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#### Page No 196:

#### Question 1:

#### Answer:

We get a triangle by joining the three non-collinear points A, B, and C.

(i) The side opposite to ∠C is AB.

(ii) The angle opposite to the side BC is ∠A.

(iii) The vertex opposite to the side CA is B.

(iv) The side opposite to the vertex B is AC.

#### Page No 196:

#### Question 2:

We get a triangle by joining the three non-collinear points A, B, and C.

(i) The side opposite to ∠C is AB.

(ii) The angle opposite to the side BC is ∠A.

(iii) The vertex opposite to the side CA is B.

(iv) The side opposite to the vertex B is AC.

#### Answer:

The measures of two angles of a triangle are 72° and 58°.

Let the third angle be *x*.

Now, the sum of the measures of all the angles of a triangle is 180^{o}â€‹.

$\therefore $ *x *+ 72^{o }+ 58^{o} = 180^{o}

⇒ *x* + 130^{o }= 180^{o}

⇒ *x *= 180^{o}â€‹ $-$ 130^{o}

⇒ *x = *50^{o}

â€‹The measure of the third angle of the triangle is 50^{o}â€‹.

#### Page No 196:

#### Question 3:

The measures of two angles of a triangle are 72° and 58°.

Let the third angle be *x*.

Now, the sum of the measures of all the angles of a triangle is 180^{o}â€‹.

$\therefore $ *x *+ 72^{o }+ 58^{o} = 180^{o}

⇒ *x* + 130^{o }= 180^{o}

⇒ *x *= 180^{o}â€‹ $-$ 130^{o}

⇒ *x = *50^{o}

â€‹The measure of the third angle of the triangle is 50^{o}â€‹.

#### Answer:

The angles of a triangle are in the ratio 1:3:5.

Let the measures of the angles of the triangle be (1x), (3x) and (5x)

Sum of the measures of the angles of the triangle = 180^{o}

∴ 1x + 3x + 5x = 180^{o}

⇒ 9x = 180^{o}

⇒ x = 20^{o}^{ }1x = 20^{o}

3x = 60^{o}

â€‹5x = 100^{o}

The measures of the angles are 20^{o}, 60^{o} and 100^{o}.

#### Page No 196:

#### Question 4:

The angles of a triangle are in the ratio 1:3:5.

Let the measures of the angles of the triangle be (1x), (3x) and (5x)

Sum of the measures of the angles of the triangle = 180^{o}

∴ 1x + 3x + 5x = 180^{o}

⇒ 9x = 180^{o}

⇒ x = 20^{o}^{ }1x = 20^{o}

3x = 60^{o}

â€‹5x = 100^{o}

The measures of the angles are 20^{o}, 60^{o} and 100^{o}.

#### Answer:

In a right angle triangle, one of the angles is 90^{o}.

It is given that one of the acute angled of the right angled triangle is 50^{o}.

We know that the sum of the measures of all the angles of a triangle is 180^{o}.

Now, let the third angle be *x*.

â€‹Therefore, we have:

90^{o}â€‹ + 50^{o} + *x *= 180^{o}

⇒ 140^{o }+ *x *= 180^{o}

â€‹ ⇒ *x =* 180^{o}$-$ 140^{o}

⇒ *x = * 40^{o}^{â€‹} The third acute angle is 40^{o}â€‹.

#### Page No 196:

#### Question 5:

In a right angle triangle, one of the angles is 90^{o}.

It is given that one of the acute angled of the right angled triangle is 50^{o}.

We know that the sum of the measures of all the angles of a triangle is 180^{o}.

Now, let the third angle be *x*.

â€‹Therefore, we have:

90^{o}â€‹ + 50^{o} + *x *= 180^{o}

⇒ 140^{o }+ *x *= 180^{o}

â€‹ ⇒ *x =* 180^{o}$-$ 140^{o}

⇒ *x = * 40^{o}^{â€‹} The third acute angle is 40^{o}â€‹.

#### Answer:

Given:

∠A = 110^{o} and â€‹∠B = ∠C

Now, the sum of the measures of all the angles of a traingle is 180^{o} .

∠A + ∠B + ∠C = 180^{o}

⇒ 110^{o} + â€‹∠B + ∠B = 180^{o}

â€‹ ⇒ 110^{o }+ 2â€‹∠B = 180^{o
â€‹ }⇒ 2â€‹∠B = 180^{o }$-$ 110^{o}^{ } ⇒ 2∠B = 70^{o}

⇒ ∠B = 70^{o }/ 2

⇒ ∠B = 35^{o}

∴ â€‹∠C = 35^{o}

The measures of the three angles:

∠A = 110^{o}, ∠B = 35^{o}, â€‹∠C = 35^{o}

#### Page No 196:

#### Question 6:

Given:

∠A = 110^{o} and â€‹∠B = ∠C

Now, the sum of the measures of all the angles of a traingle is 180^{o} .

∠A + ∠B + ∠C = 180^{o}

⇒ 110^{o} + â€‹∠B + ∠B = 180^{o}

â€‹ ⇒ 110^{o }+ 2â€‹∠B = 180^{o
â€‹ }⇒ 2â€‹∠B = 180^{o }$-$ 110^{o}^{ } ⇒ 2∠B = 70^{o}

⇒ ∠B = 70^{o }/ 2

⇒ ∠B = 35^{o}

∴ â€‹∠C = 35^{o}

The measures of the three angles:

∠A = 110^{o}, ∠B = 35^{o}, â€‹∠C = 35^{o}

#### Answer:

Given:

∠A = ∠B + ∠C

â€‹We know:

∠A + ∠B + ∠C = 180^{o}

⇒ ∠B +∠C + ∠B + ∠C = 180^{o}

â€‹ ⇒ 2∠B + 2∠C = 180^{o}

â€‹ ⇒ 2(∠B +∠C) = 180^{o}

â€‹ ⇒ ∠B + ∠C = 180/2

⇒ â€‹∠B + ∠C = 90^{o}

$\therefore $ ∠A = 90^{o}

This shows that the triangle is a right angled triangle.

#### Page No 196:

#### Question 7:

Given:

∠A = ∠B + ∠C

â€‹We know:

∠A + ∠B + ∠C = 180^{o}

⇒ ∠B +∠C + ∠B + ∠C = 180^{o}

â€‹ ⇒ 2∠B + 2∠C = 180^{o}

â€‹ ⇒ 2(∠B +∠C) = 180^{o}

â€‹ ⇒ ∠B + ∠C = 180/2

⇒ â€‹∠B + ∠C = 90^{o}

$\therefore $ ∠A = 90^{o}

This shows that the triangle is a right angled triangle.

#### Answer:

Let 3∠*A* = 4 ∠*B* = 6 ∠*C* = x

Then, we have:* $\angle \mathrm{A}=\frac{\mathrm{x}}{3},\angle \mathrm{B}=\frac{\mathrm{x}}{4},\angle \mathrm{C}=\frac{\mathrm{x}}{6}\phantom{\rule{0ex}{0ex}}\mathrm{But},\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{x}}{3}+\frac{\mathrm{x}}{4}+\frac{\mathrm{x}}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{4\mathrm{x}+3\mathrm{x}+2\mathrm{x}}{12}=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}9\mathrm{x}=180\xb0\times 12=2160\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{x}=240\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{A}=\frac{240}{3}=80\xb0,\angle \mathrm{B}=\frac{240}{4}=60\xb0,\angle \mathrm{C}=\frac{240}{6}=40\xb0$*

#### Page No 196:

#### Question 8:

Let 3∠*A* = 4 ∠*B* = 6 ∠*C* = x

Then, we have:* $\angle \mathrm{A}=\frac{\mathrm{x}}{3},\angle \mathrm{B}=\frac{\mathrm{x}}{4},\angle \mathrm{C}=\frac{\mathrm{x}}{6}\phantom{\rule{0ex}{0ex}}\mathrm{But},\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{x}}{3}+\frac{\mathrm{x}}{4}+\frac{\mathrm{x}}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{4\mathrm{x}+3\mathrm{x}+2\mathrm{x}}{12}=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}9\mathrm{x}=180\xb0\times 12=2160\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{x}=240\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{A}=\frac{240}{3}=80\xb0,\angle \mathrm{B}=\frac{240}{4}=60\xb0,\angle \mathrm{C}=\frac{240}{6}=40\xb0$*

#### Answer:

(i) It is an obtuse angle triangle as one angle is 130^{o}, which is greater than 90^{o}.

(ii) It is an acute angle triangle as all the angles in it are less than 90^{o}.

(iii) It is a right angle triangle as one angle is 90^{o}.

(iv) It is an obtuse angle triangle as one angle is 92^{o}, which is greater than 90^{o}.

#### Page No 197:

#### Question 9:

(i) It is an obtuse angle triangle as one angle is 130^{o}, which is greater than 90^{o}.

(ii) It is an acute angle triangle as all the angles in it are less than 90^{o}.

(iii) It is a right angle triangle as one angle is 90^{o}.

(iv) It is an obtuse angle triangle as one angle is 92^{o}, which is greater than 90^{o}.

#### Answer:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60^{o}.

Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.

Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.

(i) Isosceles

AC = CB = 2 cm

(ii) Isosceles

DE = EF = 2.4 cm

(iii) Scalene

All the sides are unequal.

(iv) Equilateral

XY = YZ = ZX = 3 cm

(v) Equilateral

All three angles are 60^{o}.

(vi) Isosceles

Two angles are equal in measure.

(vii) Scalene

All the angles are unequal.

#### Page No 197:

#### Question 10:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60^{o}.

Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.

Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.

(i) Isosceles

AC = CB = 2 cm

(ii) Isosceles

DE = EF = 2.4 cm

(iii) Scalene

All the sides are unequal.

(iv) Equilateral

XY = YZ = ZX = 3 cm

(v) Equilateral

All three angles are 60^{o}.

(vi) Isosceles

Two angles are equal in measure.

(vii) Scalene

All the angles are unequal.

#### Answer:

In âˆ†ABC, if we take a point D on BC, then we get three triangles, namely âˆ†ADB, âˆ†ADC and âˆ†ABC.

#### Page No 197:

#### Question 11:

In âˆ†ABC, if we take a point D on BC, then we get three triangles, namely âˆ†ADB, âˆ†ADC and âˆ†ABC.

#### Answer:

(i) No

If the two angles are 90^{o} each, then the sum of two angles of a triangle will be 180^{o}â€‹, which is not possible.

(ii) No

For example, let the two angles be 120^{o} and 150^{o}. Then, their sum will be 270^{o}â€‹, which cannot form a triangle.

(iii) Yes

For example, let the two angles be 50^{o}^{}and 60^{o}â€‹, which on adding, gives 110^{o}. They can easily form a triangle whose third angle is 180^{o}$-$ 110^{o} = 70^{o}â€‹.

(iv) No

For example, let the two angles be 70^{o}â€‹ and 80^{o}, which on adding, gives 150^{o}. They cannot form a triangle whose third angle is 180^{o}â€‹ $-$ 150^{o }= 30^{o}, which is less than 60^{o}.

(v) No

For example, let the two angles be 50^{o}^{ }and 40^{o}, which on adding, gives 90^{o}^{}. Thus, they cannot form a triangle whose third angle is 180^{o}^{ }$-$ 90^{o} = 90^{o}â€‹, which^{ }is greater than 60^{o}.

(vi) Yes

Sum of all angles = 60^{o} + 60^{o} + 60^{o}â€‹ = 180^{o}

#### Page No 197:

#### Question 12:

(i) No

If the two angles are 90^{o} each, then the sum of two angles of a triangle will be 180^{o}â€‹, which is not possible.

(ii) No

For example, let the two angles be 120^{o} and 150^{o}. Then, their sum will be 270^{o}â€‹, which cannot form a triangle.

(iii) Yes

For example, let the two angles be 50^{o}^{}and 60^{o}â€‹, which on adding, gives 110^{o}. They can easily form a triangle whose third angle is 180^{o}$-$ 110^{o} = 70^{o}â€‹.

(iv) No

For example, let the two angles be 70^{o}â€‹ and 80^{o}, which on adding, gives 150^{o}. They cannot form a triangle whose third angle is 180^{o}â€‹ $-$ 150^{o }= 30^{o}, which is less than 60^{o}.

(v) No

For example, let the two angles be 50^{o}^{ }and 40^{o}, which on adding, gives 90^{o}^{}. Thus, they cannot form a triangle whose third angle is 180^{o}^{ }$-$ 90^{o} = 90^{o}â€‹, which^{ }is greater than 60^{o}.

(vi) Yes

Sum of all angles = 60^{o} + 60^{o} + 60^{o}â€‹ = 180^{o}

#### Answer:

(i) A triangle has __3__ sides __3__ angles and __3__ vertices.

(ii) The sum of the angles of a triangle is__ 180 ^{o}__.

(iii) The sides of a scalene triangle are of

__different__lengths.

(iv) Each angle of an equilateral triangle measures

__60__.

^{o}(v) The angles opposite to equal sides of an isosceles triangle are

__equal__.

(vi) The sum of the lengths of the sides of a triangle is called its

__perimeter__.

#### Page No 197:

#### Question 1:

(i) A triangle has __3__ sides __3__ angles and __3__ vertices.

(ii) The sum of the angles of a triangle is__ 180 ^{o}__.

(iii) The sides of a scalene triangle are of

__different__lengths.

(iv) Each angle of an equilateral triangle measures

__60__.

^{o}(v) The angles opposite to equal sides of an isosceles triangle are

__equal__.

(vi) The sum of the lengths of the sides of a triangle is called its

__perimeter__.

#### Answer:

Correct option: (c)

A triangle has 6 parts: three sides and three angles.

#### Page No 197:

#### Question 2:

Correct option: (c)

A triangle has 6 parts: three sides and three angles.

#### Answer:

Correct option: (b)

(a) Sum = 30° + 60° + 70° = 160^{o}

This is not equal to the sum of all the angles of a triangle.

(b) Sum = 50° + 70° + 60° = 180^{o}

Hence, it is possible to construct a triangle with these angles.

(c) Sum = 40° + 80° + 65° = 185^{o}^{ } This is not equal to the sum of all the angles of a triangle.

(d) Sum = 72° + 28° + 90° = 190^{o}â€‹

This is not equal to the sum of all the angles of a triangle.

#### Page No 197:

#### Question 3:

Correct option: (b)

(a) Sum = 30° + 60° + 70° = 160^{o}

This is not equal to the sum of all the angles of a triangle.

(b) Sum = 50° + 70° + 60° = 180^{o}

Hence, it is possible to construct a triangle with these angles.

(c) Sum = 40° + 80° + 65° = 185^{o}^{ } This is not equal to the sum of all the angles of a triangle.

(d) Sum = 72° + 28° + 90° = 190^{o}â€‹

This is not equal to the sum of all the angles of a triangle.

#### Answer:

(b) 80^{o}

Let the measures of the given angles be (2x)^{o}, (3x)^{o}â€‹ and (4x)^{o}.

$\therefore $ (2x)^{o} + (3x)^{o }+ (4x)^{o }= 180^{o}

⇒ (9x)^{o }= 180^{o}

â€‹ ⇒ x = 180 / 9

⇒ x = 20^{o}

$\therefore $ â€‹2x = 40^{o}, 3x = 60^{o}, 4x = 80^{o}

Hence, the measures of the angles of the triangle are 40^{o}â€‹, 60^{o}, 80^{o}.

Thus, the largest angle is 80^{o}.

#### Page No 198:

#### Question 4:

(b) 80^{o}

Let the measures of the given angles be (2x)^{o}, (3x)^{o}â€‹ and (4x)^{o}.

$\therefore $ (2x)^{o} + (3x)^{o }+ (4x)^{o }= 180^{o}

⇒ (9x)^{o }= 180^{o}

â€‹ ⇒ x = 180 / 9

⇒ x = 20^{o}

$\therefore $ â€‹2x = 40^{o}, 3x = 60^{o}, 4x = 80^{o}

Hence, the measures of the angles of the triangle are 40^{o}â€‹, 60^{o}, 80^{o}.

Thus, the largest angle is 80^{o}.

#### Answer:

Correct option: (d)

The measure of two angles are complimentary if their sum is 90^{o} degrees.

Let the two angles be *x* and *y, *such that *x + y = *90^{o}^{ }.

Let the third angle be *z.*

Now, we know that the sum of all the angles of a triangle is 180^{o}â€‹.

*x + y + z*â€‹ = 180^{o}

⇒ 90^{o} + *z = *180^{o}

⇒ *z * = 180^{o }$-$ 90^{o }

= 90^{o}

â€‹The third angle is 90^{o}.

#### Page No 198:

#### Question 5:

Correct option: (d)

The measure of two angles are complimentary if their sum is 90^{o} degrees.

Let the two angles be *x* and *y, *such that *x + y = *90^{o}^{ }.

Let the third angle be *z.*

Now, we know that the sum of all the angles of a triangle is 180^{o}â€‹.

*x + y + z*â€‹ = 180^{o}

⇒ 90^{o} + *z = *180^{o}

⇒ *z * = 180^{o }$-$ 90^{o }

= 90^{o}

â€‹The third angle is 90^{o}.

#### Answer:

Correct option: (c)

Let ∠A = 70^{o}

The triangle is an isosceles triangle.

We know that the angles opposite to the equal sides of an isosceles triangle are equal.

$\therefore $ â€‹∠B = 70^{o}

â€‹We need to find the vertical angle â€‹∠C.

Now, sum of all the angles of a triangle is 180^{o}.

∠A + ∠B + ∠C = 180^{o}

⇒â€‹ 70^{o} + 70^{o}â€‹ + ∠C = 180^{o}

⇒ 140^{o }+ â€‹∠C = 180^{o}

⇒ ∠C = 180^{o}^{ }$-$ 140^{o}

⇒ ∠C = 40^{o}

#### Page No 198:

#### Question 6:

Correct option: (c)

Let ∠A = 70^{o}

The triangle is an isosceles triangle.

We know that the angles opposite to the equal sides of an isosceles triangle are equal.

$\therefore $ â€‹∠B = 70^{o}

â€‹We need to find the vertical angle â€‹∠C.

Now, sum of all the angles of a triangle is 180^{o}.

∠A + ∠B + ∠C = 180^{o}

⇒â€‹ 70^{o} + 70^{o}â€‹ + ∠C = 180^{o}

⇒ 140^{o }+ â€‹∠C = 180^{o}

⇒ ∠C = 180^{o}^{ }$-$ 140^{o}

⇒ ∠C = 40^{o}

#### Answer:

Correct option: (c)

A triangle having sides of different lengths is called a scalene triangle.

#### Page No 198:

#### Question 7:

Correct option: (c)

A triangle having sides of different lengths is called a scalene triangle.

#### Answer:

Correct option: (a)

In the isosceles ABC, â€‹the bisectors of ∠B and ∠C meet at point O.

Since the triangle is isosceles, the angles opposite to the equal sides are equal.

∠B = ∠C

$\therefore $ ∠A + ∠B + ∠C = 180^{o}

⇒ 40^{o}^{ }+ 2∠B = 180^{o}

⇒ 2∠B = 140^{o}

⇒ ∠B = 70^{o}

Bisectors of an angle divide the angle into two equal angles.

So, in âˆ†BOC:

∠OBC = 35^{o} and ∠OCB = 35^{o}

∠BOC + ∠OBC + ∠OCB = 180â€‹^{o}

⇒ ∠BOC + 35^{o} + 35^{o}^{ }= 180^{o}

⇒ ∠BOC = 180^{o}â€‹ - 70^{o}

⇒ ∠BOC = 110^{o}

#### Page No 198:

#### Question 8:

Correct option: (a)

In the isosceles ABC, â€‹the bisectors of ∠B and ∠C meet at point O.

Since the triangle is isosceles, the angles opposite to the equal sides are equal.

∠B = ∠C

$\therefore $ ∠A + ∠B + ∠C = 180^{o}

⇒ 40^{o}^{ }+ 2∠B = 180^{o}

⇒ 2∠B = 140^{o}

⇒ ∠B = 70^{o}

Bisectors of an angle divide the angle into two equal angles.

So, in âˆ†BOC:

∠OBC = 35^{o} and ∠OCB = 35^{o}

∠BOC + ∠OBC + ∠OCB = 180â€‹^{o}

⇒ ∠BOC + 35^{o} + 35^{o}^{ }= 180^{o}

⇒ ∠BOC = 180^{o}â€‹ - 70^{o}

⇒ ∠BOC = 110^{o}

#### Answer:

Correct option: (b)

The sides of a triangle are in the ratio 3:2:5.

Let the lengths of the sides of the triangle be (3x), (2x), (5x).

We know:

Sum of the lengths of the sides of a triangle = Perimeter

(3x) + (2x) + (5x) = 30

⇒ 10x = 30

⇒ x = __ 30 __

10

⇒ x = 3

First side = 3x = 9 cm

Second side = 2x = 6 cm

Third side = 5x = 15 cm

The length of the longest side is 15 cm.

#### Page No 198:

#### Question 9:

Correct option: (b)

The sides of a triangle are in the ratio 3:2:5.

Let the lengths of the sides of the triangle be (3x), (2x), (5x).

We know:

Sum of the lengths of the sides of a triangle = Perimeter

(3x) + (2x) + (5x) = 30

⇒ 10x = 30

⇒ x = __ 30 __

10

⇒ x = 3

First side = 3x = 9 cm

Second side = 2x = 6 cm

Third side = 5x = 15 cm

The length of the longest side is 15 cm.

#### Answer:

Correct option: (d)

Two angles of a triangle measure 30° and 25°, respectively.

Let the third angle be x.

x + 30^{o}^{ }+ 25^{o}^{ }= 180^{o}

â€‹ x = 180^{o}$-$ 55^{o}

x = 125^{o}

#### Page No 198:

#### Question 10:

Correct option: (d)

Two angles of a triangle measure 30° and 25°, respectively.

Let the third angle be x.

x + 30^{o}^{ }+ 25^{o}^{ }= 180^{o}

â€‹ x = 180^{o}$-$ 55^{o}

x = 125^{o}

#### Answer:

Correct option: (c)

Each angle of an equilateral triangle measures 60^{o}.

#### Page No 198:

#### Question 11:

Correct option: (c)

Each angle of an equilateral triangle measures 60^{o}.

#### Answer:

Correct option: (c)

Point P lies on âˆ†ABC.

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