MCQ Questions on Geometry for Competitive Exams

71

A is a point at a distance 26 cm from the centre O of a circle of radius 10 cm. AP and AQ are the tangents to the circle at the point of contacts P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B, AQ at C, then the perimeter of ΔABC is:

A) 40 cm B) 48 cm C) 46 cm D) 42 cm

Discuss

Answer & Solution

Answer: Option B

No explanation is given for this question. Let's Discuss on Board

72

In a circle, PQ and RS are two diameters that are perpendicular to each other. Find the length of the chord PR.

A) $$\frac{{{\text{PQ}}}}{2}$$ B) $$\sqrt 2 {\text{PQ}}$$ C) $$2{\text{PQ}}$$ D) $$\frac{{{\text{PQ}}}}{{\sqrt 2 }}$$

Discuss

Answer & Solution

Answer: Option D

Solution:

Let, PO = RO = 1
PQ = 2
PQ = $$\sqrt 2 \times \sqrt 2 $$
PQ = PR × $$\sqrt 2 $$
PR = $$\frac{{{\text{PQ}}}}{{\sqrt 2 }}$$

73

Let AX ⊥ BC of an equilateral triangle ABC. Then the sum of the perpendicular distances of the sides of ΔABC from any point inside the triangle is:

A) Greater than AX B) Less than AX C) Equal to BC D) Equal to AX

Discuss

Answer & Solution

Answer: Option D

Solution:

Let side = 2 units
Side = $$\frac{2}{{\sqrt 3 }}$$(PT + QT + TR)
2 = $$\frac{2}{{\sqrt 3 }}$$(PT + QT + TR)
∴ PT + QT + TR = $${\sqrt 3 }$$
& AX $$ = \frac{{\sqrt 3 }}{2} \times 2 = \sqrt 3 $$
So it is equal to AX

74

In the figure below, AB is a chord of a circle with centre O. A tangent AT is drawn at point A so that ∠BAT = 50°. Then ∠ADB = ?
Geometry mcq question image

A) 120° B) 130° C) 140° D) 150°

Discuss

Answer & Solution

Answer: Option B

Solution:

Let take a point 'C' on circumference of circle
Then ∠BAT = ∠BCA = 50° (Alternate segment theorem)
Geometry mcq question image

In Cyclic Quadrilateral $$\square $$ ACBD
∠D = 180° - ∠C
(In a cyclic quadrilateral the sum of opposite angles is 180°)
Then ∠D = 180° - 50° = 130°

75

Two circles with the same centre P have radii 7.5 cm and 4.4 cm. Through a point A of the larger circle, a tangent is drawn to the smaller circle touching it at B. Find AC (Approximate in cm).

A) 14 cm B) 12.14 cm C) 14.27 cm D) 13 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

(7.5)² - (4.4)² = a²
56.25 - 19.36 = a²
36.89 = a²
6.07 = a
2a = 12.14 cm

76

In ΔABC, E and D are points on sides AB and AC, respectively, such that ∠ABC = ∠ADE. If AE = 6 cm, AD = 4 cm and EB = 4 cm, then the length of DC is:

A) 11 cm B) 8 cm C) 9.5 cm D) 10 cm

Discuss

Answer & Solution

Answer: Option A

No explanation is given for this question. Let's Discuss on Board

77

A secant PAB is drawn from an external point P to the circle with centre O, intersecting it at A and B. If OP = 17 cm, PA = 12 cm and PB = 22.5 cm, then the radius of the circle is:

A) $$2\sqrt 3 {\text{ cm}}$$ B) $$\sqrt {19} {\text{ cm}}$$ C) $$\sqrt {17} {\text{ cm}}$$ D) $$3\sqrt 2 {\text{ cm}}$$

Discuss

Answer & Solution

Answer: Option B

No explanation is given for this question. Let's Discuss on Board

78

In the given, figure, CD and AB are diameters of circle and AB and CD are perpendicular to each other, LQ and SR are perpendiculars to AB and CD respectively. Radius of circle is 5 cm, PB : PA = 2 : 3. What is the length (in cm) of SM?

Geometry mcq question image

A) $$\left[ {\left( {5\sqrt 3 } \right) - 3} \right]$$ B) $$\left[ {\left( {4\sqrt 3 } \right) - 2} \right]$$ C) $$\left[ {\left( {2\sqrt 5 } \right) - 1} \right]$$ D) $$\left[ {\left( {2\sqrt 6 } \right) - 1} \right]$$

Discuss

Answer & Solution

Answer: Option D

Solution:

O is the center of circle and radius is 5 cm.
PB : PA = 2 : 3 (Given)
PB + PA = 10 cm (Diameter)
∴ 2 + 3 = 5 units → 10 cm
1 unit → 2 cm
∴ PB = 4 cm
PA = 6 cm
OP = AP - OA
AP = 6 cm, OA = 5 cm (Radius)
∴ OP = 6 - 5 = 1 cm
Similarly ON = 1 cm
Geometry mcq question image

In ΔONS
Use pythagoras theorem
SO² = NO² + NS²
25 = 1 + NS²
NS² = 24
NS = $$\sqrt {24} $$
∴ SM = NS - MN
∴ MN = OP = 1 cm
SM = $$\sqrt {24} - 1$$
SM = $$\left[ {\left( {2\sqrt 6 } \right) - 1} \right]$$ cm

79

In the following figure (not to scale), at the centre O, if the chord AB students double the angle that is subtended by chord CD and the angle ∠AEB = 2∠AOB, then ∠COD is equal to:
Geometry mcq question image

A) 30° B) 45° C) 60° D) 75°

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & 4x + 2x = {180^ \circ } \cr & 6x = {180^ \circ } \cr & \angle COD = x = \frac{{{{180}^ \circ }}}{6} = {30^ \circ } \cr} $$

80

A circle inscribed in a triangle ABC touches its sides AB, BC and AC at the points D, E and F, respectively. If AB = 18 cm, BC = 15 cm and AC = 13 cm, then the value of AD + BE + CF is:

A) 25 cm B) 33 cm C) 23 cm D) 20 cm

Discuss

Answer & Solution

Answer: Option C

Solution:

2(x + y + z) = 46
x + y + z = 23
AD + BE + CF = 23
Note:-
Geometry mcq question image

Perimeter of Δ = 2(x + y + z)
Then, AD + BE + CF = x + y + z
If a circle is inscribe into a triangle touches the side AB. BC & CA at D, E & F.
Then AD + BE + CF = $$\frac{1}{2}$$ × Perimeter of ΔABC