Selina Solutions Concise Maths Class 7 Chapter 15 Triangles provides students with a clear idea about the basic concepts covered in this chapter. The solutions are prepared by faculty after conducting research on each topic, keeping in mind the understanding capacity of students. Here, the students can download Selina Solutions Concise Maths Class 7 Chapter 15 Triangles free PDF, from the links which are provided here.

Chapter 15 helps students understand the types of triangles based on the angles and length of sides. The solutions improve problem solving and analytical thinking skills among students, which are important from the exam point of view.

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

### Exercises of Selina Solutions Concise Maths Class 7 Chapter 15 – Triangles

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#### Exercise 15A page: 176

**1. State, if the triangles are possible with the following angles :
(i) 20Â°, 70Â° and 90Â°
(ii) 40Â°, 130Â° and 20Â°
(iii) 60Â°, 60Â° and 50Â°
(iv) 125Â°, 40Â° and 15Â°**

**Solution:**

In a triangle, the sum of three angles is 180^{0}

(i) 20Â°, 70Â° and 90Â°

Sum = 20^{0} + 70^{0} + 90^{0} = 180^{0}

Here the sum is 180^{0} and therefore it is possible.

(ii) 40Â°, 130Â° and 20Â°

Sum = 40Â° + 130Â° + 20Â° = 190^{0}

Here the sum is not 180^{0} and therefore it is not possible.

(iii) 60Â°, 60Â° and 50Â°

Sum = 60Â° + 60Â° + 50Â° = 170^{0}

Here the sum is not 180^{0} and therefore it is not possible.

(iv) 125Â°, 40Â° and 15Â°

Sum = 125Â° + 40Â° + 15Â° = 180^{0}

Here the sum is 180^{0} and therefore it is possible.

**2. If the angles of a triangle are equal, find its angles.**

**Solution:**

In a triangle, the sum of three angles is 180^{0}

So each angle = 180^{0}/3 = 60^{0}

**3. In a triangle ABC, âˆ A = 45Â° and âˆ B = 75Â°, find âˆ C.**

**Solution:**

In a triangle, the sum of three angles is 180^{0}

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

45^{0} + 75^{0} + âˆ C = 180^{0}

By further calculation

120^{0} + âˆ C = 180^{0}

So we get

âˆ C = 180^{0} â€“ 120^{0} = 60^{0}

**4. In a triangle PQR, âˆ P = 60Â° and âˆ Q = âˆ R, find âˆ R.**

**Solution:**

Consider âˆ Q = âˆ R = x

âˆ P = 60Â°

We can write it as

âˆ P + âˆ Q + âˆ R = 180^{0}

Substituting the values

60^{0} + x + x = 180^{0}

By further calculation

60^{0} + 2x = 180^{0}

2x = 180^{0} â€“ 60^{0} = 120^{0}

So we get

x = 120^{0}/2 = 60^{0}

âˆ Q = âˆ R = 60^{0}

Therefore, âˆ R = 60^{0}.

**5. Calculate the unknown marked angles in each figure:**

**Solution:**

In a triangle, the sum of three angles is 180^{0}

(i) From figure (i)

90^{0} + 30^{0} + x = 180^{0}

By further calculation

120^{0} + x = 180^{0}

So we get

x = 180^{0} â€“ 120^{0} = 60^{0}

Therefore, x = 60^{0}.

(ii) From figure (ii)

y + 80^{0} + 20^{0} = 180^{0}

By further calculation

y + 100^{0} = 180^{0}

So we get

y = 180^{0} â€“ 100^{0} = 80^{0}

Therefore, y = 80^{0}.

(iii) From figure (iii)

a + 90^{0} + 40^{0} = 180^{0}

By further calculation

a + 130^{0} = 180^{0}

So we get

a = 180^{0} â€“ 130^{0} = 50^{0}

Therefore, a = 50^{0}.

**6. Find the value of each angle in the given figures:**

**Solution:**

(i) From the figure (i)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

5x^{0} + 4x^{0} + x^{0} = 180^{0}

By further calculation

10x^{0} = 180^{0}

x = 180/10 = 18^{0}

So we get

âˆ A = 5x^{0} = 5 Ã— 18^{0} = 90^{0}

âˆ B = 4x^{0} = 4 Ã— 18^{0} = 72^{0}

âˆ C = x = 18^{0}

(ii) From the figure (ii)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

x^{0} + 2x^{0} + 2x^{0} = 180^{0}

By further calculation

5x^{0} = 180^{0}

x^{0} = 180^{0}/5 = 36^{0}

So we get

âˆ A = x^{0} = 36^{0}

âˆ B = 2x^{0} = 2 Ã— 36^{0} = 72^{0}

âˆ C = 2x^{0} = 2 Ã— 36^{0} = 72^{0}

**7. Find the unknown marked angles in the given figure:**

**Solution:**

(i) From the figure (i)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

b^{0} + 50^{0} + b^{0} = 180^{0}

By further calculation

2b^{0} = 180^{0} â€“ 50^{0} = 130^{0}

b^{0} = 130^{0}/2 = 65^{0}

Therefore, âˆ A = âˆ C = b^{0} = 65^{0}.

(ii) From the figure (ii)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

x^{0} + 90^{0} + x^{0} = 180^{0}

By further calculation

2x^{0} = 180^{0} â€“ 90^{0} = 90^{0}

x^{0} = 90^{0}/2 = 45^{0}

Therefore, âˆ A = âˆ C = x^{0} = 45^{0}.

(iii) From the figure (iii)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

k^{0} + k^{0} + k^{0} = 180^{0}

By further calculation

3k^{0} = 180^{0}

k^{0} = 180^{0}/3 = 60^{0}

Therefore, âˆ A = âˆ B = âˆ C = 60^{0}.

(iv) From the figure (iv)

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

(m^{0} â€“ 5^{0}) + 60^{0} + (m^{0} + 5^{0}) = 180^{0}

By further calculation

m^{0} â€“ 5^{0} + 60^{0} + m^{0} + 5^{0} = 180^{0}

2m^{0} = 180^{0} â€“ 60^{0}Â = 120^{0}

m^{0} = 120^{0}/2 = 60^{0}

Therefore, âˆ A = m^{0} â€“ 5^{0} = 60^{0} â€“ 5^{0} = 55^{0}

âˆ C = m^{0} + 5^{0} = 60^{0} + 5^{0} = 65^{0}

**8. In the given figure, show that: âˆ a = âˆ b + âˆ c**

**(i) If âˆ b = 60Â° and âˆ c = 50Â°; find âˆ a.
(ii) If âˆ a = 100Â° and âˆ b = 55Â°; find âˆ c.
(iii) If âˆ a = 108Â° and âˆ c = 48Â°; find âˆ b.**

**Solution:**

From the figure

AB || CD

b = âˆ C and âˆ A = c are alternate angles

In triangle PCD

Exterior âˆ APC = âˆ C + âˆ D

a = b + c

(i) If âˆ b = 60Â° and âˆ c = 50Â°

âˆ a = âˆ b + âˆ c

Substituting the values

âˆ a = 60 + 50 = 110^{0}

(ii) If âˆ a = 100Â° and âˆ b = 55Â°

âˆ a = âˆ b + âˆ c

Substituting the values

âˆ c = 100 â€“ 55 = 45^{0}

(iii) If âˆ a = 108Â° and âˆ c = 48Â°

âˆ a = âˆ b + âˆ c

Substituting the values

âˆ b = 108 â€“ 48 = 60^{0}

**9. Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.**

**Solution:**

In a triangle, the sum of angles of a triangle is 180^{0}

âˆ A + âˆ B + âˆ C = 180^{0}

It is given that

âˆ A: âˆ B: âˆ C = 4: 5: 6

Consider âˆ A = 4x, âˆ B = 5x and âˆ C = 6x

Substituting the values

4x + 5x + 6x = 180^{0}

By further calculation

15x = 180^{0}

x = 180^{0}/15 = 12^{0}

So we get

âˆ A = 4x = 4 Ã— 12^{0} = 48^{0}

âˆ B = 5x = 5 Ã— 12^{0} = 60^{0}

âˆ C = 6x = 6 Ã— 12^{0} = 72^{0}

**10. One angle of a triangle is 60Â°. The, other two angles are in the ratio of 5 : 7. Find the two angles.**

**Solution:**

From the triangle ABC

Consider âˆ A = 60^{0}, âˆ B: âˆ C = 5:7

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

60^{0} + âˆ B + âˆ C = 180^{0}

By further calculation

âˆ B + âˆ C = 180^{0} â€“ 60^{0} = 120^{0}

Take âˆ B = 5x and âˆ C = 7x

Substituting the values

5x + 7x = 120^{0}

12x = 120^{0}

x = 120^{0}/12 = 10^{0}

So we get

âˆ B = 5x = 5 Ã— 10^{0} = 50^{0}

âˆ C = 7x = 7 Ã— 10^{0} = 70^{0}

**11. One angle of a triangle is 61Â° and the other two angles are in the ratio 1 Â½ : 1 1/3. Find these angles.**

**Solution:**

From the triangle ABC

Consider âˆ A = 61^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

61^{0} + âˆ B + âˆ C = 180^{0}

By further calculation

âˆ B + âˆ C = 180^{0} â€“ 61^{0} = 119^{0}

âˆ B: âˆ C = 1 Â½: 1 1/3 = 3/2: 4/3

Taking LCM

âˆ B: âˆ C = 9/6: 8/ 6

âˆ B: âˆ C = 9: 8

Consider âˆ B = 9x and âˆ C = 8x

Substituting the values

9x + 8x = 119^{0}

17x = 119^{0}

x = 119^{0}/ 17 = 7^{0}

So we get

âˆ B = 9x = 9 Ã— 7^{0} = 63^{0}

âˆ C = 8x = 8 Ã— 7^{0} = 56^{0}

**12. Find the unknown marked angles in the given figures:**

**Solution:**

In a triangle, if one side is produced

Exterior angle is the sum of opposite interior angles

(i) From the figure (i)

110^{0} = x^{0} + 30^{0}

By further calculation

x^{0} = 110^{0} â€“ 30^{0} = 80^{0}

(ii) From the figure (ii)

120^{0} = y^{0} + 60^{0}

By further calculation

y^{0} = 120^{0} â€“ 60^{0} = 60^{0}

(iii) From the figure (iii)

122^{0} = k^{0} + 35^{0}

By further calculation

k^{0} = 122^{0} â€“ 35^{0} = 87^{0}

(iv) From the figure (iv)

135^{0} = a^{0} + 73^{0}

By further calculation

a^{0} = 135^{0} â€“ 73^{0} = 62^{0}

(v) From the figure (v)

125^{0} = a + c â€¦â€¦ (1)

140^{0} = a + b â€¦â€¦ (2)

By adding both the equations

a + c + a + b = 125^{0} + 140^{0}

On further calculation

a + a + b + c = 265^{0}

We know that a + b + c = 180^{0}

Substituting it in the equation

a + 180^{0} = 265^{0}

So we get

a = 265 â€“ 180 = 85^{0}

If a + b = 140^{0}

Substituting it in the equation

85^{0} + b = 140^{0}

So we get

b = 140 â€“ 85 = 55^{0}

If a + c = 125^{0}

Substituting it in the equation

85^{0} + c = 125^{0}

So we get

c = 125 â€“ 85 = 40^{0}

Therefore, a = 85^{0}, b = 55^{0} and c = 40^{0}.

#### 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 + 80^{0} = 180^{0}

Substituting the values

x + x + 80^{0} = 180^{0}

By further calculation

2x = 180^{0} â€“ 80^{0} = 100^{0}

x = 100^{0}/2 = 50^{0}

Therefore, x = y = 50^{0}.

(ii) From the figure (ii)

b = 40^{0} as the angles opposite to equal sides

In a triangle

a + b + 40^{0} = 180^{0}

Substituting the values

a + 40^{0} + 40^{0} = 180^{0}

By further calculation

a = 180 â€“ 80 = 100^{0}

Therefore, a = 100^{0} and b = 40^{0}.

(iii) From the figure (iii)

x = y as the angles opposite to equal sides

In a triangle

x + y + 90^{0} = 180^{0}

Substituting the values

x + x + 90^{0} = 180^{0}

By further calculation

2x = 180 â€“ 90 = 90^{0}

x = 90/2 = 45^{0}

Therefore, x = y = 45^{0}.

(iv) From the figure (iv)

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

In a triangle

a + b + 80^{0 }= 180^{0}

Substituting the values

a + a + 80^{0 }= 180^{0}

By further calculation

2a = 180 â€“ 80 = 100^{0}

a = 100^{0}/2 = 50^{0}

Here a = b = 50^{0}

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

x = a + 80^{0}

So we get

x = 50 + 80 = 130^{0}

Therefore, a = 50^{0}, b = 50^{0} and x = 130^{0}.

(v) From the figure (v)

In an isosceles triangle consider each equal angle = x

x + x = 86^{0}

2x = 86^{0}

So we get

x = 86^{0}/2 = 43^{0}

For a linear pair

p + x = 180^{0}

Substituting the values

p + 43^{0} = 180^{0}

By further calculation

p = 180 â€“ 43 = 137^{0}

Therefore, p = 137^{0}.

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

**Solution:**

(i) a = 70^{0} as the angles opposite to equal sides are equal

In a triangle

a + 70^{0} + x = 180^{0}

Substituting the values

70^{0} + 70^{0} + x = 180^{0}

By further calculation

x = 180 â€“ 140 = 40^{0}

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

70^{0} = y + y

So we get

2y = 70^{0}

y = 70^{0}/2 = 35^{0}

Therefore, x = 40^{0} and y = 35^{0}.

(ii) From the figure (ii)

Each angle is 60^{0} in an equilateral triangle

In an isosceles triangle

Consider each base angle = a

a + a + 100^{0} = 180^{0}

By further calculation

2a = 180 â€“ 100 = 80^{0}

So we get

a = 80^{0}/2 = 40^{0}

x = 60^{0} + 40^{0} = 100^{0}

y = 60^{0} + 40^{0} = 100^{0}

(iii) From the figure (iii)

130^{0} = 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 = 60^{0} is the alternate angles and y = a

In a linear pair

a + 130^{0} = 180^{0}

By further calculation

a = 180 â€“ 130 = 50^{0}

Here x + p = 130^{0}

Substituting the values

x + 60^{0} = 130^{0}

By further calculation

x = 130 â€“ 60 = 70^{0}

Therefore, x = 70^{0}, y = 50^{0} and p = 60^{0}.

(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 + 30^{0} = 180^{0}

Substituting the values

a + a + 30^{0} = 180^{0}

By further calculation

2a = 180 â€“ 30 = 150^{0}

a = 150/2 = 75^{0}

We know that

b + y = 90^{0}

Substituting the values

y + y = 90^{0}

2y = 90^{0}

y = 90/2 = 45^{0}

where b = 45^{0}

Therefore, x = a + b = 75 + 45 = 120^{0} and y = 45^{0}.

(v) From the figure (v)

a + b + 40^{0} = 180^{0}

So we get

a + b = 180 â€“ 40 = 140^{0}

The angles opposite to equal sides are equal

a = b = 140/2 = 70^{0}

x = b + 40^{0} = 70^{0}Â + 40^{0}= 110^{0}

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

In the same way

y = a + 40^{0}

Substituting the values

y = 70^{0}Â + 40^{0} = 110^{0}

Therefore, x = y = 110^{0}.

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

**Solution:**

Consider âˆ† ABC

Here AB = AC and âˆ B = âˆ C

We know that

âˆ A = 100^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

100^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

2âˆ B = 180^{0} â€“ 100^{0} = 80^{0}

âˆ B = 80/2 = 40^{0}

Therefore, âˆ B = âˆ C = 40^{0}.

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

**Solution:**

It is given that the base angles of isosceles triangle ABC = 52^{0}

Here âˆ B = âˆ C = 52^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

âˆ A + 52^{0} + 52^{0} = 180^{0}

By further calculation

âˆ A = 180 â€“ 104 = 76^{0}

Therefore, âˆ A = 76^{0}.

**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 = 180^{0}

By further calculation

9x = 180^{0}

x = 180/9 = 20^{0}

So the vertical angle = 20^{0}

Each base angle = 4x = 4 Ã— 20^{0} = 80^{0}

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

**Solution:**

Consider the angle of the base of isosceles triangle = x^{0}

So the vertical angle = x + 15^{0}

In a triangle

x + x + x + 15^{0} = 180^{0}

By further calculation

3x = 180 â€“ 15 = 165^{0}

x = 165/3 = 55^{0}

Therefore, the base angle = 55^{0}

Vertical angle = 55 + 15 = 70^{0}.

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

**Solution:**

Consider the vertical angle of the isosceles triangle = x^{0}

Here each base angle = x + 15^{0}

In a triangle

x + 15^{0} + x + 15^{0}Â + x = 180^{0}

By further calculation

3x + 30^{0} = 180^{0}

3x = 180 â€“ 30 = 150^{0}

x = 150/3 = 50^{0}

Therefore, vertical angle = 50^{0} and each base angle = 50 + 15 = 65^{0}.

**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 = 180^{0}

By further calculation

8x = 180^{0}

x = 180/8 = 22.5^{0}

Therefore, each base angle = 22.5^{0} and vertical angle = 3 (22.5 + 22.5) = 3 Ã— 45 = 135^{0}.

**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 = 180^{0}

By further calculation

6x = 180^{0}

x = 180/6 = 30^{0}

Therefore, each base angle = x = 30^{0} and vertical angle = 4x = 4 Ã— 30^{0} = 120^{0}.

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

**Solution:**

In âˆ† ABC

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

Here AB = AC

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

We know that âˆ A = 40^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

40^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

40^{0}Â + 2âˆ B = 180^{0}

2âˆ B = 180 â€“ 40 = 140^{0}

âˆ B = 140/2 = 70^{0}

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

âˆ IBC = Â½ âˆ ABC = Â½ Ã— 70^{0} = 35^{0}

âˆ ICB = Â½ âˆ ACB = Â½ Ã— 70^{0} = 35^{0}

In âˆ† IBC

âˆ BIC + âˆ IBC + âˆ ICB = 180^{0}

Substituting the values

âˆ BIC + 35^{0} + 35^{0}Â = 180^{0}

By further calculation

âˆ BIC = 180 â€“ 70 = 110^{0}

Therefore, âˆ BIC = 110^{0}.

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

**Solution:**

From the âˆ† ABC

BC = BA

âˆ BCA = âˆ BAC

Here the exterior âˆ CBE = âˆ BCA + âˆ BAC

a = âˆ BCA + âˆ BCA

a = 2âˆ BCA â€¦â€¦ (1)

Here âˆ ACB = 180^{0} â€“ b

Where âˆ ACD and âˆ ACB are linear pair

âˆ BCA = 180^{0} â€“ b â€¦â€¦. (2)

We get

a = 2 âˆ BCA

Substituting the values

a = 2 (180^{0} â€“ b)

a = 360^{0} â€“ 2b

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

**Solution:**

(a) From the figure (i)

AB = AC and BP bisects âˆ ABC

AP is drawn parallel to BC

Here PB is the bisector of âˆ ABC

âˆ PBC = âˆ PBA

âˆ APB = âˆ PBC are alternate angles

x = âˆ PBC â€¦.. (1)

In âˆ† ABC

âˆ A = 60^{0}

Since AB = AC we get âˆ B = âˆ C

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

60^{0} + âˆ B + âˆ C = 180^{0}

We get

60^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

2âˆ B = 180 â€“ 60 = 120^{0}

âˆ B = 120/2 = 60^{0}

Â½ âˆ B = 60/2 = 30^{0}

âˆ PBC = 30^{0}

So from figure (i) x = 30^{0}

(b) From the figure (ii)

DA = DB = DC

Here BD bisects âˆ ABC and âˆ ADB = 70^{0}

In a triangle

âˆ ADB + âˆ DAB + âˆ DBA = 180^{0}

Substituting the values

70^{0} + âˆ DBA + âˆ DBA = 180^{0}

By further calculation

70^{0} + 2âˆ DBA = 180^{0}

2âˆ DBA = 180 â€“ 70 = 110^{0}

âˆ DBA = 110/2 = 55^{0}

Here BD is the bisector of âˆ ABC

So âˆ DBA = âˆ DBC = 55^{0}

In âˆ† DBC

DB = DC

âˆ DCB = âˆ DBC

Hence, x = 55^{0}.

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

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

**Solution:**

The sides of a square are equal and each angle is 90^{0}

In an equilateral triangle all three sides are equal and all angles are 60^{0}

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

(i) âˆ ABE = âˆ ABC + âˆ CBE

Substituting the values

âˆ ABE = 90^{0} + 60^{0}= 150^{0}

(ii) In âˆ† ABE

âˆ ABE + âˆ BEA + âˆ BAE = 180^{0}

Substituting the values

150^{0} + âˆ BAE + âˆ BAE = 180^{0}

By further calculation

2âˆ BAE = 180 â€“ 150 = 30^{0}

âˆ BAE = 30/2 = 15^{0}

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

(i) âˆ ABE = âˆ ABC – âˆ CBE

Substituting the values

âˆ ABE = 90^{0} â€“ 60^{0} = 30^{0}

(ii) In âˆ† ABE

âˆ ABE + âˆ BEA + âˆ BAE = 180^{0}

Substituting the values

30^{0} + âˆ BAE + âˆ BAE = 180^{0}

By further calculation

2âˆ BAE = 180 â€“ 30 = 150^{0}

âˆ BAE = 150/2 = 75^{0}

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

**Solution:**

In âˆ† ABC, BA and BC are produced

âˆ ABC = 54^{0} and AB = BC

In âˆ† ABC

âˆ BAC + âˆ BCA + âˆ ABC = 180^{0}

Substituting the values

âˆ BAC + âˆ BAC + 54^{0 }= 180^{0}

2âˆ BAC = 180 â€“ 54 = 126^{0}

âˆ BAC = 126/2 = 63^{0}

âˆ BCA = 63^{0}

In a linear pair

âˆ BAC + b = 180^{0}

Substituting the value

63^{0} + b = 180^{0}

So we get

b = 180 â€“ 63 = 117^{0}

In a linear pair

âˆ BCA + a = 180^{0}

Substituting the value

63^{0}Â + a = 180^{0}

So we get

a = 180 â€“ 63 = 117^{0}

Therefore, a = b = 117^{0}.

#### Exercise 15C page: 185

**1. Construct a âˆ†ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm**

**Solution:**

(i) Steps of Construction

1. Construct a line segment BC = 4 cm.

2. Taking B as centre and 6 cm as radius construct an arc.

3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AB and AC.

Therefore, âˆ†ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment CB = 6.5 cm

2. Taking C as centre and 4.2 cm as radius construct an arc.

3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AC and AB.

Therefore, âˆ†ABC is the required triangle.

(iii) Steps of Construction

1. Construct a line segment BC = 4 cm

2. Taking B as centre and 3.5 cm as radius construct an arc.

3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AB and AC.

Therefore, âˆ†ABC is the required triangle.

**2. Construct a **âˆ†** ABC such that:
(i) AB = 7 cm, BC = 5 cm and âˆ ABC = 60Â°
(ii) BC = 6 cm, AC = 5.7 cm and âˆ ACB = 75Â°
(iii) AB = 6.5 cm, AC = 5.8 cm and âˆ A = 45Â°**

**Solution:**

(i) Steps of Construction

1. Construct a line segment AB = 7 cm.

2. At the point B construct a ray which makes an angle 60^{0} and cut off BC = 5cm.

3. Now join AC.

Therefore, âˆ†ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment BC = 6 cm.

2. At the point C construct a ray which makes an angle 75^{0} and cut off CA = 5.7 cm.

3. Now join AB.

Therefore, âˆ†ABC is the required triangle.

(iii) Steps of Construction

1. Construct a line segment AB = 6.5 cm.

2. At the point A construct a ray which makes an angle 45^{0} and cut off AC = 5.8 cm.

3. Now join CB.

Therefore, âˆ†ABC is the required triangle.

**3. Construct a âˆ† PQR such that :
(i) PQ = 6 cm, âˆ Q = 60Â° and âˆ P = 45Â°. Measure âˆ R.
(ii) QR = 4.4 cm, âˆ R = 30Â° and âˆ Q = 75Â°. Measure PQ and PR.
(iii) PR = 5.8 cm, âˆ P = 60Â° and âˆ R = 45Â°.
Measure âˆ Q and verify it by calculations**

**Solution:**

(i) Steps of Construction

1. Construct a line segment PQ = 6 cm.

2. At point P construct a ray which makes an angle 45^{0}.

3. At point Q construct another ray which makes an angle 60^{0} which intersect the first ray at point R.

Therefore, âˆ† PQR is the required triangle.

By measuring âˆ R = 75^{0}.

(ii) Steps of Construction

1. Construct a line segment QR = 4.4 cm.

2. At point Q construct a ray which makes an angle 75^{0}.

3. At point R construct another ray which makes an angle 30^{0} which intersect the first ray at point R.

Therefore, âˆ† PQR is the required triangle.

By measuring the length, PQ = 2.1 cm and PR = 4.4 cm.

(iii) Steps of Construction

1. Construct a line segment PR = 5.8 cm.

2. At point P construct a ray which makes an angle 60^{0}.

3. At point R construct another ray which makes an angle 45^{0} which intersect the first ray at point Q.

Therefore, âˆ† PQR is the required triangle.

By measuring âˆ Q = 75^{0}.

Verification â€“

âˆ P + âˆ Q + âˆ R = 180^{0}

Substituting the values

60^{0}Â + âˆ Q + 45^{0} = 180^{0}

By further calculation

âˆ Q = 180 â€“ 105 = 75^{0}

**4. Construct an isosceles **âˆ†** ABC such that:
(i) base BC = 4 cm and base angle = 30Â°
(ii) base AB = 6.2 cm and base angle = 45Â°
(iii) base AC = 5 cm and base angle = 75Â°.
Measure the other two sides of the triangle.**

**Solution:**

(i) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment BC = 4 cm.

2. At the points B and C construct rays which makes an angle 30^{0} intersecting each other at the point A.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 2.5 cm in length approximately.

(ii) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment AB = 6.2 cm.

2. At the points A and B construct rays which makes an angle 45^{0} intersecting each other at the point C.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 4.3 cm in length approximately.

(iii) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment AC = 5 cm.

2. At the points A and C construct rays which makes an angle 75^{0} intersecting each other at the point B.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 9.3 cm in length approximately.

**5. Construct an isosceles âˆ†ABC such that:
(i) AB = AC = 6.5 cm and âˆ A = 60Â°
(ii) One of the equal sides = 6 cm and vertex angle = 45Â°. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30Â°. Measure âˆ A and âˆ C.**

**Solution:**

(i) Steps of Construction

1. Construct a line segment AB = 6.5 cm.

2. At point A construct a ray which makes an angle 60^{0}.

3. Now cut off AC = 6.5cm

4. Join BC.

Therefore, âˆ† ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment AB = 6 cm.

2. At point A construct a ray which makes an angle 45^{0}.

3. Now cut off AC = 6cm

4. Join BC.

Therefore, âˆ† ABC is the required triangle.

By measuring âˆ B and âˆ C, both are equal toÂ 67 Â½ ^{0}.

(iii) Steps of Construction

1. Construct a line segment BC = 5.8 cm.

2. At point B construct a ray which makes an angle 30^{0}.

3. Now cut off BA = 5.8cm

4. Join AC.

Therefore, âˆ† ABC is the required triangle.

By measuring âˆ C and âˆ A is equal to 75^{0}.

**6. Construct an equilateral triangle ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.**

**Solution:**

(i) Steps of Construction

1. Construct a line segment AB = 5cm.

2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C.

3. Now join AC and BC where âˆ† ABC is the required triangle.

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p.

5. Join PA, PB and PC.

By measuring each is 2.8 cm.

(ii) Steps of Construction

1. Construct a line segment AB = 6cm.

2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C.

3. Now join AC and BC

Therefore, âˆ† ABC is the required triangle.

**7. (i) Construct a âˆ† ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles âˆ†PQR such that PQ = PR = 6.5 cm and âˆ PQR = 75Â°. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.**

**Solution:**

(i) Steps of Construction

1. Construct a line segment BC = 4.5 cm.

2. Taking B as centre and 6 cm radius construct an arc.

3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A.

4. Now join AB and AC

Therefore, âˆ† ABC is the required triangle.

5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O.

6. Now join OB, OC and OA.

7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C.

This is the required circumcircle of âˆ† ABC.

(ii) Steps of Construction

1. Construct a line segment PQ = 6.5 cm.

2. At point Q, construct an arc which makes an angle 75^{0}.

3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R.

4. Join PR.

âˆ† PQR is the required triangle.

4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O.

5. Join OP, OQ and OR.

6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R.

This is the required circumcircle of âˆ† PQR.

(iii) Steps of Construction

1. Construct a line segment AB = 5.5 cm.

2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C.

3. Now join AC and BC.

âˆ† ABC is the required triangle.

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O.

5. Now join OA, OB and OC.

6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C.

This is the required circumcircle.

**8. (i) Construct a âˆ†ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles âˆ† MNP such that base MN = 5.8 cm, base angle MNP = 30Â°. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral âˆ†DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a âˆ† PQR such that PQ = 6 cm, âˆ QPR = 45Â° and angle PQR = 60Â°. Locate its incentre and then draw its incircle.**

**Solution:**

(i) Steps of Construction

1. Construct a line segment AB = 6 cm.

2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C.

3. Now join AC and BC.

4. Construct the angle bisector of âˆ A and âˆ B which intersect each other at point I.

5. From the point I construct IL which is perpendicular to AB.

6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally.

By measuring the required incircle the radius is 1.6 cm.

(ii) Steps of Construction

1. Construct a line segment MN = 5.8 cm.

2. At points M and N construct two rays which make an angle 30^{0} each intersecting each other at point P.

3. Construct the angle bisectors of âˆ M and âˆ N which intersect each other at point I.

4. From the point I draw perpendicular IL on MN.

5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ† PMB internally.

By measuring the required incircle the radius is 0.6 cm.

(iii) Steps of Construction

1. Construct a line segment BC = 5.5 cm.

2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A.

3. Now join AB and AC.

4. Construct the perpendicular bisectors of âˆ B and âˆ C which intersect each other at the point I.

5. From the point I construct IL which is perpendicular to BC.

6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally.

This is the required incircle.

(iv) Steps of Construction

1. Construct a line segment PQ = 6 cm.

2. At the point P construct rays which make an angle of 45^{0} and at point Q which makes an angle 60^{0}Â thats intersects each other at point R.

3. Construct the bisectors of âˆ P and âˆ Q which intersect each other at point I.

4. From the point I construct IL which is perpendicular to PQ.

5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†PQR internally.

This is the required incircle where the point I is incentre.