MCQ Questions on Geometry for Competitive Exams

11

In ΔABD, C is the midpoint of BD. If AB = 10 cm, AD = 12 cm and AC = 9 cm, then BD = ?

A) $$\sqrt {10} {\text{ cm}}$$ B) $${\text{2}}\sqrt {41} {\text{ cm}}$$ C) $$\sqrt {41} {\text{ cm}}$$ D) $${\text{2}}\sqrt {10} {\text{ cm}}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & A{B^2} + A{D^2} = 2\left( {A{C^2} + B{C^2}} \right) \cr & 100 + 144 = 2\left( {81 + {x^2}} \right) \cr & \frac{{244}}{2} = 81 + {x^2} \cr & 122 - 81 = {x^2} \cr & {x^2} = 41 \cr & x = \sqrt {41} \cr & BD = 2\sqrt {41} \cr} $$

12

In ΔABC, the perpendiculars drawn from A, B and C meet the opposite sides at points D, E and F respectively. AD, BE and CF intersect at point P. If ∠EPD = 110° and the bisectors of ∠A and ∠B meet at point Q, then ∠AQB = ?

A) 115° B) 110° C) 135° D) 125°

Discuss

Answer & Solution

Answer: Option D

Solution:

∠EPD = 110°
∠PEC = ∠PDC = 90°
∴ ∠DCE = 180° - 110° = 70°
AQ and BQ angle bisector of ∠A and ∠B respectively.
$$\eqalign{ & \angle {\text{AQB}} = {90^ \circ } + \frac{{\angle {\text{DCE}}}}{2} \cr & = {90^ \circ } + \frac{{{{70}^ \circ }}}{2} \cr & = {125^ \circ } \cr} $$

13

The tangent at a point A of a circle with centre O intersects the diameter PQ of the circle (when extended) at the point B. If ∠BAP = 125°, then ∠AQP is equal to:

A) 50° B) 55° C) 60° D) 45°

Discuss

Answer & Solution

Answer: Option B

Solution:

∵ BAT is a straight line
∠TAP = ∠TAB - ∠BAP
∠TAP = 180° - 125° = 55°
∠AQP = ∠TAP = 55° (by alternate segment theorem)

14

Two circles touch each other at point X. A common tangent touch them at two distinct points Y and Z. If another tangent passing through X cut YZ at A and XA = 16 cm, then what is the value (in cm) of YZ?

A) 18 B) 24 C) 16 D) 32

Discuss

Answer & Solution

Answer: Option D

Solution:

XA = YA = AZ = 16 cm
YZ = AY + AZ
= 16 + 16
= 32 cm

15

ABCD is a cyclic quadrilateral. Side AB and DC, when produced, meet at E and sides AD and BC when produced, meet at F. If ∠ADC = 76° and ∠AED = 55°, then ∠AFB is equal to:

A) 26° B) 29° C) 34° D) 27°

Discuss

Answer & Solution

Answer: Option D

Solution:

∠ADC = 76° ; ∠AED = 55°
∠ADC + ∠ABC = 180°
[cyclic quadrilateral]
∠ABC = 180° - 76° = 104°
In ΔADE
∠A + ∠D + ∠E = 180°
∠A = 180° - 76° - 55° = 49°
In ΔABF
∠A + ∠B + ∠F = 180°
∠F = 180° - 104° - 49° = 27°

16

ΔABC, AB = 20 cm, BC = 7 cm and CA = 15 cm. Side BC is produced to D such that ΔDAB ∽ ΔDCA. DC is qual to:

A) 10 cm B) 9 cm C) 8 cm D) 7 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \Delta DAB \sim \Delta DCA \cr & \frac{{DA}}{{DC}} = \frac{{AB}}{{AC}} = \frac{{BD}}{{AD}} \cr & \frac{{DA}}{{DC}} = \frac{{20}}{{15}} = \frac{4}{3} \cr & AD = \frac{4}{3} \times DC\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{{20}}{{15}} = \frac{{BD}}{{AD}} = \frac{4}{3} \cr & AD = \frac{3}{4} \times BD{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{4}{3} \times DC = \frac{3}{4} \times BD \cr & \frac{{DC}}{{BD}} = \frac{9}{{16}} \cr & \frac{{DC}}{{BC + DC}} = \frac{9}{{16}} \cr & \frac{{DC}}{{7 + DC}} = \frac{9}{{16}} \cr & 16DC = 63 + 9DC \cr & 7DC = 63 \cr & DC = 9{\text{ cm}} \cr} $$

17

If the angle between two radii of a circle be 130°, then the angle between the tangents at the end of these radii (in degrees) is:

A) 50° B) 40° C) 70° D) 90°

Discuss

Answer & Solution

Answer: Option A

Solution:

In quadrilateral OABC
Geometry mcq question image

130° + 90° + 90° + ∠ABC = 360°
∠ABC = 50°

18

PQR is a triangle such that PQ = PR. RS and QT are the median to the sides PQ and PR respectively. If the medians RS and QT intersect at right angle, then what is the value of $${\left( {\frac{{{\text{PQ}}}}{{{\text{QR}}}}} \right)^2}?$$

A) $$\frac{3}{2}$$ B) $$\frac{5}{2}$$ C) $$2$$ D) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & PQ = PR \cr & QR = \sqrt {\frac{{P{Q^2} + P{R^2}}}{5}} \,\,\left( {{\text{By Theorem}}} \right) \cr & Q{R^2} = \frac{{2P{Q^2}}}{5} \cr & \frac{{P{Q^2}}}{{Q{R^2}}} = \frac{5}{2} \cr} $$

19

ΔABC is isosceles having AB = AC and ∠A = 40°. Bisectors PO and OQ of the exterior angles ∠ABD and ∠ACE formed by producing BC on both sides, meet at O. Then the value of ∠BOC is

A) 70° B) 110° C) 80° D) 55°

Discuss

Answer & Solution

Answer: Option A

Solution:

AB = AC
∴ ∠B = ∠C
∠B = $$\frac{{{{180}^ \circ } - {{40}^ \circ }}}{2}$$
∠B = 70°
∠B = ∠C = 70°
Geometry mcq question image

∠BOC = 90° - $$\frac{{\angle {\text{A}}}}{2}$$
= 90° - 20°
= 70°

20

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If the area of triangle ABC is 136 cm², then the area of triangle BDE is equal to:

A) 36 cm² B) 34 cm² C) 38 cm² D) 24 cm²

Discuss

Answer & Solution

Answer: Option B

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