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

q 29

A - 60⁰

B - 50⁰

C - 40⁰

D - 70⁰

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

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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 - If ∆ ABC is an isosceles triangle with ∠ C = 90⁰ and AC = 5cm , Then AB =?

A - 2.5 cm

B - 5 cm

C - 10cm

D - 5√2 cm

Answer : D

Explanation

Clearly BC =AC=5cm.
AB2 = AC2+ BC2 =52 +52= 50 ⇒ AB  = √50 = 5√2 cm.

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Q 7 - A chord of length 30cm is at a distance of 8cm from the center of a circle. The radius of the circle is

A - 6 cm

B - 9 cm

C - 12 cm

D - 8 cm

Answer : A

Explanation

Let O be the centre of the circle and AB  be the chord. Draw OL ⊥  AB.
Then AL= 1/2 *AB = (1/2 *16)cm =8cm  and OA = 10cm.
OL2  = OA2 - AL2 = (10)2 - 82 = (100 -64 ) = 36.
⇒ OL = √36 = 6cm
Required distance = 6 cm

a 43

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 - The lengths of the diagonals of a rhombus are 24cm and 18cm respectively. The length of each side of the rhombus is

A - 12 cm

B - 9 cm

C - 15 cm

D - 8 cm

Answer : C

Explanation

Let ABCD be a rhombus in which diagonal AC=24 cm and diagonal BD =18 cm . We know that the diagonal of a   rhombus bisect each other at right angle.
∴ OA = 1/2 AC =(1/2 *24 ) cm =12cm
OB = 1/2 BD = (1/2 *18 ) cm =9cm 
AB2 = OA2 + OB2 = (12) 2 +  92 = (144 +81) = 225 
⇒ AB = √225 = 15 cm.
∴ Each side of the rhombus is 15 cm.

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