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

Given:

∠C = 90° 

CD ⊥ AB

∠A = 65°

Concept used:

The sum of the angles of a triangle is 180°.

Calculation:

In ΔABC,

∠BAC + ∠CBA  + ∠ACB  = 180°

⇒ 65° + 90° + ∠CBA  = 180°

⇒ ∠CBA  = 25°

Important Points

We can arrive at the solution based on the perpendicular as well. But it is more time-consuming. Since ∠A and ∠C are given, it is wise to directly apply them in the concept and get the solution.

Given:

Triangle ABC is congruent to triangle DEF.

AB = DE, BC = EF, AC = DF

Formula Used:

If two triangles are congruent, corresponding sides are equal.

Calculation:

EF corresponds to side BC in triangle ABC.

From the given options, EF = BC = 6.

EF = 6

Given:

In triangle ABC:

Points D and E lie on AB and AC respectively

DE ∥ BC

AD = 6 cm, DB = 4 cm ⇒ AB = AD + DB = 10 cm

AE = 9 cm

Concept used:

By Basic Proportionality Theorem (Thales’ Theorem):

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

Formula used:

AD / DB = AE / EC

Calculation:

6 / 4 = 9 / EC

⇒ 3 / 2 = 9 / EC

⇒ EC = (2 × 9) ÷ 3 = 18 ÷ 3 = 6 cm

∴ The length of EC is 6 cm

Given:

If the ratio of corresponding sides of two similar triangles is 1 ∶ 3

Formula used:

Ratio of areas of two similar triangles = (Ratio of sides)²

Calculation:

Ratio of areas = (1 ∶ 3)²

⇒ Ratio of areas = 1² ∶ 3²

⇒ Ratio of areas = 1 ∶ 9

∴ The correct answer is option (2).

In triangle \( ABC \), if the bisectors of \( \angle ABC \) and \( \angle ACB \) meet at point \( O \) and \( \angle A = 60^\circ \), then the angle \( \angle BOC \) is given by:

\[ \angle BOC = 90^\circ + \frac{\angle A}{2} \]

Substituting \( \angle A = 60^\circ \):

\[ \angle BOC = 90^\circ + \frac{60^\circ}{2} = 120^\circ \]

Thus, the measure of \( \angle BOC \) is:

\[ \boxed{120^\circ} \]

Given:

ABC is a right-angled triangle. A circle is inscribed in it.

The length of the two sides containing the right angle are 10 cm and 24 cm

Calculations:

Hypotenuse² = 10² + 24²    (Pythagoras theorem)

Hypotenuse = √676 = 26

Radius of the circle (incircle) inside a triangle = ( Sum of sides containing right angle – Hypotenuse)/2

⇒ (10 + 24 - 26)/2

⇒ 8/2

⇒ 4

∴ The correct choice is option 4.

Given:

ΔABC ~ ΔPQR

Area of ΔABC = 64 cm²

Area of ΔPQR = 81 cm²

PT = 10.8 cm 

AD is the median of ΔABC.

PT is the median of ΔPQR

Concept: 

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides & medians. 

\(\text{Area of (ΔABC)}\over{\text{Area of (ΔPQR)}}\) = \(({AD\over PT})^2\)

Calculation:

\(\text{Area of (ΔABC)}\over{\text{Area of (ΔPQR)}}\) = \(({AD\over PT})^2\)

⇒ \(64\over81\) = \(({AD\over PT})^2\) 

⇒​ \(AD\over PT\) = \(\sqrt{64\over81}\) = \(8\over9\)

⇒​ \(AD\over 10.8\) = \(8\over9\)

⇒​ AD = \({8\times10.8}\over9\) = 9.6 cm

∴ AD = 9.6 cm 

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:

AB (c)  = 7 cm, CA (b) = 5√2 cm, BC = a 

E and F are the mid-points of AC & AB.

∠A = 135°

Concept:

By cosine rule

Cos A = (b² + c² - a²)/ 2bc

Mid-point theorem

EF = BC/2

Calculation:

According to the cosine rule

\(cos135^0= \frac{(5\sqrt2)^2 + (7)^2 - (a)^2}{2\times7\times5\sqrt2}\)

⇒ \(\frac{-1}{\sqrt2}= \frac{(5\sqrt2)^2 + (7)^2 - (a)^2}{2\times7\times5\sqrt2}\)

⇒ \(-70= 50 + 49 - a^2 \)

⇒ a² = 169

⇒ a = 13, a ≠ - 13

⇒ BC = 13

From formula used

EF = BC/2 = 13/2 cm 

∴ EF = 6.5 cm

Given:

In ΔABC, O is the orthocenter and I is the incenter for the given triangle,  

If ∠BIC - ∠BOC = 90∘.

Formula used:

(1) In ΔABC, I is the incenter for the given triangle,

(1.1) ∠BIC = 90^∘ + \(\frac{1}{2}\)∠A

(1.2) ∠AIC = 90∘ + \(\frac{1}{2}\)∠B

(1.3) ∠AIB = 90∘ + \(\frac{1}{2}\)∠C

(2) In ΔABC, O is the orthocenter for the given triangle,

(2.1) ∠BOC = 180^∘ - ∠A

(2.2) ∠AOB = 180∘ - ∠C

(3.3) ∠AOC = 180∘ - ∠B

Calculation:

According to the question, the required image is:

As we know,

∠BOC = 180∘ - ∠A     ----(1)

∠BIC = 90∘ + \(\frac{1}{2}\)∠A    ----(2)

Now, subtract equation (1) from (2).

⇒ ∠BIC - ∠BOC = 90∘ + \(\frac{1}{2}\)∠A  - (180∘ - ∠A )

⇒ 90^∘ = 90∘ + \(\frac{1}{2}\)∠A  - 180∘ + ∠A

⇒ 90∘ = \(\frac{3}{2}\)∠A  - 90∘

⇒ 180∘ = \(\frac{3}{2}\)∠A

⇒ ∠A = 120^∘

∴ The required answer is 120∘.

Additional Information

(1) Incenter - It is the intersection point of all three angle bisectors of a triangle.

(1.1) Angle bisector cuts the angle into two equal half.

(2) Orthocenter - It is the intersection point of all three altitudes drawn from the vertex to the opposite side of the triangle.

(2.1) The altitude of a triangle is perpendicular to the opposite side.

Given:

In ΔABC, AB = 8cm

∠A bisected internally to intersect BC at D.

AB = 8 cm, BD = 6 cm and DC = 7.5 cm

Concept used:

In a triangle, the angle bisector of an angle is divides the opposite side to the angle in the ratio of the remaining two sides.

\(\frac{AB}{AC} = \frac{BD}{DC}\)

Calculation:

AB = 8 cm, ∠A is bisected internally to intersect BC at D,

⇒ AB/AC = BD/CD

⇒ 8/AC = 6/7.5

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.

Given:

In ΔPQR,

PQ = 12 cm, QR = 16 cm and PR = 20 cm

Concept used:

a, b and c denote the sides of the triangle and A denotes the area of the triangle,

Then the measure of the circumradius(r) is.

r = [abc/4A]

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

Hypotenuse² = Perpendicular² + Base² 

Calculation:

Here, we can see that 

(20)2 = (16)2 + (12)2  = 400 

⇒ PR² = QR² + PQ²

So, ΔPQR is a right angle triangle.

Area of ΔPQR = (½ ) × base × Perpendicular

⇒ Area of ΔPQR = (½ ) × 16 × 12

⇒ Area of ΔPQR = 96 cm²

Circumradius (r) = [abc/4 × Area] 

Circumradius (r) = [(12 × 16 × 20)/4 × 96] = 10 cm

∴ The circumradius of a triangle is 10 cm.

For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse. All the vertices of a triangle are at an equal distance from the circumcenter.

PO = QO = OR = r

⇒ PO = PR/2

⇒ PO = 20/2 = 10 cm

∴ The circumradius of a triangle is 10 cm.

∵ 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