Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Geometry (Except Circle) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II


Internal bisectors of ∠B and ∠C of ΔABC intersect at O. If ∠BOC = 102, then the value of ∠BAC is ?

A) 24
B) 42
C) 55
D) 60

In ΔBOC, ∠OBC +∠OCB = 180 – ∠BOC = 180 – 102 = 78
But ∠OBC + ∠OCB = 1/2(∠ABC + ∠ACB) (because of angle bisectors)
=> ∠ABC + ∠ACB = 78*2 = 156
Therefore ∠BAC = 180 – 156 = 24 (option ‘A’)


If the lengths of the side of a triangle are in ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm then the altitude of the triangle corresponding to the largest side is?

A) 10
B) 8
C) 7.5
D) 6

When the incircle of the triangle and its side are given the area of the triangle = r*s, where ‘r’ is the incircle and ‘s’ is the semi-perimeter of the triangle

Now let the sides of the triangle be 4x, 5x, 6x respectively
So the semi-perimeter (s) = (4x+5x+6x)/2 = 7.5
r = 3 (given)
So the area = 3*7.5 = 22.5

But also the area of the triangle = (6x* corresponding altitude)/2
=> 3*altitude
So the altitude = 22.5/3 = 7.5 (option ‘C’)


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

A) 16 cm, 20 cm, 26 cm
B) 18 cm, 24 cm, 30 cm
C) 20 cm, 24 cm, 30 cm
D) 17 cm, 25 cm, 31 cm

The radius of the circumcircle of a right triangle is always the bisector of its hypotenuse (let it be ‘c’); so the hypotenuse = 15*2 = 30

The sum of the other two sides (let them ‘a’ and ‘b’ respectively) of the right triangle is equal to twice of the sum of the incircle and circumcircle
So, a + b = 2*(15*6) = 42 ——(i)
and the sum of all the three sides = 42 + 30 = 72

Product of the sides other than the hypotenuse is equal to the product of the radius of the incircle and sum of all the three sides
So ab = 6*72 = 432 — ——(ii)

From (i) and (ii)
‘a’ and ‘b’ are 18 and 24 or vice versa

So all the sides of the triangle are 18, 24 and 30 (option ‘B’)


Adjacent sides of a parallelogram are 18 and 16. If one of the diagonal is 20 find the other. All units are in cm.

A) 2√180
B) 180
C) 2√190
D) 2√203

In a parallelogram sum of the squares of the diagonals = 2(a² + b² ), where a and b are the lengths of the sides of the parallelogram.

Now let the other diagonal be x cm
Therefore x² + 20² = 2(18² + 16² )

=> x² = 2(324 + 256) – 400

=> x = √(1160 – 400) = √760

=> x = 2√190 (option ‘C’)


ABCD is a rhombus having area √3 unit². BD is joined and O is the point of intersection of the bisectors of ∠ADB and ∠ABD. From O a perpendicular is drawn to BD. Its length is 4cm. Then BC =? Given ∠DCB = 60°.

A) 8√3
B) 3√3
C) 8
D) 48√3

Draw the figure according to the question. We know that a diagonal of a rhombus divides it into two triangles of equal areas, hence area of the ΔBCD = 96√/2 = 48√3. But it’s an equilateral triangle as each of the angles in it equals°. See how:

∠DCB = 60° (given). As a diagonal of a rhombus bisects the angles it meets, so each of the ∠CBD and ∠CDB = 60°. Being such each of its side is equal to the others.

Now we have to find the length of BC, which is eventually is a side of equilateral ΔBCD, and we know that in an equilateral triangle, area = (√3/4)*side², therefore (√3/4)BC² = 48√3.

=> BC² = 192

=> BC = 8√3 (option ‘A’)

NOTE: Actually either the area given or the fact of the perpendicular given is not needed here as this question can be solved without one of them been given.

Trigonometric Solution


The lateral side of an isosceles triangle is 15 cm and the altitude is 8 cm. What is the radius of the circumscribed circle?

A) 16 cm
B) 225/16 cm
C) 325/16 cm
D) 17 cm

The circum-radius of an isosceles triangle is:

2√(a² – b²/4), where a is the lateral side and b the base.

Lateral side i.e, ‘a’ is already given, so we need to find only the base. Of course the altitude will help.

We know the altitude of an isosceles triangle bisects the base at 90°. So by Pythagoras: half base = √(lateral side² – altitude²) = √(15² – 8²) = √161
So, the base i.e. ‘b’ = 2√161

Now the circum-radius = 15²[2√{15² – (2√161)²}/4}]

= 2252√(225 – 161)

= 2252√64

= 22516 cm (option ‘B’)

NOTE: Inradius of an isosceles triangle :[ab – (b²)/2]2√[a² – (b²)/4]


In the figure given below, XYZ is a right angle triangle in which ∠Y is 45° and ∠X is 90°. ABCD is a square inscribed in it whose area is 64 cm². What is the are of the ΔXYZ?

A) 12 cm²
B) 140 cm²
C) 144 cm²
D) 64 cm²


As ABCD is a square, ∠ADY = 90°; ∠AYD = 45° (already given. According to the angle sum property of the triangle, therefore, ∠YAD also 45°. Hence ΔAYD is an isosceles right angle triangle, AD and DY being the equal sides.

Now area of the given square is 64, hence AD = 8

We know that area of a isosceles right angle triangle = side²/2, so area of ΔADY = 8²/2 = 32

Similarly area of ΔBCZ also is 32

As ∠YAD = 45° (shown above) you can easily deduce that ∠XAB and and ∠XBA each is 45°. So ΔAXB too is an isosceles right angle triangle, where the hypotenuse AB is 8 cm.

Thus, by pythagoras AX² + BX² = AB²
=> AX² + AX² = 8²
=> 2AX² = 64
=> AX² = 32
But AX is one of the equal sides here
So the area of this triangle = AX²/2 = 32/2 = 16

It’s obvious that area of the ΔXYZ = Area of all the smaller triangles + area of the square = 32 + 32 + 16 + 64 = 144 cm² (option ‘C’)

Area of a isosceles right triangle = 2.25 of the area of the square inscribed in it.

See 2.25*64 = 144 cm² (option ‘C’)

NOTE: Also you should remember that area of each of the equal triangles inside like this is 1/2 the area of the square and area of the smallest triangle is 1/4 of the area of the square.


The perimeter of a rhombus is 40 cm and the measure of an angle is 60°, then the area is?

A) 50√3 cm²
B) 100√3 cm²
C) 100√3/2 cm²
D) 55 √3 cm²

Area of parallelogram = product of any two adjacent sides*sine of the included angles.

Rhombus also is a parallelogram whose all sides are equal. So here adjacent sides are 40/4 = 10 cm each.

Therefore its area =10*10*sin 60
= 100√3/2
= 50√3 (option ‘A’)

Another Method
Since diagonals of rhombus bisect each other at right angles, and also bisect the vertex angles, so we have 4 right angle triangles whose other angles are 30° and 60° each and one of the sides in each being PERIMETER/4 = 40/4 = 10 cm.

Now using the trigonometry both the diagonals of the rhombus can easily be found, and they thus are 10 cm and 10√3 cm.

Now area of rhombus = product of diagonals/2 = (10*10√3)/2 = 50√3 (option ‘A’)


In ΔABC, AB = 3, AC = 4. The bisector of ∠A divides the side BC at D, then find  BD : DC.

A) 1 : 2
B) 4 : 3
C) 3 : 4
D) 3 : 7


The bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle
Therefore BD : DC = 3 : 4 (option ‘C’)


An isosceles ΔABC is right angled at B. D is a point inside the ΔABC. P and Q are the feet of the perpendiculars drawn from D on the sides AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, find sin 75.




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Maha Gupta

Maha Gupta

Founder of and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)