MCQ Questions on Geometry for Competitive Exams

41

In the figure, chords AB and CD of a circle intersect externally at P. If AB = 4 cm, CD = 11 cm and PD = 15 cm, then the length of PB is:
Geometry mcq question image

A) 10 cm B) 8 cm C) 14 cm D) 12 cm

Discuss

Answer & Solution

Answer: Option A

Solution:

Let, PA = x
PA × PB = PC × PD
x × (x + 4) = 4 × 15
x × (x + 4) = 60
x² + 4x - 60 = 0
x² + 10x - 6x - 60 = 0
x(x + 10) - 6(x + 10) = 0
x = 6
PB = x + 4 = 6 + 4 = 10 cm

42

In the given figure, if AC, DE are parallel and ∠CAB = 38°, then the value of ∠ABC + 5∠CBD is:
Geometry mcq question image

A) 158° B) 178° C) 218° D) 196°

Discuss

Answer & Solution

Answer: Option C

Solution:

If AC || BD,
2a + b + a = 180° . . . . . . (1)
2a = 38°
a = 19°
From equation (1)
2 × 19° + b + 19° = 180°
b = 123°
Now,
∠ABC + 5∠CBD
= 123° + 5 × 19°
= 218°

43

A, B and C are three points on the circle. If AB = AC = 7√2 cm and ∠BAC = 90° men the radius is equal to:

A) 14 cm B) 7 cm C) 7√2 cm D) 6 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

$${\text{R}} = \frac{{{\text{Hypotenuse}}}}{2} = \frac{{14}}{2} = 7$$

44

In the given figure, PQ is a diameter of the semicircle PABQ and O is its center. ∠AOB = 64°. BP cuts AQ at X. What is the value (in degrees) of ∠AXP?
Geometry mcq question image

A) 36 B) 32 C) 58 D) 54

Discuss

Answer & Solution

Answer: Option C

Solution:

∠AOB = 64°
Then,
∠BPA = 32°
∠PAQ = 90°
[∴ PQ is a diameter]
So, ∠PXA = 90 - 32 = 58°

45

In the given figure, AQ = 4√2 cm, QC = 6√2 cm and AB = 20 cm. If PQ is parallel to BC, then what is the value (in cm) of PB?
Geometry mcq question image

A) 8 B) 12 C) 6 D) 15

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{In }}\Delta APQ\,\& \,\Delta ABC \cr & \frac{{AP}}{{AB}} = \frac{{AQ}}{{AC}} \cr & \frac{{AP}}{{20}} = \frac{{4\sqrt 2 }}{{10\sqrt 2 }} \cr & AP = \frac{{20 \times 2}}{5} = 8{\text{ cm}} \cr & PB = AB - AP = 20 - 8 = 12{\text{ cm}} \cr} $$

46

Let ABC be an equilateral triangle and AD perpendicular to BC. Then AB² + BC² + CA² =?

A) 3AD² B) 5AD² C) 2AD² D) 4AD²

Discuss

Answer & Solution

Answer: Option D

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

47

In ΔABC, ∠A = 90°, AD is the bisector of ∠A meeting BC at D, and DE ⊥ AC at E. If AB = 10 cm and AC = 15 cm, then the length of DE, in cm, is:

A) 8 B) 6 C) 6.25 D) 7.5

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & CD:DB = 3:2 \cr & CD = 5\sqrt {13} \times \frac{3}{5} \cr & CD = 3\sqrt {13} \cr & \sin \theta = \frac{{3\sqrt {13} }}{{DE}} = \frac{{5\sqrt {13} }}{{10}} \cr & DE = 6{\text{ cm}} \cr} $$

48

In the given figure, AB = DB and AC = DC. If ∠ABD = 58° and ∠DBC = (2x - 4)°, ∠ACB = (y + 15)° and ∠DCB = 63° then the value of 2x + 5y is:
Geometry mcq question image

A) 325 B) 273 C) 259 D) 268

Discuss

Answer & Solution

Answer: Option B

Solution:

ΔABC ≅ ΔDBC
y + 15 = 63°
y = 48°
2x - 4 = 29°
2x = 33°
2x + 5y
= 33° + 5 × 48°
= 33° + 240°
= 273°

49

ΔABC and ΔDBC are on the same base BC but on opposite sides of it. AD and BC intersect each other at O. If AO = a cm, DO = b cm and the area of ΔABC = x cm², then what is the area (in cm²) of ΔDBC?

A) $$\frac{a}{b}x$$ B) $$\frac{{ab}}{2}x$$ C) $$\frac{{bx}}{a}$$ D) $$\frac{{\left( {a + b} \right)}}{2}x$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Area of ΔABC : Area of ΔBDC = a : b = ak : bk
$$\eqalign{ & ak = x \cr & k = \frac{x}{a} \cr & bk = b \times \frac{x}{a} = \frac{{bx}}{a} \cr} $$
Geometry mcq question image

50

Triangle ABC is similar to triangle PQR and AB : PQ= 2 : 3. AD is median to the side BC in triangle ABC and PS is the median to side QR in triangle PQR. What is the value of $${\left( {\frac{{{\text{BD}}}}{{{\text{QS}}}}} \right)^2}?$$

A) $$\frac{3}{5}$$ B) $$\frac{4}{9}$$ C) $$\frac{2}{3}$$ D) $$\frac{4}{7}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\frac{{AB}}{{PQ}} = \frac{2}{3}\left( {{\text{Given}}} \right)$$
Geometry mcq question image

$$\eqalign{ & \frac{{{\text{area of }}\Delta ABC}}{{{\text{area of }}\Delta PQR}} = {\left( {\frac{{AB}}{{PQ}}} \right)^2} = \frac{4}{9} \cr & \left[ {{\text{Ratio of similar }}\Delta } \right] \cr & \frac{{{\text{area of }}\Delta ABD}}{{{\text{area of }}\Delta PQS}} = \frac{{\frac{1}{2} \times {\text{area of }}\Delta ABC}}{{\frac{1}{2} \times {\text{area of }}\Delta PQR}} \cr & = \frac{4}{9} = \frac{{B{D^2}}}{{Q{S^2}}} \cr & \left[ {\therefore AD\,\& \,PS\,{\text{are median}}} \right] \cr} $$