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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 = ?
Answer : B
Explanation
COD is a straight line ∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.
Q 2 - Two lines intersect
Answer : A
Explanation
Two lines intersect at a point.
Q 3 - In the given figure , AB || CD, ∠ABE =35⁰, ∠CDE = 65⁰ and ∠BED =x⁰. Then, x= ?
Answer : B
Explanation
Draw GEH ||AB||CD. ∠ BHE =∠ ABE = 35⁰ (alt .∠s) ∠ DEH =∠ CDE = 65⁰ (alt .∠s) ∴∠x=∠ BEH + ∠DEH = (35⁰ +65⁰)=100⁰.
Q 4 - The angles of a triangle are in the ratio 2:3:7. The measure OF the smallest angle is:
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
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 - 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?
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
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 =?
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 fig. PAB is a secant and PT is a tangent to the circle from P. If PT = 4cm and AB = xcm , Then PB = ?
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= ?
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.
