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 ,∠A+∠B=65⁰ and∠B +∠C = 140⁰ . Then ∠B =?

A - 25⁰

B - 35⁰

C - 40⁰

D - 45⁰

Answer : A

Explanation

(∠A+∠B) +(∠B+∠C) =(65⁰+140⁰)= 205⁰
⇒ (∠A+∠B+∠C) +∠B =205⁰ ⇒ 180⁰ +∠B=205⁰
⇒ ∠B =(205-180)⁰ =25⁰ 

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

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 the given figure , O is the center of a circle and arc ABC subtends an angle of 130⁰ at O. AB is extended to P. Then ∠PBC= ?

q 48

A - 75⁰

B - 70⁰

C - 65⁰

D - 80⁰

Answer : C

Explanation

Take a point D on the remaining part of circumference of the circle. Join DA and DC
 ∠ADC = 1/2  ∠AOC = 1/2 *130⁰ = 65⁰.
Now DABC is a cyclic quadrilateral. 
&There4; ∠ADC + ∠ABC = 180⁰⇒ 65⁰+ ∠ABC = 180⁰ ⇒ ∠ABC = 115⁰.
⇒∠ PBC= (180⁰ - 115⁰) =65⁰.

Q 9 - In the given fig. PAB is a secant and PT is a tangent to the circle from P. If PT = 4cm and AB = xcm , Then PB = ?

q 51

A - 2.5 cm

B - 2.6 cm

C - 2.25 cm

D - 2.75 cm

Answer : C

Explanation

PA *PB = PT2 ⇒ 4*(4+x)=25 ⇒ 4=x = 25/4 = 6.25 ⇒ x= 2.25 cm.

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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