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 here 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.

**Below are the MCQs for Triangles**

**1.Which triangles are similar?**

(a)Scalene

(b)Isosceles

(c)Equilateral

(d)None of these

Answer: (c)

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

(a)2.5

(b)3

(c)5

(d)6

Answer: b

Explanation: By midpoint theorem,

DE=½ BC

DE = ½ of 6

DE=3cm

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

(a)20cm

(b)8cm

(c)10cm

(d)9cm

Answer: c

Explanation: Here half of diagonals of rhombus are the sides of triangle and side of 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=10cm

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

(a)230 sq.cm.

(b)106 sq.cm

(c)107 sq.cm.

(d)108 sq.cm

Answer: d

Solution: Let A_{1} and A_{2} are areas of 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 100cm and the length of two sides are 30cm and 40cm, the length of third side will be:**

(a)30cm

(b)40cm

(c)50cm

(d)60cm

Answer: a

Solution: Perimeter of triangle = sum of all its sides

P = 30+40+x

100=70+x

x=30cm

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

(a)22cm

(b)20cm

(c)21cm

(d)18cm

Answer: d

Explanation: ABC ~ DEF

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

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6cm

AC = (12.4)/6 = 8cm

Perimeter = AB+BC+AC

= 4+6+8

=18cm

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

(a)4.33

(b)3.9

(c)5

(d)4

Answer: a

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

Therefore,

BD=DC= 2.5cm

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

**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

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

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Thank u so much sir!!