MCQ Questions on Geometry for Competitive Exams

31

What can be the maximum number of common tangent which can be drawn to two non-intersecting circles?

A) 2 B) 4 C) 3 D) 6

Discuss

Answer & Solution

Answer: Option B

Solution:

Maximum 4 tangents can be drawn to two non-intersecting circles.

32

Astha cuts a triangle out of a cardboard and tries to balance the triangle horizontally at the tip of her finger. On what point will she be able to balance the shape for any kind of triangle?

A) Incentre B) Circumcentre C) Centroid D) Orthocentre

Discuss

Answer & Solution

Answer: Option C

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

33

In an isosceles triangle ABC with AB = AC and AD is perpendicular to BC. If AD = 6 cm and the perimeter of ΔABC is 36 cm, then the area of ΔABC is:

A) 45 cm² B) 48 cm² C) 54 cm² D) 64 cm²

Discuss

Answer & Solution

Answer: Option B

Solution:

Perimeter of triangle ABC = 36 cm
By Pythagorean triplets 6, 8, 10
Now,
a = 8 cm
x = 10 cm
Area of triangle ABC = $$\frac{1}{2}$$ × 16 × 6 = 48 cm²

34

Two circles of radii 15 cm and 10 cm intersect each other and the length of their common chord is 16 cm. What is the distance (in cm) between their centres?

A) $$15 + 2\sqrt {161} $$ B) $$12 + 3\sqrt 7 $$ C) $$6 + \sqrt {161} $$ D) $$10 + \sqrt {161} $$

Discuss

Answer & Solution

Answer: Option C

Solution:

So, OO' = $$6 + \sqrt {161} $$

35

The lengths of two sides of a parallelogram are 3 cm and 10 cm. What is the sum of the squares of the diagonals of the parallelogram?

A) 218 cm² B) 109 cm² C) 169 cm² D) 206 cm²

Discuss

Answer & Solution

Answer: Option A

Solution:

AC² + BD² = 2(AB² + BC²)
AC² + BD² = 2(10² + 3²) = 218

36

In the given figure, PQR is a triangle in which angle P : angle Q : angle R = 3 : 2 : 1, and PR is perpendicular to RS. What will be the measure of angle TRS?

A) 60° B) 45° C) 50° D) 30°

Discuss

Answer & Solution

Answer: Option A

Solution:

P : Q : R = 3 : 2 : 1
3 + 2 + 1 = 6u → 180°
1u → 30°
P = 90° ∠PRS = 90°
Q = 60°
R = 30°
∠P + ∠Q = ∠PRT
90° + 60° = ∠PRS + ∠TRS
150° = 90° + ∠TRS
∠TRS = 60°

37

Two circles touch each other at point X. Two common tangents of the circles meet at point P and none of the tangents passes through X. These tangents touch the larger circle at points B and C. If the radius of the larger circles 15 cm and CP = 20 cm, then what is the radius (in cm) of the smaller circle?

A) 3.5 B) 3.75 C) 4.25 D) 4.45

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{In }}\Delta {O^1}CP \cr & {O^1}P = \sqrt {{{20}^2} + {{15}^2}} = 25{\text{ cm}} \cr & OP = 25 - 15 - r = 10 - r \cr & {\text{In }}\Delta OPD\,\& \,\Delta {O^1}CP \cr & \frac{{OP}}{{{O^1}P}} = \frac{{OD}}{{{O^1}C}} \cr & \frac{{10 - r}}{{25}} = \frac{r}{{15}} \cr & 150 - 15r = 25r \cr & r = \frac{{150}}{{40}} \cr & r = 3.75{\text{ cm}} \cr} $$

38

XY and XZ are tangents to a circle. ST is another tangent to the circle at the point R on the circle which intersects XY and XZ at S and T respectively, If XY = 9 cm and TX = 15 cm, then RT is:

A) 4.5 cm B) 3 cm C) 7.5 cm D) 6 cm

Discuss

Answer & Solution

Answer: Option D

Solution:

XY = 9 cm, TX = 15 cm
⇒ We know
Length of tangents drawn from a point to the circle are equal
Therefore
XY = XZ = 9 cm, TZ = RT
TX = 15 cm
XZ + ZT = 15
ZT = 15 - 9 = 6
RT = ZT = 6 cm

39

Each of the circles of equal radii with centres A and B pass through the centre of one another. They cut at C and D then ∠DBC is equal to?

A) 60° B) 100° C) 120° D) 140°

Discuss

Answer & Solution

Answer: Option C

Solution:

According to question
Geometry mcq question image

AB = AD = DB = r
∴ ΔADB is a equilateral triangle
∠DBA = 60°
Similar in ΔABC
∠ABC = 60°
∠DBC = 60° + 60°
∴ ∠DBC = 120°

40

In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?

A) 54° B) 81° C) 74° D) 72°

Discuss

Answer & Solution

Answer: Option D

Solution:

Let ∠DAC = x
∠ADB = 2x
∠ACD + ∠DAC = ∠ADB
∠ACD + x = 2x
∠ACD = 2x - x = x
In ΔABC
∠A + ∠B + ∠C = 180°
70° + 56° + ∠C = 180°
∠C = 54°
∠ADC = 180° - 2x
= 180° - 2 × 54°
= 180° - 108°
= 72°