MCQ: Geometry - 3 - SSC CGL MCQ

# MCQ: Geometry - 3 - SSC CGL MCQ

Test Description

## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Geometry - 3

MCQ: Geometry - 3 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Geometry - 3 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Geometry - 3 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Geometry - 3 below.
Solutions of MCQ: Geometry - 3 questions in English are available as part of our Quantitative Aptitude for SSC CGL for SSC CGL & MCQ: Geometry - 3 solutions in Hindi for Quantitative Aptitude for SSC CGL course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt MCQ: Geometry - 3 | 15 questions in 15 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study Quantitative Aptitude for SSC CGL for SSC CGL Exam | Download free PDF with solutions
MCQ: Geometry - 3 - Question 1

### In the given figure, find the length of BD.

Detailed Solution for MCQ: Geometry - 3 - Question 1

∠A = ∠A [common]
∴ ΔADE ∼ ΔACB (AA Similarly)

Hence BD = AB - AD = 19.5 - 6 = 13.5 cm.

MCQ: Geometry - 3 - Question 2

### A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Find the height of the tower.

Detailed Solution for MCQ: Geometry - 3 - Question 2

In ΔACB and PCQ
∠C = ∠C (common)
∠ABC = ∠PQC (each 90°)
∴ ΔACB ∼ ΔPC (AA Similarity)

 1 Crore+ students have signed up on EduRev. Have you?
MCQ: Geometry - 3 - Question 3

### ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120° and ∠BAC = 30°, then ∠BCD is :

Detailed Solution for MCQ: Geometry - 3 - Question 3

As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O

Given , ∠COD = 120°
∠BAC = 30°

∴ ∠BCD = 180° - ∠BAD
∴ ∠BCD = 180° – 90° = 90°

MCQ: Geometry - 3 - Question 4

The distance between the centres of the two circles of radii r1 and r2 is d. They will touch each other internally if

Detailed Solution for MCQ: Geometry - 3 - Question 4

According to question , we can draw a figure

Hence , option C is correct answer .

MCQ: Geometry - 3 - Question 5

If angles of measure (5y + 62°) and (22° + y) are supplementary, then value of y is :

Detailed Solution for MCQ: Geometry - 3 - Question 5

As we know that Sum of two supplementary angles = 180°
∴ (5y + 62°) + (22° + y) = 180°
⇒ 6y + 84° = 180°
⇒ 6y = 180° – 84° = 96°

MCQ: Geometry - 3 - Question 6

The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle be a tangent to the smaller circle, then the length of that chord is

Detailed Solution for MCQ: Geometry - 3 - Question 6

According to question , we draw a figure of a circle with centre O,

Here , BO = OC = 15 cm and OD = 9 cm.
From ∆ BDO,

∴ BC = 2 × 12 = 24 cm.

MCQ: Geometry - 3 - Question 7

Two chords AB and CD of a circle with centre O, intersect each other at P. If ∠AOD = 100° and ∠BOC = 70°, then the value of ∠APC is

Detailed Solution for MCQ: Geometry - 3 - Question 7

On the basis of question we draw a figure of a circle with centre O ,

Given , ∠AOD = 100°
We know that the angle subtended at the centre is twice to that of angle at the circumference by the same arc.

Again, ∠BOC = 70°

∴ ∠APC = 180° – 50° – 35° = 95°

MCQ: Geometry - 3 - Question 8

AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. The distance between them is 1 cm. The radius of the circle is

Detailed Solution for MCQ: Geometry - 3 - Question 8

On the basis of question we draw a figure of a circle with centre O ,

Let OE = y cm
then OF = (y +1) cm
OA = OC = r cm
AE = 4 cm; CF = 3 cm
From ∆ OAE,
OA² = AE² + OE²
⇒ r² = 16 + y²
⇒ y² = r² – 16 ......(i)
From ∆OCF,
(y + 1)² = r² – 9 ..... (ii)
By equation (ii) – (i),
(y + 1)² – y² = r² – 9 – r² + 16
⇒ 2y + 1 = 7
⇒ y = 3 cm
∴ From equation (i),
9 = r² – 16
⇒ r² = 25
⇒ r = 5 cm

MCQ: Geometry - 3 - Question 9

The orthocentre of an obtuse angled triangle lies

Detailed Solution for MCQ: Geometry - 3 - Question 9

As we know that the orthocentre of an obtuse angled triangle lies outside the triangle. Hence , correct answer is option B.

MCQ: Geometry - 3 - Question 10

If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is

Detailed Solution for MCQ: Geometry - 3 - Question 10

Given , The sum of the interior angles of a regular polygon = 1080°
Sum of the interior angles of a regular polygon of n sides = (2n – 4) × 90°
∴ (2n – 4) × 90° = 1080°
⇒ 2n – 4 = 1080 ÷ 90 = 12
⇒ 2n = 12 + 4 = 16
∴ n = 8

MCQ: Geometry - 3 - Question 11

The centroid of a triangle is the point where

Detailed Solution for MCQ: Geometry - 3 - Question 11

As we know the centroid of a triangle is the point where the point of intersection of medians of a triangle is called centroid.

MCQ: Geometry - 3 - Question 12

ABC is an isosceles triangle such that AB = AC and ∠B = 35°. AD is the median to the base BC. Then ∠BAD is:

Detailed Solution for MCQ: Geometry - 3 - Question 12

Firstly, We draw a figure of an isosceles triangle ABC,

Given that , AB = AC and ∠B = 35°
⇒ ∠ABC = ∠ACB = 35°
⇒ ∠ BAD + 90° + 35° = 180°

MCQ: Geometry - 3 - Question 13

In a triangle, if three altitudes are equal, then the triangle is

Detailed Solution for MCQ: Geometry - 3 - Question 13

We know that in a triangle, if three altitudes are equal, then the triangle is called equilateral.

MCQ: Geometry - 3 - Question 14

If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is :

Detailed Solution for MCQ: Geometry - 3 - Question 14

As per the given in question , we draw a figure of an isosceles triangle ABC,

In isosceles triangle ,
AC = BC = 5 cm

MCQ: Geometry - 3 - Question 15

A circle is touching the side BCof Δ ABC at P and is also touching AB and AC produced at Q and R respectively. If AQ = 6 cm, then the perimeter of the triangle ABC is:

Detailed Solution for MCQ: Geometry - 3 - Question 15

A circle is touching the side AQ and AR,

Then AR = AQ = 6 cm
Now AQ + AR = AB + BQ + AC + CR
⇒ 6 + 6 = AB + BQ + AC + CR
⇒ 6 + 6 = AB + BQ + AC + CR
[here BQ = BP and CR = CP]
12 = AB + AC + BP + PC
12 = AB + AC + BC
Hence, perimeter of a triangle ABC is 12 cm.

## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests
Information about MCQ: Geometry - 3 Page
In this test you can find the Exam questions for MCQ: Geometry - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for MCQ: Geometry - 3, EduRev gives you an ample number of Online tests for practice

## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests