Geometry - Online Quiz


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Following quiz provides Multiple Choice Questions (MCQs) related to Geometry. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - The sum of all angles around a point is

A - 0⁰

B - 90⁰

C - 180⁰

D - 360⁰

Answer : D

Explanation

The sum of all angle around a point is 360⁰ .

Q 2 - The shortest distance between two intersecting lines is

A - 0

B - 1

C - 2

D - None of these

Answer : A

Explanation

The shortest distance between two intersecting lines is 0.

Q 3 - In the given figure L₁ || L₂ and ∠A = 65⁰. Then ∠C= ?

q 26

A - 110⁰

B - 115⁰

C - 120⁰

D - 135⁰

Answer : B

Explanation

∠B = ∠A = 65⁰ (corr. ∠s).
∴ ∠B +∠C= 180⁰ ⇒ 65⁰ +∠C = 180⁰ ⇒ ∠C = ( 180⁰ - 65⁰) = 115⁰.

Q 4 - The angles of a triangle are in the ratio 2:3:7. The measure OF the smallest angle is:

A - 90⁰

B - 60⁰

C - 45⁰

D - 30⁰

Answer : D

Explanation

let the angles be (2x)⁰,(3x)⁰ and (7x)⁰. Then,
2x+3x+7x =180 ⇒ 12x =180 ⇒ x=15
Smallest angle = (2x)⁰ = 30⁰

Q 5 - In a ∆ ABC, AB= BC,∠ B= x⁰ and ∠A = (2x-20)⁰. Then, ∠ B= ?

q 34

A - 30⁰

B - 40⁰

C - 44⁰

D - 64⁰

Answer : C

Explanation

AB = BC ⇒ ∠C = ∠A = (2x - 20)⁰. 
∠A+ ∠B + ∠C =180⁰ ⇒ (2x - 20) + x + (2x - 20 ) = 180
⇒ 5x - 40 =180 ⇒ 5x = 220 ⇒ x=44.
∴ ∠B = 44⁰.

Q 6 - A ladder is placed in such a way that its foot is 15m away from a wall and its top reaches a window 20m above the ground. The length of the ladder is:

A - 35m

B - 17.5m

C - 25 m

D - 18 m

Answer : C

Explanation

Let BC be the wall and AB be the ladder.
Then , BC = 20 m and AC =15m
∴ AB2= BC2 +AC2 = (20)2 + (15)2 = (400 + 225) = 625
⇒ AB = √625 = 25m.

a 40

Q 7 - The radius of a circle is 13cm and AB is a chord which is at a distance of 12cm from the center. The length of the ladder is:

A - 35 cm

B - 17.5 cm

C - 25 cm

D - 10 cm

Answer : D

Explanation

Let O be the  center of the circle and AB be the chord . Form  O, draw OL ⊥ AB. join OA.
Then, oA = 13 cm and OL = 12cm.
∴ AL2 = OA2 -OL2=(13)2 - (12)2= (169-144) =25.
=.> AL= √25 =5 cm
⇒ AB = 2 * AL =(2*5) cm = 10 cm.

a 41

Q 8 - In a cyclic quad. ABCD, ∠A=80⁰. Then ∠c =?

q 45

A - 80⁰

B - 160⁰

C - 100⁰

D - 120⁰

Answer : C

Explanation

Opposite angles of a cyclic quadrilateral are supplementary.
∴ ∠A + ∠C = 180 ⁰⇒ 80⁰ + C =180⁰ ⇒ C = 100⁰.

Q 9 - AB and CD are two parallel chords on the opposite sides of the center of the circle. If AB = 10cm , CD= 24cm and the radius of the circle is 13cm, the distance between the chords is

q 50

A - 17 cm

B - 15 cm

C - 16 cm

D - 18 cm

Answer : A

Explanation

From O draw OL⊥  AB and OM   CD. Join OA and OC.
AL = 1/2 AB = 5cm , OA = 13 cm.
OL2 = OA2 - AL2 = (13) 2 - 52 = (169 - 25) = 144 ⇒  OL = √144 = 12 cm.
Now ,CM =1/2 * CD =12 cm  and  OC =13cm.
∴ OM2 = OC2 - CM2 = (13) 2 - (12) 2 = (169 - 144) = 25
⇒ OM =√ 25 = 5cm.
∴ ML = OM + OL = (5+12 ) cm =17cm.

Q 10 - In the given figure, measure of ∠ ABC is

q 56

A - 20⁰

B - 40⁰

C - 60⁰

D - 80⁰

Answer : C

Explanation

∠ADC +∠ EDC = 180⁰ ⇒ ∠ADC + 120⁰ = 180⁰ ⇒ ∠ADC= 60⁰
∠ABC = ∠ADC = 60⁰  ( s in the same segment).

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