- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude
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
x+y = 180 ⇒ x+ 4/5 x = 180 ⇒ 5x +4x = 900 ⇒ 9x=900 ⇒ x = 100.
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.
Q 4 - In the given figure , AB ll CD, ∠ABE =120⁰, ∠DCE = 100⁰ and ∠BEC =x⁰. Then, x= ?
Answer : C
Explanation
Through E draw GEH ∥ AB ∥CD AB∥ EG and BE is the transversal. ∠ ABE +∠ GEB = 180⁰ ⇒ 120⁰ +∠GEB =180⁰ ⇒ ∠GEB = 60⁰ CD ∥EH and CE is the transversal. ∴∠DCE +∠CEH = 180⁰ ⇒ 100⁰ + ∠CEH =180⁰ ⇒ CEH = 80⁰ NOW ∠GEB+ ∠BEC +∠CEH = 180⁰ ⇒ 60+x+80 =180 ⇒ x = 40
Answer : B
Explanation
By Pythagoras theorem , we have BC2 = AB2 + AC2 = (24)2 + 72 = 576 + 49 = √625 ⇒ BC = 625 = 25mm.
Q 6 - The perimeters of two similar triangles ∆ ABC and∆ PQR are 36cm and 24 cm . If PQ= 10cm, Then AB=?
Answer : C
Explanation
we have AB/PQ = 36/24 ⇒ AB / 10 = 3/2 ⇒ AB = (3/2 * 10) = 15 cm.
Q 7 - The angle in a semi circle is
Answer : C
Explanation
The angle in a semi-circle is a right angle.
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, chords AB and CD of a circle intersect externally at P. If AB =6cm, CD = 3cm and PD= 5cm then PB= ?
Answer : D
Explanation
PA * PB + PC *PD ⇒ (x+6 ) * x=8* 5 ⇒ x2 +6x - 40 =0 ⇒ (x+10) (x-4) =0 ⇒ x=4 ∴ PB= 4 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.
