3} \times 60$ = 20 minutes.

Hence, Option 4 is the correct option.

The lengths of the sides of triangle P are 3, 4 and 5 units. Another triangle Q, which is similar to P, has one side of length 60 units, what is the smallest possible perimeter of triangle Q?

1. 120 units

2. 144 units

3. 180 units

4. 240 units

Answer

Given,

Triangle P is similar to triangle Q

Perimeter of Triangle P = 3 + 4 + 5 = 12 units

Since the triangles are similar, the corresponding sides have a scale factor of k . One of the length of side of triangle Q is 60 units. The Corresponding scale factors will be:

Case 1 :

k = $\dfrac{60}{3}$ = 20

Case 2 :

k = $\dfrac{60}{4}$ = 15

Case 3 :

k = $\dfrac{60}{5}$ = 12

Among the three cases, smallest sclae factor = 12.

Smallest possible perimeter of Triangle (Q) = 12 × 12 = 144 units.

Hence, option 2 is the correct option.

Which of the following statements is correct?

1. If D is a point on side AB of ΔABC such that AD : DB = 5 : 2 and E is a point on BC such that DE ∥ AC, then ar(ΔABC) : ar(ΔDBE) = 9 : 4.

2. If the areas of two similar triangles are in the ratio 25 : 64, then their perimeters are in the ratio 5 : 8.

In the adjoining figure if AB ∥ CD, then ΔAOB ∼ ΔCOD.

In the adjoining figure, if D and E are the mid-points of AB and AC respectively, then ar(ΔADE) = $\dfrac{1}{2}$ × ar(ΔABC).

Answer

Solving option 2,

Since the triangles are similar,

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

$\Rightarrow \dfrac{Area_1}{Area_2} = \Big(\dfrac{\text{Perimeter}_1}{\text{Perimeter}_2}\Big)^2 \\[1em] \Rightarrow \dfrac{25}{64} = \Big(\dfrac{\text{Perimeter}_1}{\text{Perimeter}_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{25}{64}} = \dfrac{\text{Perimeter}_1}{\text{Perimeter}_2} \\[1em] \Rightarrow \dfrac{\text{Perimeter}_1}{\text{Perimeter}_2} =\dfrac{5}{8} \\[1em]$

Hence proved.

Hence, option 2 is the correct option.

In the given figure, D, E and F are the mid-points of the sides BC, AC and AB respectively of ΔABC. Then which of the following does not hold true?

1. ΔAFE ∼ ΔABC

2. ΔFBD ∼ ΔABC

3. ΔEDC ∼ ΔABC

4. ΔDFE ∼ ΔABC

Answer

Given,

Since D, E, F are midpoints of BC, AC and AB respectively.

According to the Mid-point Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and is half its length.

Thus, by mid-point theorem,

DE ∥ AB

DE = $\dfrac{1}{2}$AB

EF ∥ BC

EF = $\dfrac{1}{2}$BC

FD ∥ AC

DF = $\dfrac{1}{2}$AC

Ratios can be written as,

$\Rightarrow \dfrac{DF}{AC} = \dfrac{EF}{BC} = \dfrac{DE}{AB} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}AC}{AC} = \dfrac{\dfrac{1}{2}BC}{BC} = \dfrac{\dfrac{1}{2}AB}{AB} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}$

∴ ΔDFE ∼ ΔACB is true, but ΔDFE ≁ ΔABC.

Hence, option 4 is the correct option.

A triangle with sides 6, 9 and 12 units has area A sq. units. What is the area (in sq. units) of a triangle with sides 8, 12 and 16 units in terms of A?

1. $\dfrac{3}{2}$ A

2. $\dfrac{4}{3}$ A

3. $\dfrac{16}{9}$ A

4. $\dfrac{25}{16}$ A

Answer

We compare the ratios of the corresponding sides of the first triangle with sides 6, 9, 12 and the second triangle with sides 8, 12, 16.

k = $\dfrac{6}{8} = \dfrac{9}{12} = \dfrac{12}{16} = \dfrac{3}{4}$.

Therefore, the two triangles are similar with scale factor of k = $\dfrac{3}{4}$. The ratio of Areas of two triangle is equal to square of it's scale factor.

$\Rightarrow \dfrac{\text{Area}_1}{\text{Area}_2} = k^2 \\[1em] \Rightarrow \dfrac{A}{\text{Area}_2} = \Big(\dfrac{3}{4}\Big)^2 \\[1em] \Rightarrow \dfrac{A}{\text{Area}_2} = \dfrac{9}{16} \\[1em] \Rightarrow \text{Area}_2 = \dfrac{16}{9}A.$

Hence, option 3 is the correct option.

In the adjoining diagram, ST is not parallel to PQ. The necessary and sufficient conditions for △ PQR ~ △ TSR is :

1. ∠PQR = ∠STR

2. ∠QPR = ∠TSR

3. ∠PQR = ∠TSR

4. ∠PRQ = ∠RST

Answer

Given,

△ PQR ~ △ TSR.

From figure,

⇒ ∠PRQ = ∠SRT (Common angle)

The order of vertices of two similar triangles are written in such a way that the corresponding vertices occupy the same position.

∴ ∠PQR = ∠TSR

∴ △ PQR ~ △ TSR (By A.A. axiom)

Hence, Option 3 is the correct option.

Directions (Q. 44 to 47): These questions are based on the following information :

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. It is given that AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm and PQ = 2 cm.

The perimeter of ΔABC is:

1. 12 cm

2. 16 cm

3. 18 cm

4. 24 cm

Answer

From figure,

AB = AP + PB = 1 + 3 = 4 cm

AC = AQ + QC = 1.5 + 4.5 = 6 cm

In ΔAPQ and ΔABC,

$\Rightarrow \dfrac{AP}{AB} = \dfrac{1}{4}$

$\Rightarrow \dfrac{AQ}{AC} = \dfrac{1.5}{6} = \dfrac{1}{4}$

∠PAQ = ∠BAC [Common angles]

ΔAPQ ∼ ΔABC [By SAS Similarity]

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{PQ}{BC} = \dfrac{AQ}{AC} \\[1em] \Rightarrow \dfrac{PQ}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow BC = 2 \times 4 \\[1em] \Rightarrow BC = 8 \text{ cm}.$

Perimeter = AB + AC + BC

= 4 + 6 + 8

= 18 cm.

Hence, option 3 is the correct option.

The ratio of the areas of ΔAPQ and ΔABC is:

1. 1 : 3

2. 1 : 4

3. 1 : 9

4. 1 : 16

Answer

We know that,

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

$\Rightarrow \dfrac{\text{ Area(ΔAPQ)}}{\text{ Area(ΔABC)}} = \Big(\dfrac{AP}{AB}\Big)^2 \\[1em] = \Big(\dfrac{1}{4}\Big)^2 \\[1em] = \dfrac{1}{16} = 1 : 16.$

Hence, option 4 is the correct option.

Which of the following holds true?

1. ΔAPQ ∼ ΔABC

2. ΔAQP ∼ ΔABC

3. ΔAPQ ∼ ΔACB

4. none of these

Answer

In ΔAPQ and ΔABC,

$\dfrac{AP}{AB} = \dfrac{AQ}{AC} = \dfrac{1}{4}$

∠PAQ = ∠BAC [Common angles]

∴ ΔAPQ ∼ ΔABC by SAS Similarity.

Hence, option 1 is the correct option.

Which axiom of similarity applies in the above case?

1. AAA

2. AA

3. SSS

4. SAS

Answer

Two pairs of corresponding sides are proportional ($\dfrac{AP}{AB} = \dfrac{AQ}{AC}$) and icluded angle is also equal (angle A).

This is the S.A.S. (Side-Angle-Side) Similarity Axiom.

Hence, option 4 is the correct option.

Directions (Q. 48 to 51): Using the given diagram answer the following questions.

In ΔPQR, AB ∥ QR, QP ∥ CB and AR intersects CB at O.

The triangle similar to ΔARQ is:

1. ΔORC

2. ΔARP

3. ΔOBR

4. ΔQRP

Answer

In ΔARQ and ΔORC,

∠BRC = ∠ARQ [Common angle]

∠AQR = ∠BCR [Corresponding angles, AB ∥ QR, BC is the transversal]

ΔARQ ∼ ΔORC [By AA Similarity]

Hence, option 1 is the correct option.

ΔPQR ∼ ΔBCR by axiom:

1. SAS

2. AAA

3. SSS

4. AAS

Answer

In ΔPQR and ΔBCR,

∠PRQ = ∠BRC [Common angle]

∠PQR = ∠BCR [Corresponding angles, CB ∥ QP, BC is the transversal]

ΔPQR ∼ ΔBCR [By AA or AAA similarity]

Hence, option 2 is the correct option.

If QC = 6 cm, CR = 4 cm, BR = 3 cm, then the length of RP is:

1. 4.5 cm

2. 5 cm

3. 7.5 cm

4. 8 cm

Answer

RQ = RC + CQ = 4 + 6 = 10 cm.

RB = 3 cm

RC = 4 cm

From 46 que we have ΔPQR ∼ ΔBCR,

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{RP}{RB} = \dfrac{RQ}{RC} \\[1em] \Rightarrow \dfrac{RP}{3} = \dfrac{10}{4} \\[1em] \Rightarrow RP = \dfrac{5}{2} \times 3 \\[1em] \Rightarrow RP = 7.5 \text{ cm.}$

Hence, option 3 is the correct option.

The ratio PQ : BC is:

1. 2 : 3

2. 3 : 2

3. 2 : 5

4. 5 : 2

Answer

Using ΔPQR ∼ ΔBCR,

$\Rightarrow \dfrac{PQ}{BC} = \dfrac{RQ}{RC} \\[1em] = \dfrac{10}{2} \\[1em] = \dfrac{5}{2} = 5:2$

Hence, option 4 is the correct option.

Directions (Q. 52 to 55): Answer these questions on the basis of the following information:

Through the mid-point M of the side CD of a ∥gm ABCD, the line BM is drawn, intersecting AC on L and AD produced in E.

Which of the following is true for triangles BMC and EMD ?

I. They are similar to each other.

II. They are congruent to each other.

III. Their areas are equal.

IV. Their perimeters are equal.

1. I and II

2. II and III

3. I and III

4. I, II, III and IV

Answer

In ΔBMC and ΔEMD,

∠EDM = ∠MCB [Alternate angles are equal]

∠EMD = ∠BMC [Vertically opposite angles are equal]

CM = DM (As M is the mid-point of CD)

ΔBMC ≅ ΔEMD [By A.A. axiom]

Congruent triangles are similar, have equal areas and perimeters.

∴ All statements are correct

Hence, option 4 is the correct option.

ΔAEL is similar to:

1. ΔCBL

2. ΔCML

3. ΔDME

4. ΔBMC

Answer

In ΔAEL and ΔCBL,

∠EAL = ∠BCL[Alternate angles are equal]

∠ALE = ∠CLB [Vertical opposite angles are equal]

ΔAEL ∼ ΔCBL [By A.A. axiom]

Hence, option 1 is the correct option.

By which axiom are the above triangles similar?

1. AA

2. SSS

3. SAS

4. ASA

Answer

The similarity ΔAEL ∼ ΔCBL was established by using two pairs of corresponding angles.

Hence, option 1 is the correct option.

EL : BL is equal to :

1. 1 : 2

2. 2 : 1

3. 1 : 3

4. 3 : 1

Answer

AD = BC [ABCD is a parallelogram]

BC = DE [ΔBMC ≅ ΔEMD]

From figure,

AE = AD + DE

AE = BC + BC

AE = 2BC

Using ΔAEL ∼ ΔCBL,

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{EL}{BL} = \dfrac{AE}{CB} \\[1em] \Rightarrow \dfrac{EL}{BL} = \dfrac{2BC}{CB} \\[1em] \Rightarrow \dfrac{EL}{BL} = \dfrac{2}{1}\\[1em] \Rightarrow EL : BL = 2 : 1.$

Hence, option 2 is the correct option.

Directions (Q. 56 to 59): Study the following diagram carefully and answer the given questions:

In ΔABC, D and E are points on AB and AC respectively such that AD = a, DB = 3a, AE = b and EC = 3b. DQ ∥ EA and EP ∥ DA are drawn. QP is joined.

ΔADE is similar to which of the following triangles?

I. ΔABC

II. ΔDAQ

III. ΔADQ

IV. ΔEPA

V. ΔEAP

1. I, II and IV only

2. I, III and V only

3. I, II and V only

4. I, III and IV only

Answer

In ΔADE and ΔABC,

∠A is common.

$\dfrac{AD}{AB} = \dfrac{a}{a + 3a} = \dfrac{1}{4}$

$\dfrac{AE}{AC} = \dfrac{b}{b + 3b} = \dfrac{1}{4}$

By SAS Similarity of Triangles, ΔADE ∼ ΔABC.

In ΔADE and ΔDAQ,

∠DQA = ∠ADE [Corresponding angle]

∠QDA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔDAQ by AA similarity

In ΔADE and ΔEPA,

∠EPA = ∠ADE [Corresponding angle]

∠PEA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔEPA by AA similarity

Hence, option 1 is the correct option.

If DE = 2 cm, then BC is equal to:

1. 4 cm

2. 6 cm

3. 7 cm

4. 8 cm

Answer

Since ΔADE ∼ ΔABC, corresponding sides are proportional.

$\dfrac{DE}{BC} = \dfrac{AD}{AB}$

Given AD = a and DB = 3a, so AB = 4a.

$\Rightarrow \dfrac{2}{BC} = \dfrac{a}{4a} \\[1em] \Rightarrow \dfrac{2}{BC} = \dfrac{1}{4}\\[1em] \Rightarrow BC = 8 \text{ cm.}$

Hence, option 4 is the correct option.

The ratio of the perimeters of ΔADE and ΔABC is:

1. 1 : 2

2. 1 : 3

3. 1 : 4

4. 1 : 6

Answer

$\Rightarrow \dfrac{\text{ Perimeter(ΔADE)}}{\text{ Perimeter(ΔABC)}} = \dfrac{AD}{AB} \\[1em] = \dfrac{a}{4a} \\[1em] = \dfrac{1}{4} = 1:4.$

Hence, option 3 is the correct option.

The ratio of the areas of ΔADE and trapezium DBCE is:

1. 1 : 8

2. 1 : 9

3. 1 : 15

4. 1 : 16

Answer

$\Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = \Big(\dfrac{AB}{AD}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\dfrac{4a}{a}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\dfrac{4}{1}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = 16 \\[1em] \Rightarrow \text{ Area(ΔABC)} = 16 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \text{ Area quad. DBCE} = \text{ Area(ΔABC)} - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 16 (\text{ Area(ΔADE)}) - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 15 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \dfrac{\text{ Area(ΔADE)}}{\text{ Area quad. DBCE}} = \dfrac{1}{15} \\[1em]$

Hence, option 3 is the correct option.

Directions (Q. 60 to 63): Study the given information and answer the questions that follow:

In the given figure ABCD is a trapezium in which DC is parallel to AB. AB = 16 cm and DC = 8 cm, OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x − 1) cm.

From the given figure name the pair of similar triangles :

1. ΔOAB, ΔOBC

2. ΔCOD, ΔAOB

3. ΔADB, ΔACB

4. ΔCOD, ΔCOB

Answer

Given,

In ΔAOB and ΔCOD,

∠AOB = ∠COD [Vertically opposite angle are equal]

∠OAB = ∠OCD [Alternate interior angle are equal]

∴ ΔAOB ∼ ΔCOD (By A.A. axiom)

Hence, option 2 is the correct option.

The corresponding proportional sides with respect to the pair of similar triangles obtained above is :

1. $\dfrac{CD}{AB} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

2. $\dfrac{AD}{BC} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

3. $\dfrac{AD}{BC} = \dfrac{BD}{AC} = \dfrac{AB}{DC}$

4. $\dfrac{OD}{OB} = \dfrac{CD}{CB} = \dfrac{OC}{OA}$

Answer

Given,

ΔAOB ∼ ΔCOD.Since the triangles are similar, the ratios of the corresponding sides are equal,

$\dfrac{CD}{AB} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

Hence, option 1 is the correct option.

The ratio of the sides of the pair of similar triangles is:

1. 1 : 3

2. 1 : 2

3. 2 : 3

4. 3 : 1

Answer

Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

Ratio = $\dfrac{CD}{AB} = \dfrac{8}{16} = \dfrac{1}{2}$ = 1 : 2.

Hence, option 2 is the correct option.

Using the ratio of sides of the pair of similar triangles, the values of x and y are respectively :

1. x = 4.6, y = 7

2. x = 7, y = 7

3. x = 6.5, y = 7

4. x = 6.5, y = 2

Answer

Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

$\dfrac{CD}{AD} = \dfrac{OC}{OA} = \dfrac{OD}{OB} = \dfrac{1}{2}$

Solving,

$\dfrac{OC}{OA} = \dfrac{1}{2}$

Substituting values we get :

$\Rightarrow \dfrac{(x - 1)}{11} = \dfrac{1}{2} \\[1em] \Rightarrow 2 \times (x - 1) = 11 \\[1em] \Rightarrow 2x - 2 = 11 \\[1em] \Rightarrow 2x = 11 + 2 \\[1em] \Rightarrow 2x = 13 \\[1em] \Rightarrow x = \dfrac{13}{2} \\[1em] \Rightarrow x = 6.5$

Solving,

$\dfrac{OD}{OB} = \dfrac{1}{2}$

Substituting values we get :

$\Rightarrow \dfrac{5}{(y + 3)} = \dfrac{1}{2} \\[1em] \Rightarrow 2 \times 5 = (y + 3) \\[1em] \Rightarrow 10 = y + 3 \\[1em] \Rightarrow y = 10 - 3 \\[1em] \Rightarrow y = 7.$

x = 6.5, y = 7

Hence, option 3 is the correct option.

In the given diagram ΔABC ∼ ΔEFG. If ∠ABC = ∠EFG = 60°, then the length of the side FG is:

1. 15 cm

2. 20 cm

3. 25 cm

4. 30 cm

Answer

Given, ΔABC ∼ ΔEFG

So, the corresponding sides are in proportion.

$\Rightarrow \dfrac{AB}{EF} = \dfrac{BC}{FG}\\[1em] \Rightarrow \dfrac{15}{75} = \dfrac{3}{FG}\\[1em] \Rightarrow \dfrac{1}{5} = \dfrac{3}{FG}\\[1em] \Rightarrow FG = 3 \times 5\\[1em] \Rightarrow FG = 15\text{ cm}$

Hence, Option 1 is the correct option.

In the figure given below, AC is a diameter of the circle.

AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of ∆APQ is 18 cm², then the area of shaded portion QPBC is :

1. 32 cm²

2. 49 cm²

3. 80 cm²

4. 98 cm²

Answer

Given,

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In ∆APQ and ∆ABC,

⇒ ∠APQ = ∠ABC [Both equal to 90°]

⇒ ∠BAC = ∠PAQ [Common angles]

∴ ∆APQ ∼ ∆ABC (By A.A. axiom)

We know that,

The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.

$\Rightarrow \dfrac{\text{Area of ∆APQ}}{\text{Area of ∆ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{18}{\text{Area of ∆ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of ∆ABC} = \dfrac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC}= \dfrac{(4 + 3)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \dfrac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \dfrac{49}{9} \times 18 = 98 \text{ cm}^2.$

From figure,

Area QPBC = Area of ∆ABC − Area of ∆APQ

= 98 - 18 = 80 cm².

Hence, Option 3 is the correct option.