Selina Solutions for Class 6, maths Chapter 26

Class 6: Maths Chapter 26 solutions. Complete Class 6 Maths Chapter 26 Notes.

Contents

1 Selina Class 6 ICSE Solutions Mathematics : Chapter 26- Triangles
- 1.1 Exercise 26(A)
- 1.2 Exercise 26(B)
2 Download PDF
3 Chapterwise Selina Publishers ICSE Solutions for Class 6 Mathematics :
4 About Selina Publishers ICSE

Selina Class 6 ICSE Solutions Mathematics : Chapter 26- Triangles

Selina 6th Maths Chapter 26, Class 6 Maths Chapter 26 solutions

Exercise 26(A)

1. In each of the following, find the marked unknown angles:

Selina Solutions Concise Mathematics Class 6 Chapter 26 - 1

Solution:

(i) We know that,

Sum of all angles of triangle = 180⁰

Therefore,

70⁰ + 72⁰ + z = 180⁰

142⁰ + z = 180⁰

z = 180⁰ – 142⁰

We get,

z = 38⁰

(ii) We know that,

Sum of all angles of a triangle = 180⁰

First triangle

50⁰ + 80⁰ + b = 180⁰

130⁰ + b = 180⁰

b = 180⁰ – 130⁰

We get,

b = 50⁰

Second triangle

40⁰ + 45⁰ + a = 180⁰

85⁰ + a = 180⁰

a = 180⁰ – 85⁰

We get,

a = 95⁰

(iii) 60⁰ + 45⁰ + 20⁰ + x = 180⁰

125⁰ + x = 180⁰

x = 180⁰ – 125⁰

We get,

x = 55⁰

2. Can a triangle together have the following angles?

(i) 55⁰, 55⁰ and 80⁰

(ii) 33⁰, 74⁰ and 73⁰

(iii) 85⁰, 95⁰ and 22⁰

Solution:

(i) Sum of all angles of a triangle = 180⁰

Here,

55⁰ + 55⁰ + 80⁰ = 180⁰

We get,

190⁰ ≠ 180⁰

Therefore, it cannot form a triangle

(ii) 33⁰ + 74⁰ + 73⁰ = 180⁰

We get,

180⁰ = 180⁰

Therefore, it form a triangle

(iii) 85⁰ + 95⁰ + 22⁰ = 180⁰

We get,

202⁰ ≠ 180⁰

Therefore, it cannot form a triangle

3. Find x, if the angles of a triangle are:

(i) x⁰, x⁰, x⁰

(ii) x⁰, 2x⁰, 2x⁰

(iii) 2x⁰, 4x⁰, 6x⁰

Solution:

We know that,

The sum of all the angles in a triangle is 180⁰

So,

x⁰+ x⁰ + x⁰ = 180⁰

3x = 180⁰

x = 180⁰ / 3

We get,

x = 60⁰

The value of x = 60⁰

(ii) We know that,

The sum of all the angles in a triangle is 180⁰

So,

x + 2x + 2x = 180⁰

5x = 180⁰

x = 180⁰ / 5

We get,

x = 36⁰

Therefore, the value of x = 36⁰

(iii) We know that,

The sum of all the angles in a triangle is 180⁰

So,

2x + 4x + 6x = 180⁰

12x = 180⁰

x = 180⁰ / 12

We get,

x = 15⁰

Therefore, the value of x = 15⁰

4. One angle of a right-angled triangle is 70⁰. Find the other acute angle.

Solution:

We know that,

Sum of all the angles in a triangle = 180⁰

Let us consider the acute angle as x

Hence,

x + 90⁰ + 70⁰ = 180⁰

x + 160⁰ = 180⁰

x = 180⁰ – 160⁰

We get,

x = 20⁰

Therefore, the acute angle is 20⁰

5. In △ABC, ∠A = ∠B = 62⁰; find ∠C

Solution:

Given

∠A = ∠B = 62⁰

So,

∠A + ∠B + ∠C = 180⁰

62⁰ + 62⁰ + ∠C = 180⁰

124⁰ + ∠C = 180⁰

∠C = 180⁰ – 124⁰

We get,

∠C = 56⁰

Hence, ∠C = 56⁰

6. In △ABC, ∠B = ∠C and ∠A = 100⁰; find ∠B.

Solution:

Given

∠B = ∠C

We know that,

Sum of all the angles in a triangle is 180⁰

∠A + ∠B + ∠C = 180⁰

100⁰ + ∠B + ∠B = 180⁰

100⁰ + 2∠B = 180⁰

2∠B = 180⁰ – 100⁰

We get,

2∠B = 80⁰

∠B = 80⁰ / 2

∠B = 40⁰

Therefore, ∠B + ∠C = 40⁰

7. Find, giving reasons, the unknown marked angles, in each triangle drawn below:

Selina Solutions Concise Mathematics Class 6 Chapter 26 - 4

Solution:

We know that,

Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)

So,

(i) 110⁰ = x + 30⁰ [By property]

x = 110⁰ – 30⁰

We get,

x = 80⁰

(ii) x + 115⁰ = 180⁰ [By linear property of angles]

x = 180⁰ – 115⁰

We get,

x = 65⁰

By exterior angle property

x + y = 115⁰

65⁰ + y = 115⁰

y = 115⁰ – 65⁰

We get,

y = 50⁰

Therefore the value of angle x is 65⁰ and y is 50⁰

(iii) By exterior angle property,

110⁰ = 2x + 3x

5x = 110⁰

x = 110⁰ / 5

We get,

x = 22⁰

Hence,

The value of 2x = 2 × 22

= 44⁰

The value of 3x = 3 × 22

= 66⁰

8. Classify the following triangles according to angle:

Selina Solutions Concise Mathematics Class 6 Chapter 26 - 7

Solution:

(i) Since, one of the angle of a triangle is 120⁰.

Therefore, it is obtuse angled triangle

(ii) Since, all the angles of a triangle is less than 90⁰

Therefore, it is acute angled triangle

(iii) Since ∠MNL = 90⁰ and

Sum of two acute angle s,

∠M + ∠N = 30⁰ + 60⁰

= 90⁰

Therefore, it is right angled triangle

9. Classify the following triangles according to sides:

(i)

Selina Solutions Concise Mathematics Class 6 Chapter 26 - 10

(ii)

Selina Solutions Concise Mathematics Class 6 Chapter 26 - 11

Solution:

(i) In the given triangle, we find two sides are equal.

Therefore, it is isosceles triangle

(ii) In the given triangle, all the three sides are unequal.

Therefore, it is scalene triangle

(iii) In the given triangle, all the three sides are unequal.

Therefore, it is scalene triangle

(iv) In the given triangle, all the three sides are equal.

Therefore, it is equilateral triangle

Exercise 26(B)

1. Construct triangle ABC, when:

AB = 6 cm, BC = 8 cm and AC = 4 cm

Solution:

Given

AB = 6 cm

BC = 8 cm

AC = 4 cm

Now,

Steps of Construction:

(i) Draw a line AB of length 6 cm

(ii) Using compasses, take B as centre, and draw an arc of 8 cm radius

(iii) Again, taking A as centre, draw another arc of 4 cm radius, which cuts the previous arc at point C

(iv) Now join AC and BC

The obtained triangle ABC is the required triangle.

2. AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm

Solution:

Given

AB = 3.5 cm

AC = 4.8 cm

BC = 5.2 cm

Steps of Construction:

(i) Draw a line AB of length 3.5 cm

(ii) With the help of compasses, taking B as centre, draw an arc of 5.2 cm radius

(iii) Again with A as centre, draw an arc of 4.8 radius

(iv) Now, join AC and BC

3. AB = BC = 5 cm and AC = 3 cm. Measure angles A and C. Is ∠A = ∠C?

Solution:

Given

AB = BC = 5 cm

AC = 3 cm

Steps of Construction:

(i) Draw a line AB of length 5 cm

(ii) Using compasses take B as centre and draw an arc of 5 cm radius

(iii) Now, taking A as centre, draw another arc of 3 cm radius, which cuts the previous arc at point C

(iv) Now, join AC and BC

Measuring angles A and C, we get
∠A = 72.5o and ∠C = 72.5o
Hence, yes ∠A = ∠C.

4. AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal?

Solution:

Given

AB = BC = CA = 4.5 cm

Steps of Construction:

(i) Draw a line AB of length 4.5 cm

(ii) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius

(iii) Again taking AC as centre, draw another arc of 4.5 cm radius, which cuts the previous arc at point C

(iv) Now, join AC and BC

(v) All the angles in ABC i.e ∠A = ∠B = ∠C = 60⁰

Since AB = BC = CA = 4.5 cm and all the angles are equal. Hence, it is an equilateral triangle

5. AB = 3 cm, BC = 7 cm and ∠B = 90⁰

Solution:

Given

AB = 3 cm,

BC = 7 cm and

∠B = 90⁰

Steps of Construction:

(i) Draw a line segment AB of length 3 cm

(ii) Using compasses, construct ∠ABC = 90⁰

(iii) Taking B as centre, draw an arc of 7 cm length and mark as point C i.e BC = 7 cm

(iv) Now, join A and C

(v) The obtained △ABC, is the required triangle

6. AC = 4.5 cm, BC = 6 cm and ∠C = 60⁰

Solution:

Given

AC = 4.5 cm

BC = 6 cm

∠C = 60⁰

Steps of Construction:

(i) Draw a line AC of length 4.5 cm

(ii) Using compasses, construct ∠ACB = 60⁰

(iii) Draw an arc of 6 cm radius and mark it as B such that BC = 6 cm

(iv) Now, join B and A

7. AC = 6 cm, ∠A = 60⁰ and ∠C = 45⁰. Measure AB and BC.

Solution:

Given

AC = 6 cm

∠A = 60⁰

∠C = 45⁰

Steps of Construction:

(i) Draw a line segment AC of length 6 cm

(ii) With the help of Compass, construct ∠A = 60⁰

(iii) Again, using compass, construct ∠C = 45⁰

(iv) AD and CE intersect each other at point B

(v) Now, the obtained △ABC is the required triangle

(vi) Measure the side AB and BC with the help of a scale

(vii) We get, AB = 4.4 cm and BC = 5.4 cm

8. AB = 5.4 cm, ∠A = 30⁰ and ∠B = 90⁰. Measure ∠C and side BC.

Solution:

Given

AB = 5.4 cm

∠A = 30⁰ and

∠B = 90⁰

Steps of Construction:

(i) Draw a line segment AB of length 5.4 cm

(ii) With the help of compass, construct ∠A = 30⁰

(iii) Similarly, construct ∠B = 90⁰

(iv) AD and BE intersect each other at point C

(v) Hence, the obtained △ABC is the required triangle

(vi) On measuring we get, ∠C = 60 and side BC = 3.1 cm approximately

9. AB = 7 cm, ∠B = 120⁰ and ∠A = 30⁰. Measure AC and BC.

Solution:

Given

AB = 7 cm

∠B = 120⁰

∠A = 30⁰

Steps of Construction:

(i) Draw a line segment AB of length 7 cm

(ii) With the help of a compass, construct ∠A = 30⁰

(iii) AE and BD intersect each other at point C

(iv) Hence, the obtained △ABC is the required triangle

(v) On measuring the lengths, we get AC = 12 cm and BC = 7 cm respectively

10. BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle

Solution:

Given

BC = 3 cm

AC = 4 cm and

AB = 5 cm

Steps of Construction:

(i) Draw a line segment AB of length 5 cm

(ii) From B, using compass cut an arc of 3 cm radius

(iii) Similarly, from A again with the help of compass, cut an arc of 4 cm bisecting the previous arc formed from point B

(iv) Now, join point C with A and B

(v) The obtained triangle is the required △ABC

(vi) On measuring ∠ACB, we get ∠ACB = 90⁰.

Therefore, the obtained triangle ABC is a right angled triangle

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Selina Class 6 ICSE Solutions Mathematics : Chapter 26- Triangles

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Chapterwise Selina Publishers ICSE Solutions for Class 6 Mathematics :

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