Geometry Multiple Choice Questions and Answers for SSC EXAMS

 Question 1 

In a triangle ABC, AD is perpendicular to BC and AD² = BD. Dc. The measure of angle BAC is ? 

a) 60° b) 75° c)90° d) 45°

Solution

In Right Angle Triangle

Median AD² = BD.DC
hence it is Right Angle Triangle

Angle BAC = 90°

Question  2

If the perimeter of a right angle triangle is 144 cm and its circumradius is 32.5cm. Find its area. ?

A) 504 cm² b ) 520 cm² c) 512cm² d) none of these

Perimeter = 144
Then
Semiperimeter = 72
Circumradius ( R ) = 32.5
Then 2R = 65

Area of Right Angle Triangle = S( S - 2R)

Area = 72 ( 72 - 65 ) = 504 cm²

Question 3

The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its in-circle is 6 cm. Find the sides of the triangle?

a) 30,40,41 b) 18,24,30 c) 30,24,25 d) 24,36,20

If circumradius ( R ) is 15 cm then
2R or hypotenuse = 30

Incenter ( r ) = `frac{ perpendicular + base - hypotenuse }2`

6 = `frac{ perpendicular + base - 30} 2`

perpendicular + base = 42

Option b is right because 18 + 24 = 42

Question 4

In a triangle ABC angle B = 90°, if points D and E are on side BC such that BD= DE= EC. then which of the following is true?

a) 5AE² = 2AC²+ 3AD² b) 8AE² = 5AC² +3AD² c) 8AE² = 3AC² +5AD² d) 5AE² = 2AC²+ 2AD

Solution 

Let 

BD = DE = EC = 1

and 

AB = 1

Then AD, AE and AC will be `\sqrt2 , \sqrt5 and \sqrt10` respectively ( pythagoras theorem )

Now we have to put values in given expression then we get Option C as a right answer.

Question 5

dummy If the inradius and circumradius of a right angle triangle are 3cm and 10cm respectively then find the area of the triangle?

a) 75 cm² b) 65 cm² c) 69cm² d) 56 cm²

Now we have to put values in given expression then we get Option C as a right answer. Area of Right Angle Triangle = r² + 2Rr

Area = ( 3 )² + 2 ( 10 )( 3 ) = 69 cm²

Question 6

In triangle ABC, angle C = 90° and CD is perpendicular to AB at D. If AD/BD = `\sqrtk`, then AC/BC = ?

a) k b) `\sqrtk` c) `k^\frac{1}4` d) `\frac1{\sqrt k}`

Solution

Here AD = `\sqrtk` and BD = 1

BC² = BD × AB
BC² = 1 ( `\sqrtk`+ 1 )

and

AC² = AD × AB
AC² = `\sqrtk ( \sqrtk` + 1 )
`frac{AC}{BC}  =  k^\frac{1}4`

Question 7

In Triangle ABC, angle BAC 90° and AD is drawn perpendicular to BC. if BD= 7 cm and CD =28cm. what is the length (in cm) of AD?

a) 3.5 b) 7 c) 10.5 d) 14

solution Median ( AD ) = `\sqrt{BD\timesDC`

AD =  `\sqrt{7\times28`

AD = 14

Question 8

If the semiperimeter of a right angle triangle is 126 cm and length of shortest median is 56 cm. Find the area of triangle?

a) 1672 b) 1444 c) 1764 d) 1592

1 

Area of Right angle Triangle = r×s

s = `frac{a + b + c}2` ( Given)

Here

If BD is a Median then

BD = AD = DC

and

C = AD + DC
C = 56 + 56 = 112
If s (semiperimeter ) = 126
Then perimeter ( a + b + c) = 252
Then
a + b = 140 (252 - 112)

r = `frac{a + b - c}2`
r = `frac{ 140 - 112}2`

Area = 14 × 126 = 1764

Solution 2

Area = s(s - 2R)

Here c = 2R
Area = 126( 126 - 112) = 1764

Question 9

In a right-angled triangle ABC, angle ABC 90, AB= 5 cm and BC =12 cm.Find the length of BN, if N is the centroid.

a) 13/3 b) 13/2 c) 11/3 d) 6

If BD is a Median then 

BD = AD = DC

Then BD = 6.5

IF N is centroid then it divides BD in 2 : 1

BN = `frac{6.5 × 2}3 = frac{13}3`

Question 10

PQR is Right Angle Triangle at angle Q . QS is perpendicular to QR and ST is perpendicular to QR. If PQ=3cm and QR=4cm then find the length of ST?

a) 12 / 7 b) 48 / 25 c) 100 / 49 d) 96 / 41

Solution

We know that in Right Angle Triangle

`frac{PQ²}{QR²} = frac{ PS }{SR}`
`frac{PS}{SR} = frac {9}16`

PS = `frac{5 × 16}25` = 3.2

( hypotenuse is 5 by using pythagorean theorem)

In Right Angle Triangle PQR

QS = `frac{PQ × QR}{PR} = frac{ 3 × 4}5`= 2.4

In Right Angle Triangle RSQ

ST = `frac{QS × SR}{QR}`

ST = `frac{2.4 × 3.2}4 = frac{48}25`

Question 11

ABC is right angle triangle in which angle ABC-90". Incircle of ABC touches hypotenuse AC at point P. If AP =5cm, PC = 7.5cm then find incircle ( r ) ?

a) 2.5cm b) 4.5cm c) 3.5cm² d) 4cm²

Solution 

if two tangents are drawn on a circle and crossing circle. Tangents meet each other at a point then the lengths of two tangents will be the same.
So QC = 7.5 and AR = 5
Here we will use Hit and Trial method instead of basic one

AC = 12.5

Then surely the triplet must have consisting number that is multiple of 5


We know that 3 , 4 , 5 are pythagorean triplet if we multiply them by 2.5 then we get 7.5, 10, 12.5

Hense r = 2.5 cm