SSC CGL Previous Year Questions: Geometry- 5 - SSC CGL MCQ

Here BC = OD = radius {given}
In ΔBOC,
BC = OB (radius).
∴ ∠BOC = ∠BCO = 20°
∠OBA = ∠BOC + ∠BCO = 20° + 20° = 40°
Again, In DAOB,
AO = BO = radius
∠OAB = ∠OBA = 40°
∴ ∠AOB = 180° – 40° – 40° = 100°
Again,
∠AOD + ∠AOB + ∠BOC = 180°
∠AOD + 100° + 20° = 180°
∠AOD = 180° – 120° = 60°

From question ∠BDC = 50°
In ΔBDC, ∠DBC + ∠BDC + ∠BCD = 180°

∠B + ∠C = 40 × 2 = 80°
From ΔABC, ∠A + ∠B + ∠C = 180°
∠A + 80° = 180°
∠A = 180° – 80° = 100°

AB is a diameter of the circle
∴ ∠ACB = 90° {Angle made by the diameter on the semicircle
∠BAC = 30° {given)
∴ ∠BAC = ∠ACD = 30°   {As AB || CD}.
∠BCD = 90° + 30° = 120°
As ABCD is cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180°
∠BAD + 120° = 180°
∠BAD = 180° – 120° = 60°
∠BAD = ∠CAB + ∠CAD = 60°
30° + ∠CAD = 60°  ⇒ ∠CAD = 30°
Again, AB || CD, ∴ ∠BAD + ∠ADC = 180°
60° + ∠ADC = 180°
∴ ∠ADC = 180° – 60° = 120°.

As, AB = AC
∴ ∠ABC =  ∠ACB = 35°

∠BAC = 180° – 35° – 35° = 110°
∠BAC = ∠BAD + ∠CAD = 2∠BAD
{Since AD is a median}
∴ ÐBAD = 55°

AB = 10 cm, AE = 5 cm
OE = x
CD = 24 cm, DF = 12 cm
OF = 17 – x
OA = OD = radius
⇒ 5² + x² = 122 + (17 – x)²
⇒ 25 + x² = 144 + 289 – 34x + x²
⇒ 34x = 408

In quadrilateral POQS, ∠S = 20°
∠OPS = ∠OQS = 90°, ∠POQ = ?.
{Since PS and QS are tangent to the circle}
∴ Sum of angles in a quadrilateral = 360°
∠OPS + ∠S + ∠OQS + ∠POQ = 360°
90° + 20° + 90° + ∠POQ = 360°
⇒ ∠POQ = 160°

{Angle made on perimeter by the same chord is half of the angle made at the centre}

Now, In cyclic quadrilateral PRQT,
∠PTQ + ∠PRS = 180°
80° + ∠PRQ = 180°  ⇒ ∠PRQ = 180° – 80° = 100°

Here chord AB = 300 cm
AD = 15 cm
OD = 8 cm

Since, AB = AC {given}
∠ABC = ∠ACB
∠BAC = 40°  {given}
∴ ∠ABC + ∠ACB = {180° – 40°} = 140°
∴ ∠ABC  = 70°
​​​​​​​∴ ∠ABD = 180° – 70° = 110°

Let A chord AB of circle C₁, touches the concentric circle C₂ at point ‘P’.

Here OA = radius of circle C₁ = (√3 + 1)
OP = radius of circle C₂ = (√3 - 1)
As AP is a tangent to the circle C₂.
∴ ∠OPA = 90°
Now, from ΔOPA, (OA)² = (OP)² + (AP)²
(AP)² = (OA)² – (OP)²

Orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

Here in the triangle ABC, AD, BE and CF are three altitudes drawn from point A, B and C on the side BC, AC and AB respectively. All the three altitudes intersect each other at a common point ‘O’. That point ‘O’ is called ‘Orthocenter’ of the triangle ABC.

∠A + ∠B + ∠C = 180°
3∠C + 5∠C + ∠C = 180°
9∠C = 180°
∠C = 20°
∠B = 100°

Let the length of the corresponding side of other triangle is x. Then

Given, AB = 12 cm; CD = 20 cm
OE = ?

Now, AE = EB = 6cm (The line drawn from centre of circle to the chord bisect the chord)
In ΔOAE, By Pythagoras theorem
(OA)² = (OE)² + (AE)² ⇒ (10)² = (OE)² + (6)²
100 – 36 = (OE)² ⇒ 64 = OE² ⇒ OE = 8 cm

AP and PB are two tangents to the circle
∴ ∠OAP = ∠OBP = 90°
In ΔOAP, Let ∠OPA = θ.
OP = 2 × radius {given}
∴ OP = 2 × OA

∴ sinθ = sin 30° ⇒ θ = 30°
Again In ΔBOP, ∠OPA = ∠OPB = θ = 30° {By symmetry}
∴ ∠APB = 30° + 30° = 60°.

∠A + ∠B = 145°
∠C + 180° – 145° = 35°
∠C + 2∠B = 180°
⇒ 2∠B = 180° – 35° = 145°
⇒ ∠B = 145/2 = 72.5 ° = ∠A
∠B > ∠C
∴ CA > AB

∠ACB = 80°
∠ACD = 180° – 80° = 100°
∴ ∠CAD = ∠CDA
= 80/2 = 40°
∠BAC = 111° – 40°
 = 71°
∠ABC = 180° – 71° – 80° = 29°

∠AOE = 150°
∠DAO = 51°
∠EOB = 180° – 150° = 30°
OE = OB = radius
∴ ∠OEB = ∠OBE =  150/2 = 75°
∴ ∠CBE = 180° – 75° = 105°

PR² = PQ² + PR² = 3² + 4² = 25
∴ PR = √25 = 5 cm

Let side of triangle = x

Putting x in equation (i)

Radian covered in one second 
Time required to covered 55 radian 

Here RQ is the common tangent which touches circles with centre A and B at point R and Q respectively
∴ ∠ARQ = ∠BQR = 90°
On extending the line AB, tangent RQ meet the line AB at point P.
Now, In DPBQ and DPAR,
BQ || AR, ∠P = ∠P, ∠Q = ∠R  ⇒ ∠A = ∠B.
thus, DPBQ ~ DPAR {from AA theorem}

Hence, point P, divides line AB into 5 : 2 ratio externally.

In ΔAOB
AO = BO (radii of circles)
∴ ∠ABO = ∠BAO = 30º In DBOC
BO = CO (radii of circles)
∴ ∠BCO = ∠OBC = 40º
∠ABC = ∠ABO + ∠OBC
∠ABC = 30° + 40° = 70°
2 × ∠ABC = ∠AOC ⇒ x° = 140

Let O and O' be the centre of two intersecting circle, where point of intersection are P and Q and PA and PB are their diameter respectively.
∠AQP = 90° and ∠BQP = 90°
{∵ Angle in a semicircle is a right angle}
Adding both these angles,
∠AQP + ∠BQP = 180°
∴ ∠AQB = 180º

In ΔABC, D is the mid-point of side BC, since, AD divide angle A.
∴ BD = DC
and ∠ABC = ∠AEC {angle in same sector of circle}
and ∠CAE = ∠CBE
from ΔABD and ΔACE

Since AB is a diameter. Then ∠APB = 90° (angle in the semicircle)
ΔBPN ~ ΔAPB
So, BN = BP² / AB

Parallelogram Area = l × b
Rhombus Area = l × b

Therefore R = P = 2T.