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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 - Two Lines AB and CD intersect at O. If ∠AOC =50⁰,Then ∠ BOD and ∠AOD are respectively
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
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
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.
Answer : A
Explanation
(∠A+∠B) +(∠B+∠C) =(65⁰+140⁰)= 205⁰ ⇒ (∠A+∠B+∠C) +∠B =205⁰ ⇒ 180⁰ +∠B=205⁰ ⇒ ∠B =(205-180)⁰ =25⁰
Answer : B
Explanation
By Pythagoras theorem , we have BC2 = AB2 + AC2 = (24)2 + 72 = 576 + 49 = √625 ⇒ BC = 625 = 25mm.
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 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 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.
Answer : C
Explanation
∠ADC +∠ EDC = 180⁰ ⇒ ∠ADC + 120⁰ = 180⁰ ⇒ ∠ADC= 60⁰ ∠ABC = ∠ADC = 60⁰ ( s in the same segment).