# MCQ Questions for Class 10 Maths Triangles with Answers

## Class 10 Maths MCQs Chapter 6 Triangles

1. O is a point on side PQ of a APQR such that PO = QO = RO, then

(a) RS² = PR × QR

(b) PR² + QR² = PQ²

(c) QR² = QO² + RO²

(d) PO² + RO² = PR²

**Answer**

Answer: b

2. In ABC, DE || AB. If CD = 3 cm, EC = 4 cm, BE = 6 cm, then DA is equal to

(a) 7.5 cm

(b) 3 cm

(c) 4.5 cm

(d) 6 cm

**Answer**

Answer: c

3. AABC is an equilateral A of side a. Its area will be…

**Answer**

Answer: a

4. In a square of side 10 cm, its diagonal = …

(a) 15 cm

(b) 10√2 cm

(c) 20 cm

(d) 12 cm

**Answer**

Answer: b

5. In a rectangle Length = 8 cm, Breadth = 6 cm. Then its diagonal = …

(a) 9 cm

(b) 14 cm

(c) 10 cm

(d) 12 cm

**Answer**

Answer: c

6. In a rhombus if d_{1} = 16 cm, d_{2} = 12 cm, its area will be…

(a) 16 × 12 cm²

(b) 96 cm²

(c) 8 × 6 cm²

(d) 144 cm²

**Answer**

Answer: b

7. In a rhombus if d_{1} = 16 cm, d_{2} = 12 cm, then the length of the side of the rhombus is

(a) 8 cm

(b) 9 cm

(c) 10 cm

(d) 12 cm

**Answer**

Answer: c

8. If in two As ABC and DEF, \(\frac{\mathrm{AB}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{FE}}=\frac{\mathrm{CA}}{\mathrm{ED}}\), then

(a) ∆ABC ~ ∆DEF

(b) ∆ABC ~ ∆EDF

(c) ∆ABC ~ ∆EFD

(d) ∆ABC ~ ∆DFE

**Answer**

Answer: d

9. It is given that ∆ABC ~ ∆DEF and \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{1}{5}\). Then \(\frac{\operatorname{ar}(D E F)}{\operatorname{ar}(A B C)}\) is equal to

(а) 5

(b) 25

(c) \(\frac{1}{25}\)

(d) \(\frac{1}{5}\)

**Answer**

Answer: b

10. In ∠BAC = 90° and AD ⊥ BC. A Then

(а) BD.CD = BC²

(б) AB.AC = BC²

(c) BD.CD = AD²

(d) AB.AC = AD²

**Answer**

Answer: c

11. D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is

(a) 2.5

(b) 3

(c) 5

(d) 6

**Answer**

Answer: b

12. If ΔABC ~ ΔDEF and ΔABC is not similar to ΔDEF then which of the following is not true?

(a) BC.EF = AC.FD

(b) AB.ED = AC.DE

(c) BC.DE = AB.EE

(d) BC.DE = AB.FD

**Answer**

Answer: c

13. If in two triangles DEF and PQR, ZD = ZQ and ZR = ZE, then which of the following is not true?

**Answer**

Answer: b

14. If ΔABC ~ ΔPQR, \(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}\) then \(\frac{\operatorname{ar}(PQR)}{\operatorname{ar}(BCA)}\) is

(a) 9

(b) 3

(c) 1/3

(d) 1/9

**Answer**

Answer: a

15. If ΔABC ~ ΔQRP, \(\frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{PQR})}=\frac{9}{4}\), AB = 18 cm and BC = 15 cm, then PR is equal to

(a) 10 cm

(b) 12 cm

(c) \(\frac{20}{3}\)cm

(d) 8 cm

**Answer**

Answer: a

16. If in triangles ABC and DEF,\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}\) , then they will be similar, if

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

**Answer**

Answer: c

17. In the given figure, \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) and ∠ADC = 70° ∠BAC =

(a) 70°

(b) 50°

(c) 80°

(d) 60°

**Answer/Explanation**

Answer: d

Explaination:

∵ DE || BC

∴ ∠ABC = 70°.

(Corresponding angles) Using angle sum property of triangle ∠ABC + ∠BCA + ∠BAC = 180° ∠BCA = 60°.

18. In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then

(a) DE || BC

(b) DF || AC

(c) EF || AB

(d) none of these

**Answer/Explanation**

Answer: c

Explaination:

19. If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true? [NCERT Exemplar Problems] (a) BC . EF = AC . FD

(b) AB . EF = AC . DE

(c) BC . DE = AB . EF

(d) BC . DE = AB . FD

**Answer/Explanation**

Answer: c

Explaination:

20. If in two triangles ABC and PQR, \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}\), then [NCERT Exemplar Problems] (a) ΔPQR ~ ΔCAB

(b) ΔPQR ~ ΔABC

(c) ΔCBA ~ ΔPQR

(d) ΔBCA ~ ΔPQR

**Answer/Explanation**

Answer: a

Explaination:

21. Match the column:

(a) 1 → A, 2 → B, 3 → C, 4→ D

(b) 1 → D, 2 → A, 3 → C, 4 → B

(c) 1 → B, 2 → A, 3 → C, 4 → D

(d) 1 → C, 2 → B, 3 → D, 4 → A.

**Answer/Explanation**

Answer: c

Explaination:

(c) Properties of triangles

22. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are [NCERT Exemplar Problems] (a) congruent but not similar

(b) similar but not congruent

(c) neither congruent nor similar

(d) congruent as well as similar

**Answer/Explanation**

Answer: b

Explaination: (b) Similar but not congruent

23. If in triangles ABC and DEF, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}\) then they will be similar, when [NCERT Exemplar Problems] (a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

**Answer/Explanation**

Answer: c

Explaination:

24. ABC and BDE are two equilateral triangles such that D is mid-point of BC. Ratio of the areas of triangles ABC and BDE is

(a) 2 : 1

(b) 1:4

(c) 1:2

(d) 4:1

**Answer/Explanation**

Answer: d

Explaination:

25. Sides of two similar triangles are in the ratio 3 : 7. Areas of these triangles are in the ratio

(a) 9 : 35

(b) 9 : 49

(c) 49 : 9

(d) 9 : 42

**Answer/Explanation**

Answer: b

Explaination:

(b) Ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.

Required ratio = 9 : 49.

26. Sides of triangles are (i) 3 cm, 4 cm, 6 cm. (ii) 4 cm, 5 cm, 6 cm. (iii) 7 cm, 24 cm, 25 cm (iv) 5 cm, 12 cm, 14 cm. Which of these is right triangle?

(a) (i)

(b) (ii)

(c) (iii)

(d) (iv)

**Answer/Explanation**

Answer: c

Explaination:

On verification, triangle with sides 7 cm, 24 cm and 25 cm is a right triangle

∵ (25)² = (7)² + (24)².

27. If in two triangles DEF and PQR, ZD = ZQ and ZR = ZE, then which of the following is not true?

**Answer/Explanation**

Answer: b

Explaination: (b) Because ΔDEF ~ ΔQRP

28. “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.” This theorem is known as _____ .

**Answer/Explanation**

Answer:

Explaination:

Thales Theorem or Basic Proportionality Theorem

29. In ΔABC, D and E are points on sides AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 6.6 cm then find AC.

**Answer/Explanation**

Answer:

Explaination:

30. In the given figure, value of x (in cm) is _____ .

**Answer/Explanation**

Answer:

Explaination:

31. In the given figure, if ΔABC ~ ΔPQR

The value of x is_____ .

**Answer/Explanation**

Answer:

Explaination:

32. The perimeter of two similar triangles ABC and LMN are 60 cm and 48 cm respectively. If LM = 8 cm, then what is the length of AB?

**Answer/Explanation**

Answer:

Explaination:

33. In fig. ZM = ZN = 46°, express x in terms of a, b and c, where a, b and c are lengths of LM, MN and NK respectively.

**Answer/Explanation**

Answer:

Explaination:

34. ΔABC ~ ΔPQR. Area of ΔABC = 81 cm² and area of ΔPQR =121 cm². If altitude AD = 9 cm, then PM = ______ .

**Answer/Explanation**

Answer:

Explaination:

35. Given ΔABC ~ ΔPQR, if \(\frac{A B}{P Q}=\frac{1}{3}\), then find \(\frac{\operatorname{ar} \Delta \mathrm{ABC}}{\operatorname{ar} \Delta \mathrm{PQR}}\). [CBSE 2018]

**Answer/Explanation**

Answer:

Explaination:

36. In fig., DE || BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar(ΔABC) to the ar(ΔADE)? [Delhi 2019]

**Answer/Explanation**

Answer:

Explaination:

Given: In ΔABC, DE || BC

Also AD = 1 cm and BD = 2 cm

To find: ar(ΔABC) : ar(ΔADE)

As DE || BC (Given)

∠D = ∠B and ∠E = ∠C (Corresponding angles)

∴ ΔADE ~ ΔABC (By AA similarity)

ar(ΔABC) : ar(ΔADE) = 9:1.

37. In figure, S and T are points on the sides PQ and PR, respectively of APQR, such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ΔPST and ΔPQR.

**Answer/Explanation**

Answer:

Explaination:

S and T are points on the sides PQ and PR of ΔPQR and PT = 2 cm, TR = 4 cm, ST || QR

In APST and ΔPQR,

∠S = ∠Q (Corresponding angles)

∠P = ∠P (Common)

∴ ΔPST ~ ΔPQR (AA similarity)

38. In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is _____ .

**Answer/Explanation**

Answer:

Explaination:

AC² = 144 cm²

AB² = 108 cm²

BC² = 36 cm²

Now, AB² + BC²

= 144 = AC² B

∴ ∠B = 90°

39. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is

(a) 9 cm

(b) 10 cm

(c) 8 cm

(d) 20 cm

**Answer/Explanation**

Answer: b

Explaination:

Diagonals of a Rhombus are perpendicular to each other and bisect each oSo AO = 8 cm

BO = 6 cm

∠AOB = 90°

In right angled ΔAOB

AB² = AO² + OB²

AB² = 8² + 6² = 64 + 36

AB =√100 = 10 cm.

40. The sides of a triangle are 30, 70 and 80 units. If an altitude is droped upon the side of length 80 units, the larger segment cut off on this side is______.

**Answer/Explanation**

Answer:

Explaination:

Now h² = 30² – (80 – x)²

Also h² = 70² – x²

⇒ 30² – (80 – x)² = 70² – x²

⇒ 160x = 10400

⇒ x=65 units.

41. In Fig., ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.

**Answer/Explanation**

Answer:

Explaination:

Given: AC = 4 cm

For isosceles triangle,

AC = BC

∴ BC = 4 cm

Using Pythagoras theorem,

AB² = AC2 + BC²

AB² = (4)2 + (4)²

⇒ AB² = 16 + 16 = 32

⇒ AB = A√32 = 4√2 cm.