## CBSE Class 10 Triangles MCQ New Pattern By Clarify Knowledge

*CBSE Class 10 Triangles MCQ New Pattern 2022*

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## CBSE Class 10 Triangles MCQ Table

**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