MCQ Questions on Geometry for Competitive Exams

71

In a right triangle ABC, right angled at B, altitude BD is drawn to the hypotenuse AC of the triangle. If AD = 6 cm, CD = 5 cm, then find the value of AB² + BD² (in cm).

A) 30 B) 96 C) 36 D) 66

Discuss

Answer & Solution

Answer: Option B

Solution:

Since, AB² = AD × AC
AB² = 6 × 11 = 66 cm
And, BD² = AD × DC = 6 × 5 = 30
∴ AB² + BD² = 66 + 30 = 96

72

In the given figure, TB is a chord which passes through the centre of the circle. PT is a tangent to the circle at the point T on the circle. If PT = 10 cm, PA = 5 cm and AB = x cm, then the radius of the circle is:
Geometry mcq question image

A) 5√3 cm B) 6√5 cm C) 3√5 cm D) 10√3 cm

Discuss

Answer & Solution

Answer: Option A

Solution:

PA × PB = PT²
5(5 + x) = 100
x = 15
In right angle triangle PTB,
TB² = 20² - 10²
TB² = $$10\sqrt 3 $$
So, OT $$ = \frac{{10\sqrt 3 }}{2} = 5\sqrt 3 $$

73

A square is inscribed in a quarter-circle in such a manner that two of its adjacent vertices lie on the two radii at an equal distance from the centre, while the other two vertices lie on the circular arc. If the square has sides of length $$x$$. then the radius of the circle is:

A) $$\frac{{16x}}{{\pi + 4}}$$ B) $$\frac{{2x}}{{\sqrt x }}$$ C) $$\frac{{\sqrt 5 x}}{{\sqrt 2 }}$$ D) $$\sqrt 2 x$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Let ABCD is a square of x unit side
Geometry mcq question image

Then ∠AOD = 90°
Then OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
Diagonal of square ABCD = $$\sqrt 2 $$ x
Line MB || OD
i.e OD = MB = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
⇒ Then MBOD will be a rectangle become MB || OD, MB = OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$
BD || MO, MO = BD = $$\sqrt 2 $$ x
$${\text{R}} = \sqrt {{{\left( {\frac{{\text{x}}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\sqrt 2 {\text{x}}} \right)}^2}} = \frac{{\sqrt 5 {\text{x}}}}{{\sqrt 2 }}{\text{ Ans}}{\text{.}}$$

74

In ΔABC, ∠C = 90°, point P and Q are on the sides AC and BC, respectively, such that AP : PC = BQ : QC = 1 : 2. Then, $$\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}}$$ is equal to:

A) $$\frac{8}{3}$$ B) $$\frac{4}{3}$$ C) $$\frac{{13}}{9}$$ D) $$\frac{4}{9}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the side of BC & CA are 3x and 3y
QC = 2x
QB = x
PC = 2y
AP = y
In triangle AQC
AQ² = (3y)² + (2x)²
= 9y² + 4x²
BP² = (3x)² + (2y)²
= 9x² + 4y²
AB² = (3x)² + (3y)²
= 9x² + 9y²
= 9(x² + y²)
$${\text{Now, }}\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}} = \frac{{13\left( {{x^2} + {y^2}} \right)}}{{9\left( {{x^2} + {y^2}} \right)}} = \frac{{13}}{9}$$

75

In the given figure, PQRS is a cyclic quadrilateral. What is the measure of the angle PQR if PQ is parallel to SR?
Geometry mcq question image

A) 70° B) 110° C) 80° D) 100°

Discuss

Answer & Solution

Answer: Option B

Solution:

∠P + ∠R = 180° (∵ PQRS is a cyclic quadrilateral)
110° + ∠R = 180°
∠R = 70°
∠R + ∠Q = 180° (supplementary angle)
70° + ∠Q = 180°
∠PQR = 110°
Note:-
Geometry mcq question image

In any cyclic quadrilateral if two sides are parallel, then non-parallel sides are also equal in length. This type of quadrilateral is called an isosceles trapezium.
If AB || CD
Then AD = BC and ∠A = ∠B, ∠C = ∠D

76

AB is a diameter of the circle with centre O, CD is chord of the circle, If ∠BOC = 120°, then the value of ∠ADC is ?

A) 42° B) 30° C) 60° D) 35°

Discuss

Answer & Solution

Answer: Option B

Solution:

From figure ∠BOC = 120
∴ ∠AOC = 180 - 120 = 60
So, ∠ADC = $$\frac{1}{2}$$∠AOC
(Angle made on circumference is half of the angle made on centre)
= $$\frac{1}{2}$$ × 60
So, ∠ADC = 30°

77

PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 142°, then ∠OAB is equal to:

A) 58° B) 31° C) 71° D) 64°

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\angle {\text{OAB}} = \frac{{{{180}^ \circ } - {{38}^ \circ }}}{2} = {71^ \circ }$$

78

A circle is inscribed in the triangle ABC whose sides are given as AB = 10, BC = 8, CA = 12 units as shown in the figure. The value of AD × BF is:

A) 18 units B) 21 units C) 16 units D) 15 units

Discuss

Answer & Solution

Answer: Option B

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

79

In the following figure, AD bisects angle BAC. Find the length (in cm) of BD.
Geometry mcq question image

A) 4 B) 5 C) 9 D) 6

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & AD{\text{ is angle bisector of }}\angle A \cr & \therefore \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}} \cr & \Rightarrow \frac{6}{{2x - 3}} = \frac{{x - 2}}{x} \cr & \Rightarrow 6x = 2{x^2} - 4x - 3x + 6 \cr & \Rightarrow 2{x^2} - 13x + 6 = 0 \cr & \Rightarrow 2{x^2} - 12x - x + 6 = 0 \cr & \Rightarrow 2x\left( {x - 6} \right) - 1\left( {x - 6} \right) = 0 \cr & \Rightarrow \left( {x - 6} \right)\left( {2x - 1} \right) = 0 \cr & x - 6 = 0 \cr & x = 6 \cr & 2x - 1 = 0 \cr & x = \frac{1}{2}\left( {{\text{not valied}}} \right) \cr & \therefore BD = x - 2 = 6 - 2 = 4 \cr} $$

80

In the given figure, if $$\frac{{{\text{QR}}}}{{{\text{XY}}}} = \frac{{14}}{9}$$ and PY = 18 cm, then what is the value (in cm) of PQ?
Geometry mcq question image

A) 28 B) 18 C) 21 D) 24

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & \angle QXY = {120^ \circ } \cr & \angle PXY = {60^ \circ } \cr & \Delta PXY \sim \Delta PRQ \cr & \therefore \frac{{PY}}{{PQ}} = \frac{{XY}}{{QR}} \cr & \frac{{18}}{{PQ}} = \frac{9}{{14}} \cr & PQ = \frac{{18 \times 14}}{9} = 28\,{\text{cm}} \cr} $$