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#### Page No 196:

#### Question 1:

Take three noncollinear points *A*, *B* and *C* on a page of your notebook. Join *AB*, *BC* and *CA*. What figure do you get?

Name: (i) the side opposite to ∠*C*

(ii) the angle opposite to the side *BC*

(iii) the vertex opposite to the side *CA*

(iv) the side opposite to the vertex *B*

Figure

#### 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:

The measures of two angles of a triangle are 72° and 58°. Find the measure of the third angle.

#### 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}.

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

⇒ *x* + 130^{o }= 180^{o}

⇒ *x *= 180^{o} $-$ 130^{o}

⇒ *x = *50^{o}

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

#### Page No 196:

#### Question 3:

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

#### 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}

5x = 100^{o}

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

#### Page No 196:

#### Question 4:

One of the acute angles of a right triangle is 50°. Find the other acute angle.

#### 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*.

Therefore, we have:

90^{o} + 50^{o} + *x *= 180^{o}

⇒ 140^{o }+ *x *= 180^{o}

⇒ *x =* 180^{o} $-$ 140^{o}

⇒ *x = * 40^{o}

^{} The third acute angle is 40^{o}.

#### Page No 196:

#### Question 5:

One of the angles of a triangle is 110° and the other two angles are equal. What is the measure of each of these equal angles?

#### Answer:

Given:

∠A = 110^{o} and ∠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} + ∠B + ∠B = 180^{o}

⇒ 110^{o }+ 2∠B = 180^{o
}⇒ 2∠B = 180^{o }$-$ 110^{o}

^{ } ⇒ 2∠B = 70^{o}

⇒ ∠B = 70^{o }/ 2

⇒ ∠B = 35^{o}

∴ ∠C = 35^{o}

The measures of the three angles:

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

#### Page No 196:

#### Question 6:

If one angle of a triangle is equal to the sum of other two, show that the triangle is a right triangle.

#### Answer:

Given:

∠A = ∠B + ∠C

We know:

∠A + ∠B + ∠C = 180^{o}

⇒ ∠B +∠C + ∠B + ∠C = 180^{o}

⇒ 2∠B + 2∠C = 180^{o}

⇒ 2(∠B +∠C) = 180^{o}

⇒ ∠B + ∠C = 180/2

⇒ ∠B + ∠C = 90^{o}

$\therefore $ ∠A = 90^{o}

This shows that the triangle is a right angled triangle.

#### Page No 196:

#### Question 7:

In a ∆*ABC*, if 3∠*A* = 4 ∠*B* = 6 ∠*C*, calculate the angles.

#### 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:

Look at the figures given below. State for each triangle whether it is acute, right or obtuse.

Figure

#### 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:

In the given figure some triangles have been given. State for each triangle whether it is scalene, isosceles or equilateral.

Figure

#### 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:

Draw a ∆*ABC*. Take a point *D* on *BC*. Join *AD*. How many triangles do you get? Name them.

Figure

#### Answer:

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

#### Page No 197:

#### Question 11:

Can a triangle have

(i) two right angles?

(ii) two obtuse angles?

(iii) two acute angles?

(iv) each angle more than 60°?

(v) each angle less than 60°?

(vi) each angle equal to 60°?

#### Answer:

(i) No

If the two angles are 90^{o} each, then the sum of two angles of a triangle will be 180^{o}, 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}, which cannot form a triangle.

(iii) Yes

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

(iv) No

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

(vi) Yes

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

#### Page No 197:

#### Question 12:

Fill in the blanks.

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

(ii) The sum of the angles of a triangle is ...... .

(iii) The sides of a scalene triangle are of ....... lengths.

(iv) Each angle of an equilateral triangle measures ...... .

(v) The angles opposite to equal sides of an isosceles triangle are ....... .

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

#### 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:

How many parts does a triangle have?

(a) 2

(b) 3

(c) 6

(d) 9

#### Answer:

Correct option: (c)

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

#### Page No 197:

#### Question 2:

With the angles given below, in which case the construction of triangle is possible?

(a) 30°, 60°, 70°

(b) 50°, 70°, 60°

(c) 40°, 80°, 65°

(d) 72°, 28°, 90°

#### 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}

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

#### Page No 197:

#### Question 3:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is

(a) 60°

(b) 80°

(c) 76°

(d) 84°

#### Answer:

(b) 80^{o}

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

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

⇒ (9x)^{o }= 180^{o}

⇒ x = 180 / 9

⇒ x = 20^{o}

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

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

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

#### Page No 198:

#### Question 4:

The two angles of a triangle are complementary. The third angle is

(a) 60°

(b) 45°

(c) 36°

(d) 90°

#### 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}.

*x + y + z* = 180^{o}

⇒ 90^{o} + *z = *180^{o}

⇒ *z * = 180^{o }$-$ 90^{o }

= 90^{o}

The third angle is 90^{o}.

#### Page No 198:

#### Question 5:

One of the base angles of an isosceles triangle is 70°. The vertical angle is

(a) 60°

(b) 80°

(c) 40°

(d) 35°

#### 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 $ ∠B = 70^{o}

We need to find the vertical angle ∠C.

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

∠A + ∠B + ∠C = 180^{o}

⇒ 70^{o} + 70^{o} + ∠C = 180^{o}

⇒ 140^{o }+ ∠C = 180^{o}

⇒ ∠C = 180^{o}^{ }$-$ 140^{o}

⇒ ∠C = 40^{o}

#### Page No 198:

#### Question 6:

A triangle having sides of different lengths is called

(a) an isosceles triangle

(b) an equilateral triangle

(c) a scalene triangle

(d) a right triangle

#### Answer:

Correct option: (c)

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

#### Page No 198:

#### Question 7:

In an isosceles ∆*ABC*, the bisectors of ∠*B* and ∠*C* meet at a point *O*. If ∠*A* = 40°, then ∠*BOC* =

?

(a) 110°

(b) 70°

(c) 130°

(d) 150°

#### Answer:

Correct option: (a)

In the isosceles ABC, 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 ∆BOC:

∠OBC = 35^{o} and ∠OCB = 35^{o}

∠BOC + ∠OBC + ∠OCB = 180^{o}

⇒ ∠BOC + 35^{o} + 35^{o}^{ }= 180^{o}

⇒ ∠BOC = 180^{o} - 70^{o}

⇒ ∠BOC = 110^{o}

#### Page No 198:

#### Question 8:

The sides of a triangle are in the ratio 3 : 2 : 5 and its perimeter is 30 cm. The length of the longest side is

(a) 20 cm

(b) 15 cm

(c) 10 cm

(d) 12 cm

#### 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:

Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is

(a) 35°

(b) 45°

(c) 65°

(d) 125°

#### 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}

x = 180^{o} $-$ 55^{o}

x = 125^{o}

#### Page No 198:

#### Question 10:

Each angle of an equilateral triangle measures

(a) 30°

(b) 45°

(c) 60°

(d) 80°

#### Answer:

Correct option: (c)

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

#### Page No 198:

#### Question 11:

In the adjoining figure, the point *P* lies

(a) in the interior of ∆*ABC*

(b) in the exterior of ∆*ABC*

(c) on ∆*ABC*

(d) outside ∆*ABC*

Figure

#### Answer:

Correct option: (c)

Point P lies on ∆ABC.

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