Important MCQ questions Class 10 Maths Chapter 6 - Triangles - GMS - Learning Simply
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# Important MCQ questions Class 10 Maths Chapter 6 - Triangles

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# Important MCQ questions Class 10 Maths Chapter 6 - Triangles

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.

## 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.

We have provided some important questions of this chapter, along with the detailed solutions. After that, we have also provided some questions for students practice which does not have solutions. They must solve all of them to gain command over Triangles topic.

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

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)√3/2a2

(c)√3/4 a2

(d)√3/4 a

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

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

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=side2

82+62=side2

64+36=side2

side=10cm

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

Solution: Let A1 and A2 are areas of the small and large triangle.

Then,

A2/A1=(side of large triangle/side of small triangle)

A2/48=(3/2)2

A2=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

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

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

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,

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

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

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) = AB2/DE2

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

11. If △ ABC ~ △ DEF such that AB = 12 cm and DE = 14 cm. Find the ratio of areas of △ ABC and △ DEF.
1. 49/9
2. 36/49
3. 49/16
4. 25/49

Solution: We know that the ratio of areas of two similar triangles is equal

to the ratio of the squares.

Of any two corresponding sides,

area of △ ABC / area of △ DEF = (AB/DE) 2= (12/14) 2= 36/49

12. D and E are points on the sides AB and AC respectively of a △ABC such that DE || BC. Which of the following statement is true?

1. △ ADE ~ △ ABC
1. only (iii)
2. only (i)
3. only (i) and (ii)
4. all (i) , (ii) and (iii)

Solution:

In △ ADE and △ ABC, we have

[Since DE || BC ∠ ADE = ∠ B (Corresponding angles)]

and, ∠ A = △ A [Common]

Therefore, (area of △ ADE / area of △ ABC) = (AD2/AB2)

13.

In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm and area (ΔQOA) = 150 cm2, find the area of ΔPOB.

1. 233 cm2
2. 294 cm2
3. 300 cm2
4. 420 cm2

Solution: Consider Δ~QOA and Δ POB

QA || PB,

Therefore, ∠ AQO = ∠ PBO [Alternate angles]

∠ QAO = ∠ BPO  [Alternate angles]

and

∠ QOA = ∠ BOP [Vertically opposite angles]

Δs QOA ~ BOP  [by AAA similarity]

Therefore, (OQ/ OB) = (OA/OP)

Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52

Since area (QOA) =150cm2

⇒area (POB) =294cm2

14. Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The ratio of corresponding heights is:

1. 4:5
2. 5:4
3. 3:2
4. 5:7

Solution: For similar isosceles triangles,

Area (Δ1) / Area (Δ2) = (h1)2 / (h2)2

(h1 / h2) = 4/5

15. In △ABC, AB = 3 and, AC = 4 cm and AD is the bisector of ∠A. Then, BD : DC is —
1. 9: 16
2. 4:3
3. 3:4
4. 16:9

Solution:

The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides (It may be similar or may not depending on type of triangle it divides)

In △ABC

as per the statement AB/ AC= BD/DC i.e. a/b= c/d

So, BD/ DC= AB/AC= ¾

So, BD: DC = 3: 4

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…

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