- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude
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 , ∠POS = 90⁰. What Is the measure of ∠ROQ?
Answer : C
Explanation
∠ROQ = ∠POS (vert. opp. ∠s) = 90⁰.
Q 2 - Two lines intersect
Answer : A
Explanation
Two lines intersect at a point.
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= ?
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
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.
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:
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.
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= ?
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 figure, AOB is a diameter of the circle and CD || AB. If ∠DAB = 25⁰ ,Then ∠CAD=?
Answer : B
Explanation
AB DC and AC is a transversal. ∴ ∠ACD = ∠CAB = 25⁰ (alt. s ) ∠ACB = 90⁰ ( angle in a semicircle) ∴ ∠BCD =∠ACB + ∠ ACD=(90⁰ +25⁰)= 115⁰. ∠BAD + ∠BCD = 180⁰ ⇒ ∠BAC +∠CAD +∠BCD = 180⁰ ⇒ 25⁰ +∠ CAD + 115⁰ =180⁰ ⇒ ∠CAD = 40⁰
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.
