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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 - In the given figure , AB || CD, ∠ABE =35⁰, ∠CDE = 65⁰ and ∠BED =x⁰. Then, x= ?
Answer : B
Explanation
Draw GEH ||AB||CD. ∠ BHE =∠ ABE = 35⁰ (alt .∠s) ∠ DEH =∠ CDE = 65⁰ (alt .∠s) ∴∠x=∠ BEH + ∠DEH = (35⁰ +65⁰)=100⁰.
Q 4 - The angles of a triangle are in the ratio 2:3:7. The measure OF the smallest angle is:
Answer : D
Explanation
let the angles be (2x)⁰,(3x)⁰ and (7x)⁰. Then, 2x+3x+7x =180 ⇒ 12x =180 ⇒ x=15 Smallest angle = (2x)⁰ = 30⁰
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 - If ∆ ABC is an isosceles triangle with ∠ C = 90⁰ and AC = 5cm , Then AB =?
Answer : D
Explanation
Clearly BC =AC=5cm. AB2 = AC2+ BC2 =52 +52= 50 ⇒ AB = √50 = 5√2 cm.
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 : D
Explanation
Let O be the centre of the circle and AB be the chord. Draw OL ⊥ AB. Then AL= 1/2 AB = (1/2 *30)cm =15 cm and OL = 8cm. OA2 = OL2 +AL2= 82 + (15)2 = (64 + 225 ) =289 ⇒ OA = √289 = 17cm. ∴ Radius of the circle is 17 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 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.
Q 10 - In the adjoining figure,ABCD is a rhombus. If ∠A=70⁰ then ∠CDE =?
Answer : B
Explanation
Let CDB= x⁰. then , CD = CB ⇒ ∠CBD = ∠CDB = x⁰. ∠ BCD = ∠BAD = 70⁰ (opp. s of a rhombus) ∴ x+x + 70 = 180 (sum of the ∠ s of a ∆ is 180⁰) ⇒ 2x = 110 ⇒ x=55. ∴ ∠CDB= 55⁰.
