SSC CGL Previous Year Questions: Geometry- 6 - SSC CGL MCQ

PQ² = (r₁ + r₂)² – (r₁ – r₂)² = 4r₁r₂

In ΔPAC and ΔATC,
∠ATC = ∠PAC = 180° – θ.
∠PAC = ∠TCA
∴ ∠PAC ~ ΔATC

⇒ CT : CB = AC : PC

∠BAC = 50°
∴ ∠BSC = 100°
BS = SC = radius

Here we extend AG that meet BC at point D.
E and F are mid point, hence 
As, ΔAFO ~ ΔABD

∴ O be the mid point of AD i.e. AO = OD.
Now, as centroid divides the median in the ratio 2 : 1

2(AO) – 2(OG) = AO + OG.
(AO) = 3(OG)
∴ AO : OG = 3 : 1

Ratio of corresponding sides

Let altitudes drawn from vertex B and C cross each other at Point ‘O’.
then, ∠BOC = 54° = ∠EOF {vertically opposite angles}
Now, in quadrilateral AEOF
∠AEO + ∠AFO + ∠EAF + ∠EOF = 360°
90° + 90° + ∠EAF + 54° = 360°
∴ ∠EAF = 360° – 90° – 90° – 54° = 126°
∴ ∠EAF = ∠BAC = 126°

OC ⊥ AB
AC = BC = 3.5 cm OP = 13 cm
PC = 9 + 3.5 = 12.5 cm

Length of common tangent

Here, in this circle, AO = OB = CO = OD = radius
From question,
CD = CO = OD
∴ ΔOCD is an equilateral triangle.
∴ ∠OCD = ∠ODC = ∠COD = 60°
Now, consider the cyclic quadrileteral ABCD,
∠CAB + ∠BDC = ∠ACD + ∠ABC = 180° ...(i)
Let ∠ACO = x and ∠BDO = y.
then, in ∠AOC, CO = AO = radius
∴ ∠OAC = ∠ACO = x
Similarly in ΔBOD, ∠ODB = ∠BDO = y
Putting these values in equation (i)
∠CAB + ∠BDC = 180°
∠CAB + ∠BDO + ∠ODC = 180°
x + y + 60° = 180°
∴ x + y = 120° ...(ii)
Now, in ΔCPD,
∠PCD = 180° – ∠ACD = 180° – (x + 60°)
and ∠PDC = 180° – ∠BDC = 180° – (y + 60°)
Sum of angles in ΔCPD = 180°
∴ ∠PCD + ∠CPD + ∠PDC = 180°
180° – (x + 60°) + ∠CPD + 180° – (y + 60°) = 180°
60° – (x + y) + ∠CPD = 0
60° – 120° + ∠CPD = 0
∴ ∠CPD = 60°
Hence, ∠APB = 60°

Let two circles with centre A and B intersect each other such at points C and E such that CE is its common chord.
As A and B are the centre of the circles So, line AB divides chord CE at point D in two equal parts such that CD = DE and also, AB ⊥ CE
Now, Consider ΔACD and ΔBCD,
AC = 20 cm, BC = 15 cm (given)
Let, CD = x cm
and AD = y cm then BD = (25 – y) cm.
From ΔADC, (AC)² = (AD)² + (CD)²
(20)² = y² + x² ...(i)
From ΔBDC, (BC)² = (BD)² + (CD)²
(15)² = (25 – y)² + x² ...(ii)
From equation (i) and (ii), we have
(20)² – (15)² = y² – (25 – y)²
(20 + 15) (20 – 15) = (y – 25 + y) (y + 25 – y)
35 × 5 = 25 (2y – 25)
2y = 7 + 25 = 32
y = 16
Again, from equation (i),
(20)² = (16)² + x²
∴ x² = (20)² – (16)²
= 400 – 256 = 144
CD = x = 12 cm.
Common chord CE = 2 × CD = 2 × 12 = 24 cm

In ΔABC, point D and E are mid point of side AB and AC.
So, CD is the median of ΔABC.

Again, from ΔACD, ED is median of ΔACD

2x + 3x + 5x = 180°  –  45°  = 135°
⇒ 10x = 135°

In ΔABC
Given AG = BC  

i.e., GD = BD = DC
In ΔBGD
BD = DG ∴ ∠GBD = ∠DGB …(i)
In ΔCGD
GD = DC, ∴ ∠GCD = ∠DGC …(ii)
∠GBD + ∠DGB + ∠DGC + ∠DCG = 180
2(∠BGD + ∠CGD) = 180 

AB = BC = x

BD = DC = AD
∠BAD = 30° {given}
In ΔABD,
∴ ∠ABD = ∠BAD = 30°
∴ ∠ADB = 180° – 2 × 30° = 120°
∴ ∠ADC = 180° – 120° = 60°
∴ AD = DC
⇒ ∠DAC = ∠ACD = 60°

The sum of any two sides of a triangle is greater than third side and their difference is less than third side.
10 – 4 < a < 10 + 4
6 < a < 14

Angle traced by hour hand in an hour = 30°

Angle traced by minute hand in 60 minutes = 360°
∴ Angle traced in 15 minutes

From ΔPOQ,
∠OPQ = 180° – 90° – 60° = 30°


∴ QS = 2 × 3 = 6 cm

Let exterior angle = x
then, interior ∠be = 2x  
x + 2x = 180
3x = 180
x = 60°

In right Δs OAP and OPB,
AP = PB, OA = OB
OP = OP
∴ ΔOAP ≌ ΔOBP
∴ ∠AOP = ∠POB and ∠APO = ∠OPB
From ΔAOP,
∠APO = 180° – 90° – 60° = 30°
∴ ∠APB = 2 × 30 = 60°

r₁ + r₂ = 13 cm
r₂ – r₁ = 9 – 4 = 5 cm

∴ Area of square of side PQ = 12 × 12 = 144 sq. cm.

∠XOY = 90°; OX = OY = radices (r)
∴ ΔXOY is a right angled triangle.

OX is the bisector of ∠AOC.
∴ ∠AOC = 2 ∠COX
OY is the bisector of ∠BOC.
∴ ∠BOC = 2 ∠COY
∴ ∠AOC + ∠BOC
= 2∠COY + 2∠COX = 180°
⇒ 2 (∠COX + ∠YOC) = 180°
⇒ ∠XOY = 90°
∴ ∠AOX + ∠XOY + ∠BOY = 180°
∴ ∠BOY = 180° – 90° – 20° = 70°

Circumradius of a triangle

In a right angled D, the length of circumradius is half the length of hypotenuse.
∴ H² = 6² + 8²
H² = 36 + 64 ⇒ H² = 100
H = 10 cm
Hence, Circumradius = 5 cm

ΔABC is an isosceles triangle.
Therefore, AC = BC = 5 cm  
Now, AB² = AC² + BC²