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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.
Answer : D
Explanation
The sum of all angle around a point is 360⁰ .
Q 2 - Two lines intersect
Answer : A
Explanation
Two lines intersect at a point.
Q 3 - In the given figure , AB || CD, ∠BAE =110⁰ , ∠ECD = 120⁰ and ∠AEC =x⁰. Then, x= ?
Answer : A
Explanation
Draw FEG∥ AB ∥CD. AB∥ EG and AE is the transversal. ∴ ∠BAE +∠AEG = 180⁰ ⇒ 110⁰ + ∠AEG =180⁰ ⇒ ∠AEG =70⁰ Again, EG∥ CD and EC is transcersal. ∴ ∠GEC + ∠ ECD = 180⁰ ⇒ ∠GEC +120⁰ =180⁰ ⇒ ∠GEC= 60⁰ ∴ X= 70+60 =130
Answer : C
Explanation
let 2∠ A = 3∠B = 6∠ C=ℏ. Then ∠A = ℏ/2 , ∠ B = ℏ/3 and ∠ C =ℏ/6 But , ∠ A+∠B+∠C = 180⁰ ∴ ℏ/2 + ℏ/3+ ℏ/6 = 180 ⇒ 3 ℏ+2 ℏ+ ℏ = 180*6 ⇒ 6 ℏ =180*6 ⇒ ℏ=180 ⇒ ∠B = 180/3 =60⁰
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 - A chord of length 30cm is at a distance of 8cm from the center of a circle. The radius of the circle is
Answer : A
Explanation
Let O be the centre of the circle and AB be the chord. Draw OL ⊥ AB. Then AL= 1/2 *AB = (1/2 *16)cm =8cm and OA = 10cm. OL2 = OA2 - AL2 = (10)2 - 82 = (100 -64 ) = 36. ⇒ OL = √36 = 6cm Required distance = 6 cm
Q 8 - In the given figure ,ABCD is a cyclic quadrilateral in which AB || DC and ∠ BAD = 100⁰. Then , ∠ ABC=?
Answer : B
Explanation
AB DC and AD is the transversal. ∴ ∠ADC + ∠DAB=180⁰ ⇒ ADC =100⁰ =180⁰ ⇒ ADC=80⁰. Opposite angles of a cyclic quadrilateral are supplementary. ∴ ∠ADC +∠ABC = 180⁰ ⇒ 80⁰+ ∠ABC =180⁰ ⇒ ABC = 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.
