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SSC CGL Exam  >  SSC CGL Test  >  Quantitative Aptitude  >  MCQ: Triangles - 2 - SSC CGL MCQ

Triangles - 2 - Free MCQ Practice Test with solutions, SSC CGL Quant Aptitude


MCQ Practice Test & Solutions: MCQ: Triangles - 2 (15 Questions)

You can prepare effectively for SSC CGL Quantitative Aptitude for SSC CGL with this dedicated MCQ Practice Test (available with solutions) on the important topic of "MCQ: Triangles - 2". These 15 questions have been designed by the experts with the latest curriculum of SSC CGL 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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MCQ: Triangles - 2 - Question 1
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Directions: Study the following question carefully and choose the right answer:

If AD is the internal angular bisector of angle A of ΔABC with AB = 3 cm and AC= 1 cm, then what is BD : BC equal to?

Detailed Solution: Question 1

In ΔABC,
AD is the internal angle bisector of ∠A.
Using property of internal angle bisector.


Hence, option D is correct.

MCQ: Triangles - 2 - Question 2
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Directions: Study the following question carefully and choose the right answer:

In the above figure, if area of triangle ABC is 64 sq. units, then find the area of triangle PQR, where D, E and F are mid points of sides of ΔABC and P, Q and R are midpoints of sides of ΔDEF.

Detailed Solution: Question 2

Given that,
D, E and F are midpoints of BC, CA and AB and P, Q and R are midpoints of EF, FD and DE
we know that,
Area of ΔABC = 4 ΔDEF
But area of ABC = 64 sq. cm.

And area ΔDEF = 4 ΔPQR
⇒ 4 ΔPQR = 16 = 16/4 = 4 sq. units
Hence, option A is correct.

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MCQ: Triangles - 2 - Question 3
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Directions: Study the following question carefully and choose the right answer:

The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ∠ PSR is

Detailed Solution: Question 3

∠PRQ = 60° [∵ ΔPQR is an equilateral]
∠PRS = 180° – 60° = 120°
∠PSR + ∠RPS = 180° – 120° = 60° ...(i)
As QR = RS
∴ PR = RS [∵ ΔPQR is an equilateral]
∴ ∠PSR = ∠RPS
From Eq (i),
2∠PSR = 60°
∴ ∠PSR = 30°
Hence, option A is correct.

MCQ: Triangles - 2 - Question 4
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Directions: Study the following question carefully and choose the right answer:

The sides of a triangle are in geometric progression with common ratio r < 1. If the triangle is a right angled triangle, the square of common ratio is given by

Detailed Solution: Question 4

let the sides of triangle be a/r, a, ar and since r < 1.

Now, triangle is right angled.
Using Pythagoras theorem

Put r2 = x
∴ x2 + x – 1 = 0
Applying Sridharacharya rule, we get


Hence, option B is correct.

MCQ: Triangles - 2 - Question 5
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Directions: Study the following question carefully and choose the right answer:

Two medians PS and RT of ΔPQR intersect at G at right angles. If PS = 9 cm and RT = 6 cm,then the length of RS in cm is

Detailed Solution: Question 5

PS = 9 cm


RT = 6 cm

Hence, option C is correct.

MCQ: Triangles - 2 - Question 6
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Directions: Study the following question carefully and choose the right answer:

If the circumradius of an equilateral triangle be 10 cm, then the measure of its in-radius is

Detailed Solution: Question 6


Let AB = x cm

By pythagoras theorem in ΔABD,

We know that,

By pythagoras theorem in ΔBOD,

Given,
Circumradius, OB = 10 cm

Hence,

Hence, option A is correct.

MCQ: Triangles - 2 - Question 7
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Directions: Study the following question carefully and choose the right answer:

If triangles ABC and DEF are similar such that 2AB = DE and BC = 8 cm, then what is EF equal to?

Detailed Solution: Question 7


∵ ΔABC – ΔDEF


Hence, option A is correct.

MCQ: Triangles - 2 - Question 8
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Directions: Study the following question carefully and choose the right answer:

In the adjoining figure, if BC = a, AC = b, AB = c and ∠CAB = 120° , then the correct relation is :

 

Detailed Solution: Question 8

Since ∠A is an obtuse angle in Δ ABC, so


Hence, option C is correct.

MCQ: Triangles - 2 - Question 9
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Directions: Kindly study the following Question carefully and choose the right answer:

If the incentre of an equilateral triangle lies inside the triangle and its radius is 3 cm, then the side of the equilateral triangle is

Detailed Solution: Question 9


Hence, option B is correct.

MCQ: Triangles - 2 - Question 10
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Directions: Kindly study the following Question carefully and choose the right answer:

In a ΔABC, AD is perpendicular to BC and BE is perpendicular to AC. Which of the following is correct?

Detailed Solution: Question 10


From Eqs. (i) and (ii),

Hence, option C is correct.

MCQ: Triangles - 2 - Question 11
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Directions: Kindly study the following Question carefully and choose the right answer:

The length of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The value of angle C is 59°. Find the length of side AC.

Detailed Solution: Question 11

Given, AB = 12 cm, BC = 8 cm,
∠C = 59°
Let ∠A = Θ
∴ ∠B = 180° – (59° + Θ) = 121° – Θ
Now, let us see the choices. If AC = 12 cm, triangle would not be scalene. Hence, option A is ruled out. If AC =
10 cm, AB will become the largest side and ∠C the largest angle. But ∠C = 59°. Hence option B is ruled out. So,
AC is either 14 cm or 16 cm. In any case, ∠B will be the largest angle and ∠A (say Θ) the smallest:
Also, ∠B = 180° – (59° + Θ) = 121° – Θ
By sine formula

∴ sin (121° – Θ) ≈ sin (120° – Θ) = sin 120° cos Θ – cos120° sin Θ



Hence, option C is correct.

MCQ: Triangles - 2 - Question 12
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Directions: Kindly study the following Question carefully and choose the right answer:

In a triangle, if orthocentre, circumcentre, incentre and centroid coincide, then the triangle must be

Detailed Solution: Question 12

In an equilateral triangle, orthocentre, circum-cente, incentre and centroid coincide.
Hence, option C is correct.

MCQ: Triangles - 2 - Question 13
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Directions: Kindly study the following Question carefully and choose the right answer:
Let ABC is triangle right angled at B. If AB = 6 cm and BC = 8 cm, then what is the length of the circumradius of the ΔABC?

Detailed Solution: Question 13

ΔABC is right angled at B.
Using Pythagoras theorem,
AC2 = AB2 + BC2
AC = 10 cm

and in case of right angled triangle, radius lies on h hypotenuse and is the circumcircle of ΔABC.
Radius of circumcircle = 10/2 = 5 cm
Hence, option D is correct.

MCQ: Triangles - 2 - Question 14
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Directions: Study the following question carefully and choose the right answer:

The coordinates of the in centre of the triangle whose sides are 3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0, are

Detailed Solution: Question 14

3x – 4y ≡ 0 ...(i)
5x + 12y ≡ 0 ....(ii)
y – 15 ≡ 0 ...(iii)

From (i) and (ii), A = (0, 0)
From (i) and (iii), B = (20, 15)
From (ii) and (iii), C = (–36, 15)

Let (α, β) be the incentre co-ordinates of ΔABC

Hence, option B is correct.

MCQ: Triangles - 2 - Question 15
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Directions: Study the following question carefully and choose the right answer:

If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then.

Detailed Solution: Question 15

The line segments joining the mid points of the sides of a triangle form four triangles, each of
which is similar to the original triangle.
Hence, option A is correct.

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About this SSC CGL Practice Test

This test contains 15 MCQs on MCQ: Triangles - 2 with detailed solutions for each question. It is part of the Quantitative Aptitude for SSC CGL course on EduRev, designed to align with the current SSC CGL syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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