Geometry - Online Quiz

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 - Two Lines AB and CD intersect at O. If ∠AOC =50⁰,Then ∠ BOD and ∠AOD are respectively

q 17

A - 130⁰ ,50⁰

B - 50⁰, 130⁰

C - 60⁰, 120⁰

D - 40⁰, 140⁰

Answer : B

Explanation

∠ BOD=∠ AOC (vert . opp.  ∠s) =50⁰
∠AOC +∠AOD = 180⁰ ⇒ 50⁰ +∠AOD= 180⁰ ⇒∠AOD=130⁰
∴ ∠ BOD= 50⁰ and ∠AOD = 130⁰.

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 - If OE is the bisector of ∠AOD in the given figure ,then the value of X and y are respectively

q 25

A - 45⁰, 45⁰

B - 66⁰, 48⁰

C - 48⁰ ,66⁰

D - 30⁰, 60⁰

Answer : B

Explanation

∠AOC is a straight angle.
∴ 132⁰ + y⁰ = 180⁰ ⇒ y = (180 -  132 ) = 48⁰.
∠AOC = ∠BOC (vert. opp. ∠s) = 132⁰
∴ x= 1/2 ∠AOD = 1/2 * 132⁰ = 66⁰
∴ x= 66 and y = 48.

Q 4 - In a ∆ ABC, if 2∠A =3∠B =6∠C, Then ∠B= ?

A - 30⁰

B - 90⁰

C - 60⁰

D - 45⁰

Answer : C

Explanation

let 2∠ A = 3∠B = 6∠ C=ℏ. Then  ∠A = ℏ/2 , ∠ B = ℏ/3 and ∠ C =ℏ/6
But , ∠ A+∠B+∠C = 180⁰
∴  ℏ/2 + ℏ/3+ ℏ/6 = 180 ⇒ 3 ℏ+2 ℏ+ ℏ = 180*6 ⇒ 6 ℏ =180*6 ⇒ ℏ=180    ⇒ ∠B = 180/3 =60⁰

Q 5 - ∆ABC is right angled at A. If AB =24 mm and AC =7mm, BC=?

A - 31mm

B - 25mm

C - 30mm

D - 28mm

Answer : B

Explanation

By Pythagoras  theorem , we have 
 BC2 = AB2 + AC2    
= (24)2  + 72
= 576 + 49 = √625 ⇒ BC = 625 = 25mm.

Q 6 - Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12m , what is the distance between their tops?

A - 13 m

B - 14 m

C - 15 m

D - 12.8 m

Answer : A

Explanation

Let AB and CD be the poles such that
 AB = 6m , CD = 11 m and BD =12m 
Draw AE ⊥ CD . Then , AE = BD = 12m
CE = CD - DE = CD - AB = (11 - 6) m =5m.
from right   AEC we have
AC2 = AE2  + CE2 = (12)2 + 52 = (114 +25)=169
⇒ Ac =  √169 = 13m
∴ Distance between their tops= 13m

a 39

Answer : C

Explanation

The angle in a semi-circle is a right angle.

Q 8 - In the given figure ,ABCD is a cyclic quadrilateral in which AB || DC and ∠ BAD = 100⁰. Then , ∠ ABC=?

q 46

A - 80⁰

B - 100⁰

C - 75⁰

D - 150⁰

Answer : B

Explanation

AB   DC and AD  is the transversal.
∴ ∠ADC + ∠DAB=180⁰ ⇒ ADC =100⁰ =180⁰ ⇒ ADC=80⁰.
Opposite angles of a cyclic quadrilateral are supplementary.
∴ ∠ADC +∠ABC = 180⁰ ⇒ 80⁰+ ∠ABC =180⁰ ⇒ ABC = 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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