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 - 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.

Answer : A

Explanation

Two lines intersect at a point.

Q 3 - In the given figure , AB || CD, ∠BAE =110⁰ , ∠ECD = 120⁰ and ∠AEC =x⁰. Then, x= ?

q 28

A - 130⁰

B - 65⁰

C - 75⁰

D - 110⁰

Answer : A

Explanation

Draw FEG∥ AB ∥CD.
AB∥ EG and AE is the transversal.
∴ ∠BAE +∠AEG = 180⁰
⇒ 110⁰ + ∠AEG =180⁰ ⇒ ∠AEG =70⁰
Again, EG∥ CD and EC is transcersal.
∴ ∠GEC + ∠ ECD = 180⁰    ⇒ ∠GEC +120⁰ =180⁰ ⇒  ∠GEC= 60⁰
∴  X= 70+60 =130

a 28

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 ,∠A-∠B=33⁰ and∠ B -∠C = 18⁰ . Then∠ B =?

A - 35⁰

B - 55⁰

C - 45⁰

D - 57 ⁰

Answer : B

Explanation

∠ A- ∠B = 33⁰ and ∠B -∠C =18⁰
⇒ A= 33+ B and C=B -18
= (33+B) + B + (B-18) =180
⇒ 3B =165 ⇒ B 55.
 ∴  ∠B =55⁰.

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

Answer : C

Explanation

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

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 adjoining figure, ABCD is a rhombus whose diagonals intersect at O. IF ∠OAB =40⁰ and ∠ABO =x⁰, then X= ?

q 54

A - 50⁰

B - 35⁰

C - 40⁰

D - 45⁰

Answer : A

Explanation

We know that the diagonals of a  rhombus
bisect each other at right angle . So ,∠ AOB = 90⁰. 
Now ,∠ OAB + ∠ABO + ∠AOB = 180⁰
⇒ 40 +x + 90 = 180 ⇒ x=50.

a 54


aptitude_geometry.htm

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