SSC CGL Previous Year Questions: Geometry- 1 - SSC CGL MCQ

# SSC CGL Previous Year Questions: Geometry- 1 - SSC CGL MCQ

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## 19 Questions MCQ Test - SSC CGL Previous Year Questions: Geometry- 1

SSC CGL Previous Year Questions: Geometry- 1 for SSC CGL 2024 is part of SSC CGL preparation. The SSC CGL Previous Year Questions: Geometry- 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL Previous Year Questions: Geometry- 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL Previous Year Questions: Geometry- 1 below.
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SSC CGL Previous Year Questions: Geometry- 1 - Question 1

### ΔABC ~ ΔNLM and ar(ΔABC) : ar(ΔLMN) = 4 : 9. If AB = 6 cm, BC = 8 cm and AC =12 cm, then ML is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 1

We have,

SSC CGL Previous Year Questions: Geometry- 1 - Question 2

### In a circle of radius 17 cm, a chord is at a distance of 8 cm from the centre of the circle. What is the length of the chord?       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 2

Here, AB is a chord
OA = OB = 17cm (radius)
OD = 8cm

Chord AB = 2 ´ 15 = 30cm

SSC CGL Previous Year Questions: Geometry- 1 - Question 3

### In a circle with centre O, an arc ABC subtends an angle of 110° at the centre of the circle. The chord AB is produced to a point P. Then ∠CBP is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 3

Here, OA = OB (radius of the circle).
∴ ∠OAB = ∠OBA
In ΔOAB,

∠C = 180° – 55° = 125°
∠CBP = 180° - 125° = 55°

SSC CGL Previous Year Questions: Geometry- 1 - Question 4

In ΔABC, P is a point on BC such that BP : PC = 4 : 11. If Q is the midpoint of BP, then ar(ΔABQ) : ar(ΔABC) is equal to:      (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 4

Here, BC = BP + PC
= 4x + 11x = 15x
and BQ = PQ = 2x
Let AD is the height of the ΔABC,

SSC CGL Previous Year Questions: Geometry- 1 - Question 5

In ΔABC, AD is the median and G is a point on AD such that AG : GD = 2 : 1. Then ar(ΔBDG) : ar(ΔABC) is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 5

Let ΔABC is a equilateral triangle in which AD is a median.
Then, ∠ADB = 90° {∵ ΔABC is equilateral)

SSC CGL Previous Year Questions: Geometry- 1 - Question 6

From a point P outside a cir cle, PAB is a secan t and PT is a tangent to the circle, where, A, B and T are points on the circle. If PT = 5 cm, PA = 4 cm and AB = x cm, then x is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 6

From circle properties
PT2 = PA ´ PB
(5)2 = 4 x (x + 4)
x + 4 = (25/4) = 6.25
x = 6.25 – 4 = 2.25 cm

SSC CGL Previous Year Questions: Geometry- 1 - Question 7

ΔABC ~ ΔRQP and AB = 4 cm, BC = 6 cm and AC = 5 cm. If ar(ΔABC) : ar(ΔPQR) = 9 : 4, then PQ is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 7

SSC CGL Previous Year Questions: Geometry- 1 - Question 8

In a circle with centre O, AB is the diameter and CD is a chord such that ABCD is a trapezium. If ∠BAC = 40°, then ∠CAD is equal to:       (SSC CGL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 8

As AB is a diameter.
In Δ ABC, ∠BCA = 90°
∠BAC = 40°
∴ ∠ABC = 180° – 90° – 40° = 50°
and ABCD is cyclic trapezium
so, ∠ABC + ∠ADC = 180°
∴ ∠ADC = 180° – ∠ABC
= 180° – 50° = 130°
and ∠DCA = ∠BAC = 40° {∵ AB || CD}.
= 180° – (130° + 40°) = 10°

SSC CGL Previous Year Questions: Geometry- 1 - Question 9

It is given that ΔABC ~ ΔPRQ and that Area ABC : Area PRQ = 16 : 169. If AB = x, AC = y,  BC = z (all in cm), then PQ is equal to:      (SSC CHSL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 9

= 13/4 y.

SSC CGL Previous Year Questions: Geometry- 1 - Question 10

The side BC of a right-angled triangle ABC (∠ABC = 90º) is divided into four equal parts at P, Q and R respectively. If AP2 + AQ2 + AR2 = 3b2 + 17na2, then n is equal to :       (SSC CHSL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 10

In ΔABC, AB = c, BC = a, and AC= b.
and BP = PQ = QR = RC = a/4.

from (i) + (ii) + (iii)

Again from ΔABC, b2 = c2 + a2
c2 = b2 – a2
∴ AP2 + AQ2 + AR2 = 3 (b2 – a2) + 7a2/8

Hence, n = - (1/8).

SSC CGL Previous Year Questions: Geometry- 1 - Question 11

OABC is a quadrilateral, where O is the centre of a circle and A, B, C are points in the circle, such that ∠ABC = 120º. What is the ratio of the measure of ∠AOC to that of ∠OAC?       (SSC CHSL-2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 11

Ratio of

SSC CGL Previous Year Questions: Geometry- 1 - Question 12

PA and PB are two tangents from a point P outside a circle with centre O. If A and B are points on the circle such that ∠APB = 80°, then ∠OAB is equal to:       (SSC Sub. Ins. 2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 12

∠PAO = 90°
and ∠OPA = 80°/2 = 40°
In Δ AOP, ∠AOP + ∠APO + ∠PAO = 180°
∠AOP + 40° + 90° = 180°
∴∠AOP = 180° – 130° = 50°

SSC CGL Previous Year Questions: Geometry- 1 - Question 13

ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and ∠ADC = 145°. What is the measure of ∠BAC?        (SSC Sub. Ins. 2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 13

As ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
∠B + 145° = 180°
∠B = 180° – 145° = 35°
Again as AB is a diameter.
∴ ∠ACB = 90°
Now, In ΔACB, ∠ BAC + ∠ABC + ∠ACB = 180°
∠BAC + 35° + 90° = 180°
∠BAC = 180° – 90° – 35° = 55°.

SSC CGL Previous Year Questions: Geometry- 1 - Question 14

In ΔABC, ∠A = 30°. If the bisectors of the angle B and angle C meet at a point O in the interior of the triangle, then ∠BOC is equal to:      (SSC Sub. Ins. 2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 14

= 75° + ∠BOC = 180°
∠BOC = 180° – 75° = 105°

SSC CGL Previous Year Questions: Geometry- 1 - Question 15

Let ΔABC ~ ΔQPR and . If AB = 12cm, BC = 6cm and AC = 9cm. Then PR is equal to:      (SSC Sub. Ins. 2018)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 15

As ΔABC ~ ΔQPR, then

SSC CGL Previous Year Questions: Geometry- 1 - Question 16

In the given figure , BD passes through centr e O, AB = 12 and AC = 8. What is the radius of the circle?       (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 16

According to question,
AB = 12
AC = 8
As, we know that
(12)2 = 8 × (8 + x)

ΔABD is a right angle because ∠B = 90°
By Pythagorean theorem,
BD2 = (18)2 – (12)2 {Here AD = 8 + 10 = 18 cm}
= 324 – 144 = 180
∴ BD = 6√5 cm

SSC CGL Previous Year Questions: Geometry- 1 - Question 17

In the given figure, area of isosceles triangle PQT is 128 cm2 and QT = PQ and PQ = 4 PS, PT || SR, then what is the area (in cm2) of the quadrilateral PTRS?        (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 17

From question, PT = TQ = x (Let)

Again,
As PQ || RS and PS || QR (By symmetry)
∴ PQRS is a parallelogram with base (b) = 4 cm height (h) = PT = 16 cm
Area = b × h = 4 × 16 = 64 cm2

SSC CGL Previous Year Questions: Geometry- 1 - Question 18

If ΔDEF is right angled at E, DE = 15 and ∠DFE = 60°, then what is the value of EF?       (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 18

According to question,
DE = 15, EF = ?
∠DFE = 60°
∠DEF = 90°

∴

∴

SSC CGL Previous Year Questions: Geometry- 1 - Question 19

In the given figure, O is the centre of the circle, ∠DAB = 110°and ∠BEC = 100°. What is the value (in degrees) of ∠OCB?            (SSC Sub. Ins. 2017)

Detailed Solution for SSC CGL Previous Year Questions: Geometry- 1 - Question 19

∠BEC = 100° (given)
So, ∠BEC + ∠BFC = 180°
100° + ∠BFC = 180°
⇒ ∠BFC = 180° – 100° = 80°
Now, ∠BOC = 2 × ∠BFC (Angle made by the same chord at the center)
= 2 × 80° = 160°
In ΔOBC, OB = OC (radius)
∴ ∠OBC = ∠OCB
Now, ∠BOC + ∠OBC + ∠OCB = 180°
160° + 2∠OCB = 180°
⇒

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