Congruence and Similarity MCQ Quiz - Objective Question with Answer for Congruence and Similarity - Download Free PDF

Given:

Δ ABC ∼ Δ QPR

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR}=\frac{64}{25}\)

Formula used:

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

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR} = (\frac{corresponding \ side \ of \ Δ ABC}{corresponding \ side \ of \ Δ PQR})^2\)

Calculation:

\(\frac{64}{25} = (\frac{AB}{QP})^2\)

⇒ \(\frac{64}{25} = (\frac{AB}{12})^2\)

⇒ \(\frac{64}{25} = \frac{AB^2}{144}\)

⇒ \(\frac{64}{25} × 144 = AB^2\)

⇒ \(\frac{64 × 144}{25} = AB^2\)

⇒ \(\frac{9216}{25} = AB^2\)

⇒ 368.64 = \(AB^2\)

⇒ AB = \(\sqrt{368.64}\)

⇒ AB = 19.2 cm

∴ The correct answer is option (4).

Given:

The ratio of corresponding sides of two similar triangles = 5 : 7

Formula Used:

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

Calculation:

Let the ratio of the corresponding sides be 5/7.

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⇒ (5/7)²

⇒ 25/49

The ratio of the areas of the two triangles is 25 : 49.

Given:

In ΔPOR, a line joining point X on side PQ and point Y on side PR is parallel to side QR.

The ratio PY ∶ YR is 3 ∶ 5.

Length PX = 7 cm.

Formula used:

By the basic proportionality theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Calculations:

Let the length of PQ be x cm.

According to the basic proportionality theorem,

\(\dfrac{PX}{XQ} = \dfrac{PY}{YR}\)

Given PY : YR = 3 : 5,

⇒ \(\dfrac{PX}{XQ} = \dfrac{3}{5}\)

Let XQ = y cm,

⇒ \(\dfrac{7}{y} = \dfrac{3}{5}\)

⇒ 7 × 5 = 3 × y

⇒ 35 = 3y

⇒ y = \(\dfrac{35}{3} cm\)

Therefore,

PQ = PX + XQ = 7 cm + \(\dfrac{35}{3} cm\)

⇒ PQ = \(\dfrac{21}{3} + \dfrac{35}{3}\)

⇒ PQ = \(\dfrac{56}{3} cm\)

⇒ PQ = 18.66 cm

∴ The correct answer is option (3).

Given:

ΔABC ~ ΔQRP

QR = 8 cm

PR = 10 cm

PQ = 6 cm

ar(ΔABC) : ar(ΔQRP) = 1 : 4

Formula used:

If ΔABC ~ ΔQRP,

Then ar(ΔABC) / ar(ΔQRP) = (AB/QR)² = (BC/RP)² = (AC/QP)²

Calculation:

ar(ΔABC) / ar(ΔQRP) = 1/4

(AB/QR)² = 1/4

⇒ (AB/8)² = 1/4

⇒ AB/8 = √(1/4)

⇒ AB/8 = 1/2

⇒ AB = 8/2

⇒ AB = 4 cm

∴ AB = 4 cm

Given:

In ΔPQR:

PQ = 5 cm, QR = 6 cm, PR = 7 cm

In ΔXYZ:

XY = 5 cm, YZ = 6 cm, XZ = 7 cm

Formula used:

SSS (Side-Side-Side) Congruence Rule:

If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.

Calculation:

Comparing side lengths:

PQ = XY = 5 cm

QR = YZ = 6 cm

PR = XZ = 7 cm

Since all corresponding sides are equal, by the SSS Congruence Rule:

⇒ ΔPQR ≅ ΔXYZ

∴ The correct statement is ΔPQR ≅ ΔXYZ.

Given:

ΔPQR ∼ ΔDEF

The sides of ΔPQR and ΔDEF are in the ratio 5 ∶ 6.

ar(PQR) = 75 cm²

Concepts used:

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

Calculation:

ΔPQR ∼ ΔDEF

ar(PQR)/ar(DEF) = (Side of ΔPQR/Side of ΔDEF)²

⇒ 75/ar(DEF) = (5/6)²

⇒ ar(DEF) = 108 cm²

∴ Area of ΔDEF is equal to 108 cm².

Given:

 Δ ABC ∼ Δ QPR, \(\rm \frac{ar(Δ ABC)}{ar(Δ PQR)}=\frac{4}{9}\),

AC = 12 cm, AB = 18 cm and BC = 10 cm

Concept used:

Δ ABC ∼ Δ QPR 

⇒ Ratio of respective areas = Ratio of square of respective sides

that is, \(\rm \frac{ar(Δ ABC)}{ar(Δ QPR)}=\frac{AB^2}{QP^2}=\frac{BC^2}{PR^2}=\frac{AC^2}{QR^2}\)

Calculation:

⇒ 4/9 = (10)²/PR²

⇒ PR² = 900/4

⇒ PR2 = 225

⇒ PR = 15 cm

Mistake Points In this question, it is given that ΔABC is similar to ΔQPR. Do not misread as ΔPQR.

Δ ABC ∼ Δ QPR, this relation matters while applying the similarity rule, 

What is given in the ratio form, is just for your confusion.

∵ AE + EC = AC

⇒ EC = 5 cm

According to angle bisector theorem,

So, AE/EC = AB/BC

⇒ 4/5 = AB/10

∴ AB = 8 cm

Given,

the area of quadrilateral MBCN = 130 cm².

AN : NC = 4 : 5

Hence, AC = 4 + 5 = 9

In ΔABC, If MN∥BC, then

Area of ΔAMN/area of ΔABC = (AN/AC)²

Area of ΔAMN/area of ΔABC = (4/9)² = 16/81

Area of ΔAMN = 16 unit and area of ΔABC = 81 unit

Now, Area of quadrilateral MBCN = area of ΔABC - area of ΔAMN

⇒ 81 - 16 unit = 130 cm²

⇒ 65 unit = 130 cm²

⇒ 1 unit = 130/65 = 2 cm²

Given

AD = 12 cm, CD = 16 cm 

Formula used 

Angle Bisector theorem,

AD/CD = AB/BC

Perimeter of triangle = Sum of all sides

Calculation

In Δ ABC, BD is an angle bisector of ∠B

AD/CD = AB/BC

⇒ 12/16 = AB/BC

⇒ AB : BC = 3 : 4

From the triplets of right angle triangle

If AB = 3x and BC = 4x

then AC = 5x

AC = 12 + 16 = 28 cm

⇒ 5x = 28 cm

⇒ x = 5.6 cm

Perimeter of triangle = 5x + 3x + 4x = 12x

⇒ 67.2 cm

∴ The required answer is 67.2 cm

Given:

DE = 9, EF = 12 cm, and DF = 7 cm

DO is the angle bisector of ∠EDF

Concept used:

In a triangle, If AD is the angle bisector of ∠BAC intersect the opposite side BC at D then

By angle bisector theorem

AB/AC = BD/CD

Calculation:

DE = 9, EF = 12 cm, and DF = 7 cm

DO is the angle bisector of ∠EDF

By angle bisector theorem,

DE/DF = EO/OF

⇒ 9/7 = (EF - OF)/OF

⇒ 9/7 = (12 - OF)/OF

⇒ 9 × OF = 84 - 7 × OF

⇒ 16 × OF = 84

⇒ OF = 84/16

⇒ OF = 5.25 cm

∴ The value of OF is 5.25 cm.

Given :-

ΔABC and ΔPQR are similar

AB = 8 cm

PQ = 12 cm

QR = 18 cm 

RP = 24 cm

Concept Use :-

Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles. 

(1) AB/PQ = BC/QR = AC/PR

(2) angle A = angle P , angle B = angle Q , angle C = angle R

Calculation :-

⇒  AB/PQ = BC/QR = AC/PR

⇒ 8/12 = BC/18 = AC/24

⇒ BC = (18 × 8) /12 = 12cm

and,

⇒ AC = (12 × 24) /18 = 16 cm

Now,

⇒ Perimeter of triangle ABC = AB +BC + AC = 8 + 12 + 16

⇒ Perimeter of triangle ABC = 36cm

Given:

Area of ΔABC = 44 cm²

BD = CD

AE = BE

Concept used:

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

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

The ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both triangles.

Calculation:

According to the concept,

DE ∥ AC, and AC = 1/2 × DE

Let BC = 2 unit, and BD = 1 unit

Area of ΔBDE/Area of ΔBCA = (BD/BC)²

⇒ Area of ΔBDE/44 = (1/2)²

⇒ Area of ΔBDE = (1/4) × 44 = 11 cm2

∴ The area of ΔBDE is 11 cm²

GIVEN:

∠AED = ∠BAC

AD || BC

AD : BC = 7 : 9

AC = 36 cm

CONCEPT:

Similarity of triangle

CALCULATION:

Since ∠AED = ∠BAC  

∠EAD = ∠ACB (alternate interior angle)

Rule of similarity AA(Angle-Angle)

∆EDA and ∆ABC are similar.

Now,

AD/BC = AE/AC

⇒ 7/9 = AE/36

⇒ AE = 28 cm

Now,

EC = AC – AE = 36 – 28 = 8 cm

∴ The value of EC is 8 cm 

Given:

In a Δ ABC, ∠BAC = 90°, AD is drawn perpendicular from A on BC

Concept used:

Similarity of triangles

Calculation:

Since AD⊥BC, ΔBAC ~ ΔBDA

So, we know,

BC/AB = AB/BD

⇒ BC × BD = AB² 

⇒ BC : AB :: AB : BD

∴ AB is the following is the mean proportional between BD and BC.