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MCQ Questions for Class 10 Maths Triangles with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 6 Triangles Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Triangles MCQs with Answers to know their preparation level.

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 d1 = 16 cm, d2 = 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 d1 = 16 cm, d2 = 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.


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