Class 10 Maths Chapter 6 (Triangles) MCQs are provided here online for students, who are preparing for the board exam, along with their answers. The objective questions are as per the latest CBSE syllabus and NCERT curriculum. These multiple-choice questions for triangles are presented here with detailed explanations, by which students can easily score good marks. Also, check chapter-wise MCQs for 10th Maths subject.

## Class 10 Maths MCQs for Triangles

Multiple choice questions for triangles are given below for students to enhance their problem-solving abilities and confidence. Triangles are the most common concept whose applications are seen frequently in day to day life. Solve the objective questions here and learn more about triangles. Get important questions for class 10 Maths here as well.

Click here to download the PDF of additional MCQs for Practice on Arithmetic Progression, Chapter of Class 10 Maths along with answer key:

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Students can also get access to Triangles Class 10 Notes here.

**Below are the MCQs for Triangles**

**1. Which of the following triangles have the same side lengths?**

(a) Scalene

(b) Isosceles

(c) Equilateral

(d) None of these

Answer: **(c) Equilateral**

Explanation: Equilateral triangles have all its sides and all angles equal.

**2. Area of an equilateral triangle with side length a is equal to:**

(a) √3/2a

(b) √3/2a^{2}

(c) √3/4 a^{2}

(d) √3/4 a

Answer:** (c) √3/4 a ^{2}**

Area of an equilateral triangle with side length a = √3/4 a^{2}

**3. D and E are the midpoints of side AB and AC of a triangle ABC, respectively and BC = 6 cm. If DE || BC, then the length (in cm) of DE is:**

(a) 2.5

(b) 3

(c) 5

(d) 6

Answer: **(b) 3**

Explanation: By midpoint theorem,

DE=½ BC

DE = ½ of 6

DE=3 cm

**4. The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of rhombus in length is:**

(a) 20 cm

(b) 8 cm

(c) 10 cm

(d) 9 cm

Answer: **(c) 10 cm**

Explanation: Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.

By Pythagoras theorem,

(16/2)^{2}+(12/2)^{2}=side^{2}

8^{2}+6^{2}=side^{2}

64+36=side^{2}

side=10 cm

**5. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is**:

(a) 230 sq.cm.

(b) 106 sq.cm

(c) 107 sq.cm.

(d) 108 sq.cm

Answer: **(d) 108 sq.cm**

Solution: Let A_{1} and A_{2} are areas of the small and large triangle.

Then,

A_{2}/A_{1}=(side of large triangle/side of small triangle)

A_{2}/48=(3/2)^{2}

A_{2}=108 sq.cm.

**6. If perimeter of a triangle is 100 cm and the length of two sides are 30 cm and 40 cm, the length of third side will be:**

(a) 30 cm

(b) 40 cm

(c) 50 cm

(d) 60 cm

Answer: **(a) 30 cm**

Solution: Perimeter of triangle = sum of all its sides

P = 30+40+x

100=70+x

x=30 cm

**7. If triangles ABC and DEF are similar and AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm, the perimeter of triangle is:**

(a) 22 cm

(b) 20 cm

(c) 21 cm

(d) 18 cm

Answer:** (d) 18 cm**

Explanation: ABC ~ DEF

AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6 cm

AC = (12.4)/6 = 8 cm

Perimeter = AB+BC+AC

= 4+6+8

=18 cm

**8. The height of an equilateral triangle of side 5 cm is:**

(a) 4.33 cm

(b) 3.9 cm

(c) 5 cm

(d) 4 cm

Answer: **(a) 4.33 cm**

Explanation: The height of the equilateral triangle ABC divides the base into two equal parts at point D.

Therefore,

BD=DC= 2.5 cm

In triangle ABD, using Pythagoras theorem,

AB^{2}=AD^{2}+BD^{2}

5^{2}=AD^{2}+2.5^{2}

AD^{2} = 25-6.25

AD^{2}=18.75

AD=4.33 cm

**9. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if**

(a) ∠A=∠F

(b) ∠B=∠D

(c) ∠A=∠D

(d) ∠B=∠E

Answer: **(b) ∠B=∠D**

If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if ∠B=∠D.

**10. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio**

(a) 2: 3

(b) 4: 9

(c) 81: 16

(d) 16: 81

Answer: **(d) 16: 81**

Explanation: Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB^{2}/DE^{2}

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)^{2} = 16/81 = 16: 81

**11. Which of the following are not similar figures?**

(a) Circles

(b) Squares

(c) Equilateral triangles

(d) Isosceles triangles

Answer: **(d) Isosceles triangles**

Explanation:

All circles, squares, and equilateral triangles are similar figures.

**12. In triangle ABC, ∠BAC = 90° and AD ⊥ BC. Then**

(A) BD . CD = BC^{2}

(B) AB . AC = BC^{2}

(C) BD . CD = AD^{2}

(D) AB . AC = AD^{2}

Answer: **(c) BD . CD = AD ^{2}**

Explanation:

In ΔADB and ΔADC,

∠D = ∠D = 90°

∠DBA = ∠DAC

By AAA similarity criterion,

ΔADB ~ ΔADC

BD/AD = AD/CD

BD.CD = AD2

**13. If in two triangles ABC and PQR, AB/QR = BC/PR = CA/PQ, then**

(a) ΔPQR ~ ΔCAB

(b) ΔPQR ~ ΔABC

(c) ΔCBA ~ ΔPQR

(d) ΔBCA ~ ΔPQR

Answer: (a) ΔPQR ~ ΔCAB

Explanation:

Given that, in triangles ABC and PQR, AB/QR = BC/PR = CA/PQ

If sides of one triangle are proportional to the side of the other triangle, and their corresponding angles are also equal, then both the triangles are similar by SSS similarity. Therefore, ΔPQR ~ ΔCAB

**14. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are**

(a) congruent but not similar

(b) similar but not congruent

(c) neither congruent nor similar

(d) congruent as well as similar

Answer: **(b) similar but not congruent**

Explanation:

In ΔABC and ΔDEF,

∠B = ∠E, ∠F = ∠C and AB = 3 DE

By AA similarity criterion,

ΔABC ~ ΔDEF

AB = 3DE

⇒ AB/DE = 3

⇒ AB/DE = BC/EF = AC/DF = 3

For triangles to be congruent, the ratio of sides must be 1

Therefore, triangles are similar but not congruent.

**15. It is given that ΔABC ~ ΔPQR, with BC/QR = 1/4 then, ar(ΔPRQ)/ar(ABC) is equal to**

(a) 16

(b) 4

(c) 1/4

(d) 1/16

Answer: **(a) 16**

Explanation:

Given,

ΔABC ~ ΔPQR

and BC/QR = 1/4

Ratio of area of similar triangles is equal to the square of its corresponding sides.

So, ar(ΔPRQ)/ar(ABC) = (QR/BC)^{2} = (4/1)^{2} = 16

**16. ****It is given that ΔABC ~ ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:**

(a) DE = 12 cm, ∠F = 50°

(b) DE = 12 cm, ∠F = 100°

(c) EF = 12 cm, ∠D = 100°

(d) EF = 12 cm, ∠D = 30°

Answer: **(b) DE = 12 cm, ∠F = 100°**

Explanation:

Given,

ΔABC ~ ΔDFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm

In triangle ABC,

∠A + ∠B + ∠C = 180°

∠B = 180° – 30° – 50° = 100°

Since ΔABC ~ ΔDFE, the corresponding angles are equal.

Thus, ∠D = ∠A = 30°

∠F = ∠B = 100°

∠E = ∠C = 50°

And

AB/DF = AC/DE

5/7.5 = 8/DE

DE = (8 × 7.5)/5 = 12 cm

**17. If triangle ABC is similar to triangle DEF, then,**

(a) AB/FD = BC/EF = CA/DE

(b) AB/DE = BC/DF = CA/EF

(c) AB/DE = BC/EF = CA/FD

(d) AB/BC = CA/DE = EF/FD

Answer: **(c) AB/DE = BC/EF = CA/FD**

Explanation:

If two triangles are similar, i.e. when ΔABC ~ ΔDEF, then

(i) their corresponding angles are equal and

∠A = ∠D, ∠B = ∠E, ∠C = ∠F and

(ii) their corresponding sides are in the same ratio (or proportion).

AB/DE = BC/EF = CA/FD

**18. Which of the following is not a similarity criterion for two triangles?**

(a) AAA

(b) SAS

(c) SSS

(d) ASA

Answer: **(d) ASA**

Explanation:

The main criteria for similarity of two triangles are AAA, AA, SAS and SSS.

**19. The ratio of the areas of two similar triangles is equal to**

(a) square of the ratio of their corresponding sides

(b) cube of the ratio of their corresponding sides

(c) square root of the ratio of their corresponding sides

(d) twice the ratio of their corresponding sides

Answer: **(a) square of the ratio of their corresponding sides**

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

**20. In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is**

(a) 120°

(b) 60°

(c) 90°

(d) 45°

Answer: **(c) 90°**

Explanation:

Given,

In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

Here, AC is the longest side.

If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.

So, AC^{2} = AB^{2} + BC^{2}

(12)^{2} = (6√3)^{2} + (6)^{2}

144 = 108 + 36

144 = 144

∴ ∆ABC is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is ∠B and is equal to 90°.

tks for giving

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