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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.

Q 1 - In the given figure, straight line AB and CD intersect at O. IF ∠δ =3∠v, then ∠v = ?

q 18

Answer : B

Explanation

COD is a  straight line 
∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.

Q 2 - The shortest distance between two intersecting lines is

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

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 the given figure , AB ll CD, ∠ABE =120⁰, ∠DCE = 100⁰ and ∠BEC =x⁰. Then, x= ?

q 29

Answer : C

Explanation

Through E draw GEH ∥ AB ∥CD
AB∥ EG and BE is the transversal.
∠ ABE +∠ GEB = 180⁰ ⇒ 120⁰ +∠GEB =180⁰ ⇒ ∠GEB = 60⁰
CD ∥EH and CE is the transversal. 
∴∠DCE +∠CEH = 180⁰  ⇒ 100⁰ + ∠CEH =180⁰  ⇒ CEH = 80⁰
NOW ∠GEB+ ∠BEC +∠CEH = 180⁰ ⇒ 60+x+80 =180 ⇒ x = 40

a 29

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

Answer : B

Explanation

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

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:

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 angle in a semi circle is

Answer : C

Explanation

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

Q 8 - In the given figure , POQ is a diameter and PQRS is a cyclic quadrilateral. If ∠PSR =130⁰, Then ∠ RPQ =?

q 47

Answer : A

Explanation

PQRS is a cyclic quadrilateral.
∠PSR + ∠PQR = 180⁰ ⇒ 130⁰ + ∠PQR =180⁰⇒∠ PQR=50⁰.
Also PRQ = 90⁰ (angle in a semi- circle)
In PQR we have
∠PQR + ∠PRQ + ∠RPQ = 180⁰⇒ 50⁰ +90⁰+∠RPQ =180⁰ ⇒ ∠RPQ = 40⁰.

Q 9 - In the given figure, chords AB and CD of a circle intersect externally at P. If AB =6cm, CD = 3cm and PD= 5cm then PB= ?

q 52

Answer : D

Explanation

PA * PB + PC *PD ⇒ (x+6 ) * x=8* 5 ⇒ x² +6x - 40 =0
⇒ (x+10) (x-4) =0 ⇒ x=4
∴ PB= 4 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

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

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