MCQ Questions on Geometry for Competitive Exams

11

Number of circles that can be drawn through three non-collinear points are

A) exactly one B) two C) three D) more than three

Discuss

Answer & Solution

Answer: Option A

Solution:

Only one (1) circle can be drawn through three non-collinear points.
Geometry mcq question image

12

The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. The area of the field is?

A) 252 m² B) 1152 m² C) 96 m² D) 156 m²

Discuss

Answer & Solution

Answer: Option A

Solution:

Let ABCD is quadrilateral and its BD diagonal
BD = 24 metres
And, AM = 8 metres
CN = 13 metres
Geometry mcq question image

$$\eqalign{ & {\text{Area of }}\square ABCD = {\text{ar}}\left( {\Delta ABD} \right) + {\text{ar}}\left( {\Delta BCD} \right) \cr & = \left( {\frac{1}{2} \times BD \times AM} \right) + \left( {\frac{1}{2} \times BD \times CN} \right) \cr & = \frac{1}{2} \times BD\left[ {AM + CN} \right] \cr & = \frac{1}{2} \times 24\left[ {8 + 13} \right] \cr & = 12 \times 21 \cr & {\text{Area of }}\square ABCD = 252{\text{ metr}}{{\text{e}}^2} \cr} $$

13

The exterior angle of a triangle is 115° and the corresponding interior opposite angles are in the ratio 2 : 3. The measure of greatest angle of the triangle is:

A) 70° B) 65° C) 69° D) 79°

Discuss

Answer & Solution

Answer: Option C

Solution:

5x = 115° (by exterior angle theorem)
x = 23°
2x = 46°
3x = 69°
∠ABC = 65°
The greatest angle = 69°

14

For an equilateral triangle, the ratio of the in-radius and the outer-radius is

A) 1 : 2 B) 1 : 3 C) 1 : √2 D) 1 : √3

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the side of equilateral ΔABC be a & r = in-radius & R = outer radius
Geometry mcq question image

$$\eqalign{ & r = \frac{a}{{2\sqrt 3 }},\,\,\,R = \frac{a}{{\sqrt 3 }} \cr & r:R = \frac{a}{{2\sqrt 3 }}:\frac{a}{{\sqrt 3 }} = 1:2 \cr} $$

15

If in the following figure (not to the scale), ∠ACB = 135° and the radius of the circle is 2√2 cm, then the length of the chord AB is
Geometry mcq question image

A) 3√2 cm B) 4√2 cm C) 4 cm D) 6 cm

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Hence }}A{B^2} = {\left( {2\sqrt 2 } \right)^2} + {\left( {2\sqrt 2 } \right)^2} \cr & A{B^2} = 8 + 8 \cr & A{B^2} = 16 \cr & AB = \sqrt {16} \cr & AB = 4 \cr} $$

16

A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the circumference is

A) 80° B) 90° C) 60° D) 30°

Discuss

Answer & Solution

Answer: Option D

Solution:

∵ OA = AB = OB (given)
Geometry mcq question image

∴ ΔAOB is equilateral triangle
So,
∠AOB = ∠OAB = ∠ABO = 60°
∵ ∠ACB = $$\frac{1}{2}$$ ∠AOB
= $$\frac{1}{2}$$ × 60°
= 30°

17

AB is a common tangent to both the circles in the given figure. Find the distance (correct to two decimal places) between the centres of the two circles.

A) 18.98 units B) 23.58 units C) 26.59 units D) 21.62 units

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \Rightarrow {\text{By figure,}} \cr & \Delta CAE \sim \Delta DBE \cr & \frac{{CA}}{{BD}} = \frac{{AE}}{{BE}} \cr & \frac{5}{x} = \frac{8}{{12}} \cr & x = 7.5 = \left( {\frac{{15}}{2}} \right) \cr & \Rightarrow {\text{Length of T}}{\text{.L}} = \sqrt {{d^2} - {{\left( {{R_1} + {R_2}} \right)}^2}} \cr & {\left( {20} \right)^2} = {d^2} - {\left( {5 + 7.5} \right)^2} \cr & 400 = {d^2} - 156.25 \cr & {d^2} = 400 + 156.25 \cr & {d^2} = 556.25 \cr & d = \sqrt {556.25} \cr & d = 23.58 \cr} $$

18

If two circles do not touch or intersect each other and one does not lie inside the other, then find the number of common tangents.

A) 5 B) 4 C) 2 D) 3

Discuss

Answer & Solution

Answer: Option B

Solution:

4 tangents

19

If $$a$$ and $$b$$ are the lengths of the sides of a right angled triangle whose hypotenuse is 10 and whose area is 20, then the value of ($$a$$ + $$b$$)² is

A) 140 B) 120 C) 180 D) 160

Discuss

Answer & Solution

Answer: Option C

Solution:

In right ΔABC,
Geometry mcq question image

a² + b² = 10² (by pt) . . . . (i)
Area ΔABC = $$\frac{1}{2}$$ ab = 20
ab = 40
(a + b)² = a² + b² + 2ab
= 10² + 2(40)
= 180

20

In the given figure, ABC is an equilateral triangle. Two circles of radius 4 cm and 12 cm are inscribed in the triangle. What is the side (in cm) of an equilateral triangle?
Geometry mcq question image

A) $$\frac{{32}}{{\sqrt 3 }}$$ B) $$24\sqrt 3 $$ C) $$\frac{{64}}{{\sqrt 3 }}$$ D) $$64\sqrt 3 $$

Discuss

Answer & Solution

Answer: Option B

Solution:

Now, In ΔAOF,
∠AFO = 90°
∠OAF = 30°
Geometry mcq question image

⇒ AF = $$4\sqrt 3 $$ cm
⇒ AO = 8 cm
and AE = AO + OD + DO' + O'E
= 8 + 4 + 12 + 12
= 36 cm
⇒ Median = 36 cm
In equilateral Δ,
$$\eqalign{ & {\text{Median}} = \frac{{\sqrt 3 }}{2} \times {\text{Side}} \cr & {\text{36}} = \frac{{\sqrt 3 }}{2} \times {\text{Side}} \cr & {\text{Side}} = \frac{{72}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = 24\sqrt 3 {\text{ cm}} \cr} $$