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MCQ Questions on Geometry for Competitive Exams

Prepare for competitive exams with our Geometry MCQ questions. Practice essential theorems and problem-solving techniques with questions tailored to exam patterns. Page-7 section-1

Each section contains 100 MCQs on Geometry .
61
The perimeter of right-angle triangle is 60 cm and its hypotenuse is 26 cm. What is the area (in cm2) of the triangle?
Discuss
Answer & Solution
Answer: Option D
Solution:
Geometry mcq question image
Hypotenuses = 26 cm
Perimeter = 60 cm
Using by pythagoras triplet - 5, 12, 13
13 unit → 26 cm
1 unit → 2 cm
Area of triangle = $$\frac{1}{2}$$ × 10 × 24 = 120 cm2
62
In an isosceles ΔABC, AD is the median to the unequal side meeting BC at D. DP is the angle bisector of ∠ADB and PQ is drawn parallel to BC meeting AC at Q. Then the measure of ∠PDQ is
Discuss
Answer & Solution
Answer: Option B
Solution:
Geometry mcq question image
∠PDQ = 45° + 45° = 90°
63
Two concentric circles are of redii 13 cm and 5 cm. The length of the chord of the larger circle which touches the smaller circle is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Geometry mcq question image
AM = 12
AB = 2 × 12 = 24 cm
64
Two tangents AP and AQ are drawn to a circle with centre O from an external point A, where P and Q are points on the circle. If AP = 12 cm and ∠PAQ = 60°, then the length of chord PQ is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Geometry mcq question image
In ΔAPO
Geometry mcq question image
2 unit → 12
1 unit → 6
PT = 6
PQ = 2 × PT = 2 × 6 = 12 cm
65
Two tangents PA and PB are drawn from an external point P to a circle with centre O at the points A and B respectively on it, such that ∠APB = 120° and AP = 12.5 cm. The length of OP is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Geometry mcq question image
∠APO = 60°
∠POA = 30°
Geometry mcq question image
1 unit → 12.5
2 unit → 12.5 × 2 = 25
PO = 25 cm
66
In the given figure, PT is a common tangent to three circles at points A, B and C respectively. The radius of the small, medium and large circles is 4 cm, 6 cm and 9 cm. O1, O2 and O3 are the centre of the three circles what is the value (in cm) of PC?
Geometry mcq question image
Discuss
Answer & Solution
Answer: Option A
Solution:
Geometry mcq question image
ΔPO1A is similar to ΔPO2B
AO1 : BO2 = 2 : 3
O1O2 = 10 = 3 - 2 = 1 unit
            1 unit → 10
PO2 = 3 units → 30
ΔPO3C is similar to ΔPO2B.
BO2 : CO3 = 2 : 3
2 units → 30
1 unit → 15
PO3 = 45
CO3 = 9
$$\eqalign{ & {\text{PC}} = \sqrt {{{45}^2} - {9^2}} \cr & = \sqrt {2025 - 81} \cr & = \sqrt {1944} \cr & = 18\sqrt 6 \cr} $$
67
In ΔPQR, S is a point on the side QR such that ∠QPS = $$\frac{1}{2}$$ ∠PSR, ∠QPR = 78° and ∠PRS = 44°. What is the measure of ∠PSQ?
Discuss
Answer & Solution
Answer: Option C
Solution:
∠QPS = $$\frac{1}{2}$$∠PSR = x (Let)
∠PSR = 2x
∠PQS + ∠QPS = 2x
∠QPR = 78° and ∠PRS = 44°
Geometry mcq question image
In ΔPQR
∠P + ∠Q + ∠R = 180°
78° + x + 44° = 180°
x = 180° - 122°
x = 58°
∠PSQ = 180° - 2x
= 180° - 2 × 58°
= 180° - 116°
= 64°
68
A secant is drawn from a point P to a circle so that it meets the circle first at A, then goes through the centre, and leaves the circle at B. If the length of the tangent from P to the circle is 12 cm, and the radius of the circle is 5 cm, then the distance from P to A is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Geometry mcq question image
PQ = 12 cm
r = 5 cm
PQ2 = PA × PB
122 = x(x + 10)
144 = x(x + 10)
x2 + 10x - 144 = 0
x2 + (18x - 8x) -144 = 0
x(x + 18) - 8(x + 18) = 0
x = 8, x = -18
PA = 8 cm
69
In ΔABC, D and E are points on the sides AB and AC, respectively, such that DE || BC and DE : BC = 6 : 7. (Area of ΔADE) : (Area of trapezium BCED) = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Geometry mcq question image
AD : BD = 6 : 1
ΔADE : trapezium BCDE = 36 : 13
70
In the given figure, ABCD is a rectangle and P is a point on DC such that BC = 24 cm, DP = 10 cm and CD = 15 cm. If AP produced intersects BC produced at Q, then the length of AQ.
Geometry mcq question image
Discuss
Answer & Solution
Answer: Option C
Solution:
Geometry mcq question image
$$\eqalign{ & {\text{Let the length of }}CQ = x\,\& \,\angle PQC = \theta \cr & {\text{In }}\Delta ABQ, \cr & \tan \theta = \frac{{15}}{{24 + x}}.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{Again in }}\Delta PCQ \cr & \tan \theta = \frac{5}{x}\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & {\text{From equation}}\left( {\text{i}} \right)\,\& \,\left( {{\text{ii}}} \right) \cr & \frac{{15}}{{24 + x}} = \frac{5}{x} \cr & 3x = 24 + x \cr & 2x + 24 \cr & x = 12 \cr & {\text{In }}\Delta ABQ, \cr & {\text{By pythagoras theorem}} \cr & A{Q^2} = A{B^2} + B{Q^2} \cr & = {15^2} + {36^2} \cr & = 225 + 1296 \cr & = 1521 \cr & AQ = 39 \cr} $$