Triangles MCQ Quiz - Objective Question with Answer for Triangles - Download Free PDF

Last updated on Jun 3, 2025

Latest Triangles MCQ Objective Questions

Triangles Question 1:

In Δ ABC, AB = 12 cm, BC = 16 cm and AC = 20 cm. A circle is inscribed inside the triangle. What is the radius (in cm) of the circle? 

  1. 3
  2. 4
  3. 5
  4. 6

Answer (Detailed Solution Below)

Option 2 : 4

Triangles Question 1 Detailed Solution

Given:

In Δ ABC, AB = 12 cm, BC = 16 cm, AC = 20 cm

Formula used:

Area of the triangle (Δ) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Where s = semi-perimeter = \(\frac{a+b+c}{2}\)

Radius (r) of the inscribed circle = \(\frac{\Delta}{s}\)

Calculations:

qImage67b6ff6e841ba58e876ee323

a = 12 cm, b = 16 cm, c = 20 cm

s = \(\frac{12+16+20}{2}\) = 24 cm

Area (Δ) = \(\sqrt{24(24-12)(24-16)(24-20)}\)

⇒ Area (Δ) = \(\sqrt{24×12×8×4}\)

⇒ Area (Δ) = \(\sqrt{9216}\)

⇒ Area (Δ) = 96 cm2

Radius (r) = \(\frac{96}{24}\)

⇒ Radius (r) = 4 cm

∴ The correct answer is option (2).

Triangles Question 2:

In ΔABC,  AB = AC = 12 cm, BC = 5 cm and D is a point on AC such that DB = BC. What is the measure of CD?

  1. \(\frac{7}{3}cm\)
  2. \(\frac{25}{12}cm \)
  3. \(\frac{11}{6}cm\)
  4. \(\frac{29}{12}cm\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{25}{12}cm \)

Triangles Question 2 Detailed Solution

Given:

In triangle ABC, AB = AC = 12 cm.

BC = 5 cm.

D is a point on AC such that DB = BC.

Formula used:

Cosine Rule in a triangle: In any triangle with sides a, b, c and angle C opposite to side c,

c2 = a2 + b2 - 2ab cos(C).

qImage68397c93b8f7a5d59a4eba12

Calculation:

In △ABC, AB = 12 cm, AC = 12 cm, BC = 5 cm.

Since AB = AC, △ABC is an isosceles triangle.

Let's find cos(∠C) in △ABC using the Cosine Rule:

AB2 = BC2 + AC2 - 2 × BC × AC × cos(∠C)

122 = 52 + 122 - 2 × 5 × 12 × cos(∠C)

144 = 25 + 144 - 120 × cos(∠C)

0 = 25 - 120 × cos(∠C)

120 × cos(∠C) = 25

cos(∠C) = 25 / 120

cos(∠C) = 5 / 24

Now consider △DBC.

We are given DB = BC. Since BC = 5 cm, then DB = 5 cm.

We know ∠C is common to both triangles. Let CD = x cm.

Apply the Cosine Rule in △DBC to find CD:

DB2 = BC2 + CD2 - 2 × BC × CD × cos(∠C)

52 = 52 + x2 - 2 × 5 × x × (5 / 24)

25 = 25 + x2 - (50x / 24)

0 = x2 - (25x / 12)

Since x = CD cannot be 0 (as D is a point on AC), we can divide by x:

0 = x - (25 / 12)

x = 25 / 12 cm

∴ The correct answer is option 2.

Triangles Question 3:

In Δ PQR, PS is the median of the triangle. QR is the base. If the value of PQ is 20 cm and PS is 15 cm and SR is 10 cm. Then find the value of PR.

  1. 5√20
  2. 5√15
  3. 10√15
  4. 5√10
  5. 7√15

Answer (Detailed Solution Below)

Option 4 : 5√10

Triangles Question 3 Detailed Solution

F4 Savita SSC 22-11-22 D15

Given:

In Δ PQR, PS is the median of the triangle.. QR is the base.

the value of PQ is 20 cm and PS is 15 cm and SR is 10 cm.

Formula used:

As per Apollonius theorem,

PQ2 + PR2 = 2 (Qs2 + PS2)

Calculation:

Let PR = x, PS median, so QS = SR

As per the theorem,

202 + x2 = 2 (152 + 102)

⇒ 400 + x2 = 2 × 325

⇒ x2. = 650 - 400

⇒ x2 = 250

⇒ x = 5√10 cm

The value of PR is 5√10 cm.

Triangles Question 4:

In ∆LMN, medians MX and NY are perpendicular to each other and intersect at Z. If MX = 20 cm and NY = 30 cm, what is the area of ∆LMN (in cm2)?

  1. 200
  2. 400
  3. 300
  4. 450

Answer (Detailed Solution Below)

Option 2 : 400

Triangles Question 4 Detailed Solution

Given:

∆LMN with medians MX and NY

MX ⊥ NY

MX intersects NY at Z

MX = 20 cm

NY = 30 cm

Formula Used:

The centroid of a triangle divides each median in the ratio 2:1.

Area of a triangle = 1/2 × base × height

Area of ∆LMN = 3 × Area of ∆MNZ (since Z is the centroid)

Calculations:

qImage6826dc59a8cdd0dce2d3c783

Since Z is the centroid, it divides the medians in the ratio 2:1.

MZ : ZX = 2 : 1

⇒ MZ = (2/3) × MX = (2/3) × 20 = 40/3 cm

⇒ ZX = (1/3) × MX = (1/3) × 20 = 20/3 cm

NZ : ZY = 2 : 1

⇒ NZ = (2/3) × NY = (2/3) × 30 = 20 cm

⇒ ZY = (1/3) × NY = (1/3) × 30 = 10 cm

Since MX ⊥ NY, ∆MNZ is a right-angled triangle with legs NZ and MZ.

Area of ∆MNZ = 1/2 × base × height = 1/2 × NZ × MZ

⇒ Area of ∆MNZ = 1/2 × 20 × (40/3)

⇒ Area of ∆MNZ = 10 × (40/3) = 400/3 cm2

Area of ∆LMN = 3 × Area of ∆MNZ

⇒ Area of ∆LMN = 3 × (400/3)

⇒ Area of ∆LMN = 400 cm2

∴ The area of ∆LMN is 400 cm2.

Triangles Question 5:

In Δ ABC =  ∠A = 2∠B. The bisector of ∠A meets BC at D such that CD = 4 cm and BD = 6 cm. The length of side AC is:

  1. \(4 \sqrt{5}\) cm 
  2. \(2 \sqrt{10}\) cm 
  3. \(6 \sqrt{5}\) cm 
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : \(2 \sqrt{10}\) cm 

Triangles Question 5 Detailed Solution

Given:

ΔABC, ∠A = 2∠B

AD bisects ∠A

CD = 4 cm

BD = 6 cm

Image of triangle ABC

Formula used:

Congruent triangles: Corresponding sides and angles are equal.

Calculation:

qImage67becac60f61b83af8e9dcef

DC = 4 cm 

BC = 6 + 4 = 10 CM

Now in Δ ACB and Δ DCA

ΔACB, ∠A = 2∠B = 2X

∠BAC = 2X, ∠ABC = X

And, Now in Δ DCB

ADC = 2X (Sum of interior angle is equal to opposite exterior angle.)

∠DAC = X (As AD is bisector of ∠A)

And AC is common in both the triangle.

Then by ASA(angle side angle) Δ ACB and Δ DCA is congruent. 

Then, 

AC/DC = CB/CA

AC2 = DC × CB

AC2 = 4 × 10

AC = 2√10

∴ The length of side AC is 2√10 cm.

Top Triangles MCQ Objective Questions

In the triangle ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°. What is the value of the length of the side BC? 

F2 Savita SSC 1-2-23 D5

  1. 10 cm
  2. 7.13 cm
  3. 13.20 cm
  4. 11.13 cm

Answer (Detailed Solution Below)

Option 4 : 11.13 cm

Triangles Question 6 Detailed Solution

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Given:

In the triangle, ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°.

Concept used:

According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a2 = b2 + c2 - 2bc × cos∠A

 Trigo

Calculation:

​According to the concept,

BC2 = AB2 + AC2 - 2 × AB × AC × cos60°

⇒ BC2 = 122 + 102 - 2 × 12 × 10 × 1/2

⇒ BC2 = 124

⇒ BC ≈ 11.13

∴ The measure of BC is 11.13 cm.

The perimeter of a triangle with sides of integer values is equal to 13. How many such triangles are possible?

  1. 5
  2. 8
  3. 7
  4. 6

Answer (Detailed Solution Below)

Option 1 : 5

Triangles Question 7 Detailed Solution

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Concept used:

If the perimeter of the triangle is "p"

Let Total possible triangles "t"

If p = even, then

t = p2/48

If p = odd, then

t = (p + 3)2/48

Calculation:

According to the question,

Total possible triangles = (13 + 3)2/48

⇒ 5.33 ≈ 5

∴ Total possible triangles are 5.

What is the radius of circle which circumscribes the triangle ABC whose sides are 16, 30, 34 units, respectively? 

  1. 16 units
  2. 17 units
  3. 28 units
  4. 34 units

Answer (Detailed Solution Below)

Option 2 : 17 units

Triangles Question 8 Detailed Solution

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Given:

Triangle first side (a) = 16 units

Triangle second side (b) = 30 units

Triangle third side (c) = 34 units

Formula used:

Heron's formula:

Area of triangle = √{s × (s - a) × (s - b) × (s - c)}

Where, semi - perimeter (s) = (a + b + c)/2

and a, b and c are the sides of a triangle.

Circum-radius of a triangle = (a × b × c)/(4 × area of triangle)

Calculation:

qImage666d6f9b1665f80061279994

Semi - perimeter = (16 + 30 + 34)/2 = 80/2 = 40 units

Area of triangle = √{s × (s - a) × (s - b) × (s - c)}

⇒ √{40 × (40 - 16) × (40 - 30) × (40 - 34)}

⇒ √{40 × 24 × 10 × 6} = √57600 = 240 unit2

Circum-radius of a triangle = (a × b × c)/(4 × area of triangle)

⇒ (16 × 30 × 34)/(4 × 240) = 17 units

∴ The correct answer is 17 units.

Shortcut TrickCalculation:

The sides of a triangle given are Pythagorean triples.

So the hypotenuse = 34 units

and the circumradius of right angled triangle = 34/2 = 17 units

The lengths of the three sides of a triangle are 30 cm, 42 cm and x cm. Which of the following is correct?

  1. 12 ≤ x < 72
  2. 12 > x > 72
  3. 12 < x < 72
  4. 12 ≤ x ≤ 72

Answer (Detailed Solution Below)

Option 3 : 12 < x < 72

Triangles Question 9 Detailed Solution

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Given:

First side of triangle = 30 cm

Second side of triangle = x cm

Third side of triangle = 42 cm

Concept used:

(3rd side - 1st side) < second side < (3rd side + 1st side)

Calculation:

Range of second side = (42 - 30) < x < (42 + 30)

⇒ 12 < x < 72

∴ The correct option is 3.

ABC is a triangle and D is a point on the side BC. If BC = 16 cm, BD = 11 cm and ∠ADC = ∠BAC, then the length of AC is equal to:

  1. 4\(\sqrt5\) cm
  2. 4 cm
  3. 3\(\sqrt5\) cm
  4. 5 cm

Answer (Detailed Solution Below)

Option 1 : 4\(\sqrt5\) cm

Triangles Question 10 Detailed Solution

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Given:

BC = 16 cm, BD = 11 cm and ∠ADC = ∠BAC

Concept:

If two angles and a side of the two triangles are equal, then both triangles will be similar by AA property.

Calculation:

In ΔABC and ΔDAC

⇒ ∠ADC = ∠BAC

⇒ ∠C = Common angle in both triangle

So, ΔABC and ΔDAC are similar triangle.

⇒ \({BC\over AC}={AC\over DC}\)

⇒ AC2 = BC × DC

⇒ AC2 = 16 × 5 = 80

⇒ AC = 4√5

∴ The required result will be 4√5.

In a triangle ABC, angle B = 90° and p is the length of the perpendicular from B to AC. If BC = 10 cm and AC = 12 cm, then what is the value of p?

  1. \( \frac{5 \sqrt{11}}{3}\)
  2. \(\frac{10 \sqrt{11}}{3} \)
  3. \( \frac{40}{\sqrt{61}} \)
  4. \( \frac{12}{25}\)

Answer (Detailed Solution Below)

Option 1 : \( \frac{5 \sqrt{11}}{3}\)

Triangles Question 11 Detailed Solution

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Given:

ABC is right angle triangle at angle B, BC = 10 cm

 AC = 12 cm, p is length of the perpendicular from B to AC

Formula used:

ArΔ = 1/2 × base × height

Calculation:

F1 Vinanti Defence 01.12.23 D9

In an Δ ABC, by using the Pythagoras theorem

AC2 = AB2 + BC2

144 = AB2 + 100

AB2 = 44

AB = √44

Here, We can find the area in two ways,

1) By taking AC as the base & length p as the perpendicular.

2) By taking BC as base & AB as the perpendicular

As, Area (ΔABC) = Area (ΔABC)

⇒ 1/2 × 10 × √44 = 1/2 × 12 × p

⇒ 5 × 2√11 = 6p

⇒ p = (5√11)/3 cm

∴ The correct answer is (5√11)/3 cm

'O' is a point in the interior of an equilateral triangle. The perpendicular distance from 'O' to the sides are \(\sqrt3\) cm, 2\(\sqrt3\) cm, 5\(\sqrt3\) cm. The perimeter of the triangle is :

  1. 48 cm
  2. 32 cm
  3. 24 cm
  4. 64 cm

Answer (Detailed Solution Below)

Option 1 : 48 cm

Triangles Question 12 Detailed Solution

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Given:

The perpendicular distance:

P1 = √3; P2 = 2√3; P3 = 5√3

Concept used:

Height of an equilateral triangle = (√3 × side)/2 

Height of equilateral triangle = sum of perpendicular distance with point

Perimeter of an equilateral triangle = 3 × side

Calculation:

qImage64a830c6abb988593c41f7ce

Height of equilateral triangle = sum of perpendicular distance

⇒ (√3 × side)/2 = P1 + P2 + P3 

⇒ (√3 × side)/2 = √3 + 2√3 + 5√3

⇒ side = 8 × 2 = 16 cm

Perimeter of an equilateral triangle = 3 × side

⇒ 3 × 16 = 48 cm

∴ The correct answer is 48 cm.

In triangle ABC, AD is the angle bisector of angle A. If AB = 8.4 cm and AC = 5.6 cm and DC = 2.8 cm, then the length of side BC will be:

  1. 4.2 cm
  2. 5.6 cm
  3. 7 cm
  4. 2.8 cm

Answer (Detailed Solution Below)

Option 3 : 7 cm

Triangles Question 13 Detailed Solution

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Given:

AB = 8.4 cm, and AC = 5.6 cm, DC = 2.8 cm

Concept used:

The angle bisector of a triangle divides the opposite side into two parts proportional to the other two sides of the triangle.

Calculation:

 

F1 SSC Amit A 24-02-2023 D11

According to the concept,

AB/AC = BD/DC

⇒ 8.4/5.6 = BD/2.8

⇒ 8.4/2 = BD

⇒ 4.2 = BD

So, BD + DC = BC

BC = 4.2 + 2.8

⇒ 7 cm

∴ The length of side BC will be 7 cm.

In the given figure, AB = DB and AC = DC. If ∠ABD = 58° and ∠DBC = (2x - 4)°, ∠ACB = (y + 15)° and ∠DCB = 63°, then the value of 2x + 5y is :

F1 SSC Ishita 24.02.23 D1

  1. 325°
  2. 273°
  3. 259°
  4. 268°

Answer (Detailed Solution Below)

Option 2 : 273°

Triangles Question 14 Detailed Solution

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Given:

AB = DB and AC = DC.

∠ABD = 58° and ∠DBC = (2x - 4)°, 

∠ACB = (y + 15)° and ∠DCB = 63°

Concept used:

If all three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS (Side - Side - Side) rule.

Calculation:

F1 SSC Ishita 24.02.23 D2

As AB = DB, AC = DC, and BC is common for two triangle 

So, ΔABC ≅ ΔDBC

So, ∠ABC = ∠DBC = ∠ABD/2

⇒ 58°/2 = 29°

So,

(2x - 4)° = 29°

⇒ 2x = 33°

Again,

∠ACB = ∠DCB = 63°

So,

(y + 15)° = 63°

⇒ y = 48°

So,

2x + 5y = 33° + 5 × 48°

⇒ 33° + 240°

⇒ 273°

∴ The required answer is 273°.

In ΔABC, M is the midpoint of the side AB. N is a point in the interior of ΔABC such that CN is the bisector of ∠C and CN ⊥ NB. What is the length (in cm) of MN, if BC = 10 cm and AC = 15 cm?

  1. 2.5
  2. 2
  3. 5
  4. 4

Answer (Detailed Solution Below)

Option 1 : 2.5

Triangles Question 15 Detailed Solution

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Given:

In ΔABC, M is the midpoint of the side AB

N is a point in the interior of ΔABC such that CN is the bisector of ∠C and CN ⊥ NB

BC = 10 cm

AC = 15 cm

Concept used:

Midpoint theorem - The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side

Calculation:

Construction: Produce BN to P which meets AC at P.

And Join MN

F3 Savita SSC 17-5-22 D2 V2 

According to the question

In ΔNPC and ΔNBC

∠N = ∠N  [90°]

BC = PC [corresponding side]

BN = NP [corresponding angle]

 ΔNPC ≅ ΔNBC 

Hence, NB = NP (It means Point N is the midpoint of side BP)

And BC = PC = 10 cm

So, AP = AC – PC

AP = (15 – 10) cm

⇒ AP = 5 cm

Now, In ΔABP

M and N are the midpoints of AB and BP

So, According to the midpoint theorem

⇒ MN = \(\frac{AP}{{2}}\)

⇒ \(\frac{5}{{2}}\) cm

⇒ 2.5 cm

∴ The length of MN is 2.5 cm

Shortcut Trick F2 Revannath Teaching 1.11.2022 D1 F2 Revannath Teaching 1.11.2022 D2

The using mid-point theorem,

In ΔBAP

MN = \(AP\over2\) = \(\frac{5}{{2}}\) = 2.5 cm

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