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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 - 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
Q 5 - The angle of a triangle are 3x⁰, (2x-7)⁰ and (4x-11)⁰. The value of x is :
Answer : A
Explanation
The sum of the angle of a triangle is 180⁰. ∴ 3x = 2x - 7 + 4x -11 = 180 ⇒ 9x =162 ⇒ x = 18. Hence, x = 18.
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.
Answer : C
Explanation
Opposite angles of a cyclic quadrilateral are supplementary. ∴ ∠A + ∠C = 180 ⁰⇒ 80⁰ + C =180⁰ ⇒ C = 100⁰.
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.
