Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15B

Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15B are designed with the aim of improving problem solving abilities among students. The classification of triangles based on the angles and length of sides are the major concepts discussed under this exercise. Students can cross check their answers and the method of solving using the solutions created by the faculty at BYJU’S. The solved examples create better conceptual knowledge, which are important from the exam perspective. Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15B, PDF links are given below for free download.

Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15B Download PDF

Access other exercises of Selina Solutions Concise Maths Class 7 Chapter 15: Triangles

Exercise 15A Solutions

Exercise 15C Solutions

Access Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15B

Exercise 15B page: 180

1. Find the unknown angles in the given figures:

Solution:

(i) From the figure (i)

x = y as the angles opposite to equal sides

In a triangle

x + y + 80⁰ = 180⁰

Substituting the values

x + x + 80⁰ = 180⁰

By further calculation

2x = 180⁰ – 80⁰ = 100⁰

x = 100⁰/2 = 50⁰

Therefore, x = y = 50⁰.

(ii) From the figure (ii)

b = 40⁰ as the angles opposite to equal sides

In a triangle

a + b + 40⁰ = 180⁰

Substituting the values

a + 40⁰ + 40⁰ = 180⁰

By further calculation

a = 180 – 80 = 100⁰

Therefore, a = 100⁰ and b = 40⁰.

(iii) From the figure (iii)

x = y as the angles opposite to equal sides

In a triangle

x + y + 90⁰ = 180⁰

Substituting the values

x + x + 90⁰ = 180⁰

By further calculation

2x = 180 – 90 = 90⁰

x = 90/2 = 45⁰

Therefore, x = y = 45⁰.

(iv) From the figure (iv)

a = b as the angles opposite to equal sides are equal

In a triangle

a + b + 80⁰ = 180⁰

Substituting the values

a + a + 80⁰ = 180⁰

By further calculation

2a = 180 – 80 = 100⁰

a = 100⁰/2 = 50⁰

Here a = b = 50⁰

We know that in a triangle the exterior angle is equal to sum of its opposite interior angles

x = a + 80⁰

So we get

x = 50 + 80 = 130⁰

Therefore, a = 50⁰, b = 50⁰ and x = 130⁰.

(v) From the figure (v)

In an isosceles triangle consider each equal angle = x

x + x = 86⁰

2x = 86⁰

So we get

x = 86⁰/2 = 43⁰

For a linear pair

p + x = 180⁰

Substituting the values

p + 43⁰ = 180⁰

By further calculation

p = 180 – 43 = 137⁰

Therefore, p = 137⁰.

2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:

Solution:

(i) a = 70⁰ as the angles opposite to equal sides are equal

In a triangle

a + 70⁰ + x = 180⁰

Substituting the values

70⁰ + 70⁰ + x = 180⁰

By further calculation

x = 180 – 140 = 40⁰

y = b as the angles opposite to equal sides are equal

Here a = y + b as the exterior angle is equal to sum of interior opposite angles

70⁰ = y + y

So we get

2y = 70⁰

y = 70⁰/2 = 35⁰

Therefore, x = 40⁰ and y = 35⁰.

(ii) From the figure (ii)

Each angle is 60⁰ in an equilateral triangle

In an isosceles triangle

Consider each base angle = a

a + a + 100⁰ = 180⁰

By further calculation

2a = 180 – 100 = 80⁰

So we get

a = 80⁰/2 = 40⁰

x = 60⁰ + 40⁰ = 100⁰

y = 60⁰ + 40⁰ = 100⁰

(iii) From the figure (iii)

130⁰ = x + p as the exterior angle is equal to the sum of interior opposite angles

It is given that the lines are parallel

Here p = 60⁰ is the alternate angles and y = a

In a linear pair

a + 130⁰ = 180⁰

By further calculation

a = 180 – 130 = 50⁰

Here x + p = 130⁰

Substituting the values

x + 60⁰ = 130⁰

By further calculation

x = 130 – 60 = 70⁰

Therefore, x = 70⁰, y = 50⁰ and p = 60⁰.

(iv) From the figure (iv)

x = a + b

Here b = y and a = c as the angles opposite to equal sides are equal

a + c + 30⁰ = 180⁰

Substituting the values

a + a + 30⁰ = 180⁰

By further calculation

2a = 180 – 30 = 150⁰

a = 150/2 = 75⁰

We know that

b + y = 90⁰

Substituting the values

y + y = 90⁰

2y = 90⁰

y = 90/2 = 45⁰

where b = 45⁰

Therefore, x = a + b = 75 + 45 = 120⁰ and y = 45⁰.

(v) From the figure (v)

a + b + 40⁰ = 180⁰

So we get

a + b = 180 – 40 = 140⁰

The angles opposite to equal sides are equal

a = b = 140/2 = 70⁰

x = b + 40⁰ = 70⁰ + 40⁰= 110⁰

Here the exterior angle of a triangle is equal to the sum of its interior opposite angles

In the same way

y = a + 40⁰

Substituting the values

y = 70⁰ + 40⁰ = 110⁰

Therefore, x = y = 110⁰.

3. The angle of vertex of an isosceles triangle is 100°. Find its base angles.

Solution:

Consider ∆ ABC

Here AB = AC and ∠B = ∠C

We know that

∠A = 100⁰

In a triangle

∠A + ∠B + ∠C = 180⁰

Substituting the values

100⁰ + ∠B + ∠B = 180⁰

By further calculation

2∠B = 180⁰ – 100⁰ = 80⁰

∠B = 80/2 = 40⁰

Therefore, ∠B = ∠C = 40⁰.

4. One of the base angles of an isosceles triangle is 52°. Find its angle of vertex.

Solution:

It is given that the base angles of isosceles triangle ABC = 52⁰

Here ∠B = ∠C = 52⁰

In a triangle

∠A + ∠B + ∠C = 180⁰

Substituting the values

∠A + 52⁰ + 52⁰ = 180⁰

By further calculation

∠A = 180 – 104 = 76⁰

Therefore, ∠A = 76⁰.

5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.

Solution:

Consider the vertical angle of an isosceles triangle = x

So the base angle = 4x

In a triangle

x + 4x + 4x = 180⁰

By further calculation

9x = 180⁰

x = 180/9 = 20⁰

So the vertical angle = 20⁰

Each base angle = 4x = 4 × 20⁰ = 80⁰

6. The vertical angle of an isosceles triangle is 15° more than each of its base angles. Find each angle of the triangle.

Solution:

Consider the angle of the base of isosceles triangle = x⁰

So the vertical angle = x + 15⁰

In a triangle

x + x + x + 15⁰ = 180⁰

By further calculation

3x = 180 – 15 = 165⁰

x = 165/3 = 55⁰

Therefore, the base angle = 55⁰

Vertical angle = 55 + 15 = 70⁰.

7. The base angle of an isosceles triangle is 15° more than its vertical angle. Find its each angle.

Solution:

Consider the vertical angle of the isosceles triangle = x⁰

Here each base angle = x + 15⁰

In a triangle

x + 15⁰ + x + 15⁰ + x = 180⁰

By further calculation

3x + 30⁰ = 180⁰

3x = 180 – 30 = 150⁰

x = 150/3 = 50⁰

Therefore, vertical angle = 50⁰ and each base angle = 50 + 15 = 65⁰.

8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.

Solution:

Consider each base angle of an isosceles triangle = x

Vertical angle = 3 (x + x) = 3 (2x) = 6x

In a triangle

6x + x + x = 180⁰

By further calculation

8x = 180⁰

x = 180/8 = 22.5⁰

Therefore, each base angle = 22.5⁰ and vertical angle = 3 (22.5 + 22.5) = 3 × 45 = 135⁰.

9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.

Solution:

It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4

Consider base angle = x

Vertical angle = 4x

In a triangle

x + x + 4x = 180⁰

By further calculation

6x = 180⁰

x = 180/6 = 30⁰

Therefore, each base angle = x = 30⁰ and vertical angle = 4x = 4 × 30⁰ = 120⁰.

10. In the given figure, BI is the bisector of ∠ABC and CI is the bisector of ∠ACB. Find ∠BIC.

Solution:

In ∆ ABC

BI is the bisector of ∠ABC and CI is the bisector of ∠ACB

Here AB = AC

∠B = ∠C as the angles opposite to equal sides are equal

We know that ∠A = 40⁰

In a triangle

∠A + ∠B + ∠C = 180⁰

Substituting the values

40⁰ + ∠B + ∠B = 180⁰

By further calculation

40⁰ + 2∠B = 180⁰

2∠B = 180 – 40 = 140⁰

∠B = 140/2 = 70⁰

Here BI and CI are the bisectors of ∠ABC and ∠ACB

∠IBC = ½ ∠ABC = ½ × 70⁰ = 35⁰

∠ICB = ½ ∠ACB = ½ × 70⁰ = 35⁰

In ∆ IBC

∠BIC + ∠IBC + ∠ICB = 180⁰

Substituting the values

∠BIC + 35⁰ + 35⁰ = 180⁰

By further calculation

∠BIC = 180 – 70 = 110⁰

Therefore, ∠BIC = 110⁰.

11. In the given figure, express a in terms of b.

Solution:

From the ∆ ABC

BC = BA

∠BCA = ∠BAC

Here the exterior ∠CBE = ∠BCA + ∠BAC

a = ∠BCA + ∠BCA

a = 2∠BCA …… (1)

Here ∠ACB = 180⁰ – b

Where ∠ACD and ∠ACB are linear pair

∠BCA = 180⁰ – b ……. (2)

We get

a = 2 ∠BCA

Substituting the values

a = 2 (180⁰ – b)

a = 360⁰ – 2b

12. (a) In Figure (i) BP bisects ∠ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects ∠ABC and ∠ADB = 70°.

Solution:

(a) From the figure (i)

AB = AC and BP bisects ∠ABC

AP is drawn parallel to BC

Here PB is the bisector of ∠ABC

∠PBC = ∠PBA

∠APB = ∠PBC are alternate angles

x = ∠PBC ….. (1)

In ∆ ABC

∠A = 60⁰

Since AB = AC we get ∠B = ∠C

In a triangle

∠A + ∠B + ∠C = 180⁰

Substituting the values

60⁰ + ∠B + ∠C = 180⁰

We get

60⁰ + ∠B + ∠B = 180⁰

By further calculation

2∠B = 180 – 60 = 120⁰

∠B = 120/2 = 60⁰

½ ∠B = 60/2 = 30⁰

∠PBC = 30⁰

So from figure (i) x = 30⁰

(b) From the figure (ii)

DA = DB = DC

Here BD bisects ∠ABC and ∠ADB = 70⁰

In a triangle

∠ADB + ∠DAB + ∠DBA = 180⁰

Substituting the values

70⁰ + ∠DBA + ∠DBA = 180⁰

By further calculation

70⁰ + 2∠DBA = 180⁰

2∠DBA = 180 – 70 = 110⁰

∠DBA = 110/2 = 55⁰

Here BD is the bisector of ∠ABC

So ∠DBA = ∠DBC = 55⁰

In ∆ DBC

DB = DC

∠DCB = ∠DBC

Hence, x = 55⁰.

13. In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.

Find, in each case: (i) ∠ABE (ii) ∠BAE

Solution:

The sides of a square are equal and each angle is 90⁰

In an equilateral triangle all three sides are equal and all angles are 60⁰

In figure (i) ABCD is a square and ∆ BEC is an equilateral triangle

(i) ∠ABE = ∠ABC + ∠CBE

Substituting the values

∠ABE = 90⁰ + 60⁰= 150⁰

(ii) In ∆ ABE

∠ABE + ∠BEA + ∠BAE = 180⁰

Substituting the values

150⁰ + ∠BAE + ∠BAE = 180⁰

By further calculation

2∠BAE = 180 – 150 = 30⁰

∠BAE = 30/2 = 15⁰

In figure (ii) ABCD is a square and ∆ BEC is an equilateral triangle

(i) ∠ABE = ∠ABC – ∠CBE

Substituting the values

∠ABE = 90⁰ – 60⁰ = 30⁰

(ii) In ∆ ABE

∠ABE + ∠BEA + ∠BAE = 180⁰

Substituting the values

30⁰ + ∠BAE + ∠BAE = 180⁰

By further calculation

2∠BAE = 180 – 30 = 150⁰

∠BAE = 150/2 = 75⁰

14. In ∆ ABC, BA and BC are produced. Find the angles a and h. if AB = BC.

Solution:

In ∆ ABC, BA and BC are produced

∠ABC = 54⁰ and AB = BC

In ∆ ABC

∠BAC + ∠BCA + ∠ABC = 180⁰

Substituting the values

∠BAC + ∠BAC + 54⁰ = 180⁰

2∠BAC = 180 – 54 = 126⁰

∠BAC = 126/2 = 63⁰

∠BCA = 63⁰

In a linear pair

∠BAC + b = 180⁰

Substituting the value

63⁰ + b = 180⁰

So we get

b = 180 – 63 = 117⁰

In a linear pair

∠BCA + a = 180⁰

Substituting the value

63⁰ + a = 180⁰

So we get

a = 180 – 63 = 117⁰

Therefore, a = b = 117⁰.