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SSC CGL Exam  >  SSC CGL Test  >  Quantitative Aptitude  >  MCQ: Triangles - 1 - SSC CGL MCQ

Triangles - 1 - Free MCQ Practice Test with solutions, SSC CGL Quant Aptitude


MCQ Practice Test & Solutions: MCQ: Triangles - 1 (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 - 1". 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 - 1 - Question 1
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Directions: Study the following question carefully and choose the right answer:

Consider the following statements
I. If G is the centroid of ΔABC, then GA = GB = GC.
II. If H is the orthocentre of ΔABC, then HA = HB = HC.
Which of the statements given above is/are correct?

Detailed Solution: Question 1

GA = GB = GC is true only and only for equilateral triangle and here it is not given that ABC is an equilateral triangle. So, only for equilateral triangle.
Hence, it is also not correct.
Hence, option D is correct

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

The in-radius of an equilateral triangle is of length 3 cm. Then the length of each of its medians is

Detailed Solution: Question 2

In equilateral triangle centroid, incentre, orthocentre coincide at the same point.

∴ Height = 3 × in-radius = 3 × 3 = 9 cm.
Hence, option D is correct.

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MCQ: Triangles - 1 - Question 3
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Directions: Study the following question carefully and choose the right answer:

ABC is an equilateral triangle and CD is the internal bisector of ∠C. If DC is produced to E such that AC = CE, then ∠CAE is equal to

Detailed Solution: Question 3

∠BCA = 60° [∵ ΔPQR is an equilateral]
∠BCD = ∠DCA = 30° [∵ CD is bisector of ∠C of an equilateral triangle]
∠DCE = 180°
∠ACE = 180° – 30° = 150°
∠CAE + ∠CEA = 180° – 150° = 30° ...(i)
Given, AC = CE
∴ ∠CEA = ∠CAE
From Eq. (i),
2∠CAE = 30°
∴ ∠CAE = 15°
Hence, option D is correct.

MCQ: Triangles - 1 - Question 4
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Directions: Study the following question carefully and choose the right answer:

AB is a straight line, C and D are points the same side of AB such that AC is perpendicular to AB and DB is perpendicular to AB. Let AD and BC meet at E.
what is  equal to?

Detailed Solution: Question 4

Since, AB is a straight line and C and D are points such that AC ⊥ AB and BD ⊥ AB.

∴ AC || BD
So, ABCD forms trapezium.
Now, by property of trapezium diagonals intersect each other in the ratio of lengths of parallel sides.

But the value of  Can't be determined.
So, we can't find the value of 

Hence, option D is correct.

MCQ: Triangles - 1 - Question 5
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Directions: Study the following question carefully and choose the right answer:

If the orthocentre and the centroid of a triangle are the same, then the triangle is :

Detailed Solution: Question 5

In equilateral triangle orthocentre and centroid lie at the same point.
Hence, option C is correct.

MCQ: Triangles - 1 - Question 6
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Directions: Study the following question carefully and choose the right answer:

G is the centroid of the equilateral ΔABC. If AB = 10 cm then length of AG is

Detailed Solution: Question 6

AB = 10 cm
∴ BD = AB/2 = 5cm


∠ADB = 90°
By pythagoras theorem in ΔABD,

We know that,

Hence, option B is correct.

MCQ: Triangles - 1 - Question 7
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Directions: Study the following question carefully and choose the right answer:

The three sides of a triangle are 15, 25 and x units. Which one of the following is correct?

Detailed Solution: Question 7

In a triangle
Sum to two sides is always greater than 3rd side
i.e., x < 40 ..... (i)
Difference of two sides is always less than 3rd side
i.e., 10 < x ..... (7ii)
From Eqs. (i) and (ii),
10 < x < 40.
Hence, option A is correct.

MCQ: Triangles - 1 - Question 8
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Directions: Study the following question carefully and choose the right answer:

If in a triangle, the circumcentre, incentre, centroid and orthocentre coincide, then the triangle is

Detailed Solution: Question 8

In an equilateral triangle, centroid, incentre etc lie at the same point.
Hence, option D is correct

MCQ: Triangles - 1 - Question 9
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Directions: Study the following question carefully and choose the right answer:

The radius of the incircle of the equilateral triangle having each side 6 cm is

Detailed Solution: Question 9

Smart way :
Note : Radius of incircle of an equilateral triangle of side 

∴ Required radius of the incircle 

Traditional method :
AB = 6 cm
∴ BD = AB = 3 cm
2
∠ADB = 90°
By pythagoras theorem in ΔABD,

We know that,

Hence, option B is correct.

MCQ: Triangles - 1 - Question 10
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Directions: Study the following question carefully and choose the right answer:

Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units? Where, n is a positive real number.

Detailed Solution: Question 10

By hit and trial method,
Put n = 2 in option (d)
= [(2 × 2), (2)2 – 1, (2)2 + 1] = (4, 3, 5)
Which satisfy Pythagoras theorem and one side differes from hypotenuse by 2 units.
Hence, option D is correct.

MCQ: Triangles - 1 - Question 11
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Directions: Study the following question carefully and choose the right answer:

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

Detailed Solution: Question 11

Triangle will be equilateral.
Hence, option B is correct.

MCQ: Triangles - 1 - Question 12
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Directions: Study the following question carefully and choose the right answer:

If the three medians of a triangle are same then the triangle is

Detailed Solution: Question 12

The median of an equilateral triangle are equal.
Hence, option A is correct.

MCQ: Triangles - 1 - Question 13
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Directions: Study the following question carefully and choose the right answer:

The sides of a right angled triangle are equal to three consecutive numbersexpressed in centimeters. What can be the area of such a triangle?

Detailed Solution: Question 13

Since, the triangle is right angled. So, all the three consecutive sides must satisfy Pythagoras theorem.

Hence, 3, 4 and 5 are the sides of triangle which satisfy pythagoras theoram.

Hence, option A is correct.

MCQ: Triangles - 1 - Question 14
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Directions: Study the following question carefully and choose the right answer:

If ABC is an equilateral triangle and D is a point on BC such that AD ⊥ BC, then

Detailed Solution: Question 14


Let AB = BC = CA = 2x units
We know that a perpendicular from any vertex of an equilateral triangle bisects the opposite side.
∴ BD = CD = x units
∴ AB : BD = 2x : x = 2 : 1.
Hence, option C is correct.

MCQ: Triangles - 1 - Question 15
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Directions: Study the following question carefully and choose the right answer:

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

Detailed Solution: Question 15

AC = BC = 5 cm

Hence, option C is correct.

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About this SSC CGL Practice Test

This test contains 15 MCQs on MCQ: Triangles - 1 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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