MCQ Questions on Geometry for Competitive Exams

31

Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 24 cm, then what is the distance between their centres?

A) 12 cm B) 18 cm C) 16 cm D) 14 cm

Discuss

Answer & Solution

Answer: Option D

Solution:

In ΔADO
OD = $$\sqrt {{{\left( {15} \right)}^2} - {{\left( {12} \right)}^2}} $$ = 9 cm
In ΔADC
DC = $$\sqrt {{{\left( {13} \right)}^2} - {{\left( {12} \right)}^2}} $$ = 5 cm
or use Triplet
5, 12, 13 and 9, 12, 15
∴ OC = DC + OD = 5 + 9 = 14 cm

32

In the given figure, SX is tangent. SX = OX = OR. If QX = 3 cm and PQ = 9 cm. Then what is the value (in cm) of OS ?
Geometry mcq question image

A) 6 B) 5 C) 4 D) 3

Discuss

Answer & Solution

Answer: Option D

Solution:

In the given circle,
SX is a tangent
SX² = (XQ) × (PX)
SX² = 3 × [PQ+ QX]
SX² = 3 × [3 + 9]
SX² = 3 × 12
SX² = 36
SX = 6 cm
RO = 6 = OX = SX
∴ OQ = OX - QX and
PO = PQ - OQ
OQ = 6 - 3 = 3 cm and
PO = 9 - 3 = 6 cm
PQ and RS Intersects at O,
∴ PO × OQ = RO × OS
⇒ 6 × 3 = 6 × OS
⇒ OS = 3 cm

33

In a circle of radius 10 cm, with centre O, PQ and PR are two chords each of length 12 cm. PO intersects chord QR at the points S. The length of OS is:

A) 2.8 cm B) 2.5 cm C) 3.2 cm D) 3 cm

Discuss

Answer & Solution

Answer: Option A

Solution:

QS² = 12² - (10 - x)² = 10² - x²
12² - 10² = (10 - x)² - x²
22 × 2 = 10(10 - 2x)
4.4 = 10 - 2x
2x = 5.6
x = 2.8 cm

34

ABCD is a quadrilateral such that ∠D = 90°. A circle with centre O touches the sides AB, BC, CD and DA at P, Q, R and S, respectively. If BC = 40 cm, BP = 28 cm and CD = 25 cm, then what is the radius of the circle?
Geometry mcq question image

A) 12 cm B) 13 cm C) 5 cm D) 8 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

BC = 40, BP = 28, CD 25 (given)
BQ = BP = 28
(Tangent from an external point to a circle are equal)
CQ = 40 - 28 = 12
RC = CQ = 12
(Tangent from an external point to a circle are equal)
DR = 25 - 12 = 13
Join the line OR & OS (Both are radius)
∠DSO = ∠DRO = 90°
In quadrilateral DROS
∠ROS = 360° - (90° + 90° + 90°) = 90°
Hence, DROS is a square.
Hence radius of circle = 13
∠SDR = 90° (Given)
∴ ∠ROS = 90°
Hence DROS is a square.

35

In ΔABC, Perimeter of ΔABD = Perimeter of ΔBCD and AB = 60 cm, BC = 80 cm and AC = 100 cm. Then find BD.
Geometry mcq question image

A) 20 cm B) 20√5 cm C) 24√5 cm D) None of these

Discuss

Answer & Solution

Answer: Option C

Solution:

Let AD = x and BD = y
Perimeter of ΔABD = Perimeter of ΔBCD
60 + x + y = 80 + y + 100 - x
2x = 120
x = 60
CD = 100 - 60 = 40
Let ∠ACB = θ
In ΔABC,
cosθ $$ = \frac{{80}}{{100}} = \frac{4}{5}$$
In ΔCBD, by using cosine formula we get
$$\eqalign{ & \cos \theta = \frac{{{\text{B}}{{\text{C}}^2} + {\text{C}}{{\text{D}}^2} - {\text{B}}{{\text{D}}^2}}}{{2{\text{BC}}.{\text{CD}}}} \cr & \frac{4}{5} = \frac{{{{80}^2} + {{40}^2} - {{\text{y}}^2}}}{{2 \times 40 \times 80}} \cr & \frac{4}{5} \times 80 \times 80 = 6400 + 1600 - {{\text{y}}^2} \cr & 5120 = 8000 - {{\text{y}}^2} \cr & {{\text{y}}^2} = 2880 \cr & {\text{y}} = 24\sqrt 5 {\text{ cm}} \cr} $$

36

The lengths of the two sides forming the right angle of a right-angled triangle are 21 cm and 20 cm. What is the radius of the circle circumscribing the triangle?

A) 14.5 cm B) 14 cm C) 12 cm D) 15.5 cm

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & AC = \sqrt {{{21}^2} + {{20}^2}} \cr & = \sqrt {441 + 400} \cr & = \sqrt {841} \cr & = 29 \cr & R = \frac{{AC}}{2} = \frac{{29}}{2} = 14.5{\text{ cm}} \cr} $$

37

Two sides of a triangle are of length 3 cm and 8 cm. If the length of the third side is '$$x$$' cm, then:

A) 5 < $$x$$ B) 5 < $$x$$ < 11 C) 0 < $$x$$ < 11 D) $$x$$ > 11

Discuss

Answer & Solution

Answer: Option B

Solution:

8 - 3 < $$x$$ < 8 + 3
5 < x < 11

38

In a triangle PQR, ∠PQR = 90°, PQ = 10 cm and PR = 26 cm, then what is the value (in cm) of inradius of incircle?

A) 9 B) 4 C) 8 D) 6

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & QR = \sqrt {{{\left( {26} \right)}^2} - {{\left( {10} \right)}^2}} \cr & = \sqrt {676 - 100} \cr & = \sqrt {576} \cr & = 24 \cr & {\text{Incirde radius in a right angled triangle }}\left( r \right) \cr & = \frac{{{\text{P}} + {\text{B}} - {\text{H}}}}{2} \cr & = \frac{{10 + 24 - 26}}{2} \cr & = \frac{8}{2} \cr & = 4{\text{ cm}} \cr} $$

39

The circles of same radius 13 cm intersect each other at A and B. If AB = 10 cm, then the distance between their centres

A) 18 cm B) 12 cm C) 24 cm D) 26 cm

Discuss

Answer & Solution

Answer: Option C

Solution:

In right angle ΔAOC,
OC = $$\sqrt {{{13}^2} - {5^2}} $$ = 12
OO' = 12 + 12 = 24 cm

40

In ΔABC, D is a point on side BC such that ∠ADC = ∠BAC. If CA = 12 cm, CD = 8 cm, then CB (in cm) = ?

A) 18 B) 12 C) 15 D) 10

Discuss

Answer & Solution

Answer: Option A

Solution:

In ΔABC,
∠BAC = ∠ADC, (given)
In, ΔABC & ΔADC,
∠BAC = ∠ADC (given)
∠C = ∠C (common)
$$\eqalign{ & \therefore \Delta {\text{ABC}} \sim \Delta {\text{DAC}}\,\,\,\left( {{\text{A}} - {\text{A}}\,{\text{theorem}}} \right) \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} = \frac{{{\text{AC}}}}{{{\text{DC}}}} \cr & \Rightarrow \frac{{{\text{BC}}}}{{12}} = \frac{{12}}{8} \cr & \Rightarrow {\text{BC}} = \frac{{144}}{8} = 18{\text{ cm}} \cr} $$