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.
Answer : D
Explanation
The sum of all angle around a point is 360⁰ .
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
∠ 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 - 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 - The lengths of the diagonals of a rhombus are 24cm and 18cm respectively. The length of each side of the rhombus is
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.
