MCQ Questions on Geometry for Competitive Exams

1

If in the following figure (not to scale), ∠DAB + ∠CBA = 90°, BC = AD, AB = 20 cm, CD = 10 cm then the area of the quadrilateral ABCD is:
Geometry mcq question image

A) 120 cm² B) 150 cm² C) 100 cm² D) 75 cm²

Discuss

Answer & Solution

Answer: Option D

Solution:

$${\text{Area of ABCD}} = \frac{1}{2} \times 30 \times 5 = 75{\text{ c}}{{\text{m}}^2}$$

2

A circle touches the side BC of ΔABC at D and AB and AC are produced to E and F, respectively. If AB = 10 cm, AC = 8.6 cm and BC = 6.4 cm, then BE = ?

A) 2.2 cm B) 3.2 cm C) 3.5 cm D) 2.5 cm

Discuss

Answer & Solution

Answer: Option D

Solution:

10 + x = 8.6 + y (AE = AF)
x - y = -1.4 . . . . . . (i)
x + y = 6.4 . . . . . . (ii)
2x = 5.0
x = 2.5
BE = 2.5

3

In an isosceles triangle ABC, AB = AC, XY || BC. If ∠A = 30°, the ∠BXY = ?

A) 75° B) 30° C) 150° D) 105°

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{If }}\angle A = {30^ \circ } \cr & {\text{Then }}\angle ABC = \angle ACB = \frac{{{{180}^ \circ } - {{30}^ \circ }}}{2} = {75^ \circ } \cr & \angle BXY = {180^ \circ } - \angle ABC \cr & = {180^ \circ } - {75^ \circ } \cr & = {105^ \circ } \cr} $$

4

If M is the mid-point of the side BC of ΔABC, and the area of ΔABM is 18 cm², then the area of ΔABC is:

A) 30 cm² B) 34 cm² C) 36 cm² D) 32 cm²

Discuss

Answer & Solution

Answer: Option C

Solution:

BM : MC= 1 : 1
Area of ΔBMA : Area of ΔAMC
= 1 : 1
↓
18
2 unit → 18 × 2
Area of ΔABC = 36 cm²

5

In the given figure, PQ = PS = SR and ∠QPS = 40°, then what is the value of ∠QPR (in degrees)?
Geometry mcq question image

A) 45 B) 60 C) 75 D) 50

Discuss

Answer & Solution

Answer: Option C

Solution:

∠PQS = ∠PSQ = $$\frac{{{{180}^ \circ } - {{40}^ \circ }}}{2}$$ = 70°
∠PSR = 180° - 70° = 110°
∠SPR = ∠SRP = $$\frac{{{{180}^ \circ } - {{110}^ \circ }}}{2}$$ = 35°
∠QPR = ∠QPS + ∠SPR
= 40° + 35°
= 75°

6

ABC is an equilateral triangle. Points D, E and F are taken as the mid-point on sides AB, BC, CA respectively, so that AD = BE = CF. Then AE, BF, CD enclosed a triangle which is:

A) equilateral B) isosceles triangle C) right angle triangle D) None of these

Discuss

Answer & Solution

Answer: Option A

Solution:

Given in question AD = BE = CF
[DB = AF = EC] Because AB = BC = CA
So, Triangle is equilateral

7

Incentre of ΔABC is I. ∠ABC = 90° and ∠ACB = 70°. Then ∠BIC is

A) 115° B) 100° C) 110° D) 105°

Discuss

Answer & Solution

Answer: Option B

Solution:

∵ Sum of all angles of a triangle = 180°
So, ∠BAC = 180° - (90° + 70°) = 20°
Geometry mcq question image

So, ∠BIC = 90° + $$\frac{1}{2}$$∠A
∠BIC = 90° + $$\frac{1}{2}$$ × 20°
∠BIC = 100°

8

In ΔABC, the bisector of ∠A intersect side BC at D. If AB = 12 cm, AC = 15 cm and BC = 18 cm, then the length of BD is:

A) 7.5 cm B) 8 cm C) 9.6 cm D) 9 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

BD : DC = 12 : 15 = 4 : 5
BD = 18 × $$\frac{4}{9}$$ = 8 cm

9

ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°, then the value of ∠GAB is:

A) 100° B) 80° C) 130° D) 90°

Discuss

Answer & Solution

Answer: Option A

Solution:

We know that
⇒ Add of total exterior angle of a triangle (polygon) = 360°
Geometry mcq question image

⇒ So, 130° + 130° + x° = 360°
⇒ x° = 100°

10

A triangle ABC is inscribed in a circle with centre O. AO is produced to meet the circle at K and AD ⊥ BC. If ∠B = 80° and ∠C = 64°, then the measure of ∠DAK is:

A) 10° B) 16° C) 12° D) 20°

Discuss

Answer & Solution

Answer: Option B

Solution:

∠ACK = 90° (In semicircle)
∠B = ∠K = 80°
In ΔAKC, ∠A + ∠K + ∠C = 180°
∠A + 80° + 90° = 180°
∠A = 10°
In ΔABD, ∠A = 90° - 80° = 10°
∠CAD = 90° - 64° = 26°
∠DAK = 26° - 10° = 16°