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CBSE Class 10  >  Class 10 Test  >  Olympiad Preparation  >  Math Olympiad Test: Triangles - 2 - Class 10 MCQ

Math Olympiad Test: Triangles - 2 - Free MCQ with solutions Class 10


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

You can prepare effectively for Class 10 Olympiad Preparation for Class 10 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Math Olympiad Test: Triangles - 2". These 15 questions have been designed by the experts with the latest curriculum of Class 10 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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Math Olympiad Test: Triangles - 2 - Question 1
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In an isosceles triangle AB = AC = 25 cm BC = 14 cm. What is the length of altitude from A on BC?

Detailed Solution: Question 1


Here altitude from A to BC is AD, we know in isosceles altitude on non-equal sides also median, 
so, BD = CD = BC​/2 = 14​/2 = 7cm
Now, we applying Pythagoras theorem in △ABD,
AB2 = BD2 + AD2
(25)2 = (7)2 + AD2
AD2 = 625 − 49
AD2 = 576
AD = 24cm.

Math Olympiad Test: Triangles - 2 - Question 2
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The area of two equilateral triangle are in the ratio 196 : 169. What is the ratio between their perimeters?

Detailed Solution: Question 2

Ratio of perimeters = ratio of sides 

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Math Olympiad Test: Triangles - 2 - Question 3
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D and E are points on the sides AB and AC respectively of a DABC such that DE || BC find value of x if AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x - 4) cm, EC = 3x cm

Detailed Solution: Question 3

∵ DE ║ BE
∴ DADE ~ DABC

⇒
⇒ adjusting this expression, we have
⇒ 
⇒ 
⇒ 21x2 - 12x = 15x2 − 6x − 8 + 20 x
⇒ 6x2 − 26x + 8 = 0
⇒ 6x2 −24x − 2x + 8 = 0
⇒ 6x(x − 4) − 2(x − 4) = 0

When, 
∴ 

Math Olympiad Test: Triangles - 2 - Question 4
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In the given figure in DPQR, ST || QR, so that PS = (7x - 4) cm, PT = (5x - 2) cm, QS = (3x + 4) cm, RT = 3x cm. 

What is the value of x?

Detailed Solution: Question 4

In ΔPQR and ΔPST

⇒ According to Thales’ theorem

Subtracting unity from both sides, we have,

⇒ 21x2 - 12x = 15x2 - 6x + 20x - 8
⇒ 6x2 - 26x + 8 = 0
⇒ 3x2 - 13x + 4 = 0
⇒ 3x2 - 12x - x + 4 = 0
⇒ 3x(x - 4) - 1(x - 4) = 0
⇒ (x - 4) (3x - 1) = 0

∴ x = 4

Math Olympiad Test: Triangles - 2 - Question 5
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A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. What is the width of the street?

Detailed Solution: Question 5


In ΔABE,
AE2 + AB2 = BE2
⇒ 92 + AB2 = 152
⇒ AB2 = 225 - 81
= 144
⇒ AB = 12 m
Similarly,
In DBCD
⇒ BC2 = BD2 - DC2 = 152 - 122 = 92
BC = 9m
∴ width of street = (AB + BC) = 12 + 9 = 21 m

Math Olympiad Test: Triangles - 2 - Question 6
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The corresponding altitude of two similar triangle are 6 cm and 9 cm respectively. What is ratio of their areas?

Detailed Solution: Question 6

Ratio of areas = (ratio of altitudes)2

Math Olympiad Test: Triangles - 2 - Question 7
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In the given figure, ΔODC ~ ΔOBA, ∠BOC = 115°, ∠CDO = 70°. What is the value of ∠OAB?

Detailed Solution: Question 7

∠ CDO + ∠DCO = 115°
⇒ ∠DCO = 115° - 70° = 45°
∠OAB = 45° (∴ DODC ~ DOBA)

Math Olympiad Test: Triangles - 2 - Question 8
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A tree is broken of 6 m from the ground and its top touches the ground at a distance of 8 m from the base of the tree. What is the original height of the tree?

Detailed Solution: Question 8


A is the point of cleavage C is the point where top of three touches the ground
In ΔABC
AB2 + BC2 = AC2 (Pythagoras’ Theorem)
62 + 82 = AC2
= AC = 10 m
∴ Total (original) height = 10 m + 6 m = 16 m

Math Olympiad Test: Triangles - 2 - Question 9
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An exterior angle of a triangle measures 110° and its interior opposite angles are in the ratio 2 : 3. Which is the largest angle of the triangle?

Detailed Solution: Question 9

Let the angles be 2x and 3x respectively
∴ 2x + 3x = 110°
⇒ 5x = 110°
⇒ x = 22°
∴ Angles are 44°, 66°, and 70°

Math Olympiad Test: Triangles - 2 - Question 10
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What is the perimeter of a rhombus the length of whose diagonal are 16 cm and 30 cm?

Detailed Solution: Question 10

Let the side of rhombus be x cm
Now, we know that the diagonals of rhombus bisect each other at right angles.
∴ By using Pythagoras’ Theorem

= 64 + 225 = 289
⇒ x = 17 cm
∴ Perimeter =  4 × x = 4 × 17 = 68 cm.

Math Olympiad Test: Triangles - 2 - Question 11
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In the given figure BC is parallel to DE. Area of DABC is 25 cm2. Area of trapezium BCED is 24 cm2, if DE = 14 cm

What is the length of BC?

Detailed Solution: Question 11

∵  BC ║ DE
∴ 


[Ar(DADE) = Ar(DABC) + Ar(¨BCED)

⇒ BC = 10 cm

Math Olympiad Test: Triangles - 2 - Question 12
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A man goes 10 m due south and then 24 m due west. How far is he from the starting point?

Detailed Solution: Question 12

In ΔABC
AC2 = AB2 + BC2 (Pythagoras’ theorem)

⇒ AC = 

Math Olympiad Test: Triangles - 2 - Question 13
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A vertical pole 12 m long casts a shadow of 8 m long on the ground. At the same time, a tower casts the shadow 40 m long on the ground. What is the height of the tower?

Detailed Solution: Question 13

Both Ds are right angled, and also the elevation of coming sunlight will be same.
∴ All the angles of the ΔABC and ΔPQR will be correspondingly identical.
∴ ΔABC ~ ΔPQR
⇒ AB/PQ = BC/QR 
⇒ 12/PQ = 8/40 
⇒ PQ = 12 × 5 = 60m

Math Olympiad Test: Triangles - 2 - Question 14
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In an isosceles triangle ΔABC if AB = AC = 13 cm and altitude from A on BC is 5 cm. What is the length of BC?

Detailed Solution: Question 14

In ΔABD

AB2 = BD2 + AD2
⇒ BD = 

= 12 cm
∵ D bisects BC
∴ BC = 2 × BD
= 2 × 12 = 24 cm

Math Olympiad Test: Triangles - 2 - Question 15
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The Perimeter of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm. Then what is the corresponding side of the second circle?

Detailed Solution: Question 15

We know that if, ΔABC ~ ΔPQR, then

or, Applying componendo and dividendo, it can be clearly seen that,


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About this Class 10 Practice Test

This test contains 15 MCQs on Math Olympiad Test: Triangles - 2 with detailed solutions for each question. It is part of the Olympiad Preparation for Class 10 course on EduRev, designed to align with the current Class 10 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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