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#### Page No 196:

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:

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 180o​.
$\therefore$    + 72+ 58o = 180o
⇒ x + 130= 180o
⇒ x = 180o$-$ 130o
⇒ x = 50o
​The measure of the third angle of the triangle is 50o​.

#### Page No 196:

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 = 180o
∴ 1x + 3x + 5x = 180o
⇒ 9x = 180o
⇒ x = 20o
1x = 20o
3x = 60o
​5x = 100o
The measures of the angles are 20o, 60o and 100o

#### Page No 196:

In a right angle triangle, one of the angles is 90o.
It is given that one of the acute angled of the right angled triangle is 50o.
We know that the sum of the measures of all the angles of a triangle is 180o.
Now, let the third angle be x.
​Therefore, we have:
90o​ + 50o= 180o
⇒        140o = 180o
​   ⇒                    x = 180o $-$ 140o
⇒                     x =  40o
The third acute angle is 40o​.

#### Page No 196:

Given:
∠A = 110o and ​∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180o .
∠A + ∠B + ∠C = 180o
⇒    110o + ​∠B + ∠B = 180o
​      ⇒    110o  + 2​∠B = 180o
​
⇒                2​∠B = 180$-$ 110o
⇒                2∠B =  70o
⇒                  ∠B = 70/ 2
⇒                  ∠B = 35o

∴ ​∠C = 35o
The measures of the three angles:
∠A = 110o, ∠B = 35o, ​∠C = 35o

#### Page No 196:

Given:
∠A = ∠B + ∠C
​We know:
∠A + ∠B + ∠C = 180o
⇒ ∠B +∠C + ∠B + ∠C = 180o
​    ⇒ 2∠B + 2∠C = 180o
​    ⇒ 2(∠B +∠C) = 180o
​    ⇒ ∠B + ∠C = 180/2
⇒ ​∠B + ∠C = 90o
$\therefore$ ∠A = 90o
This shows that the triangle is a right angled triangle.

#### Page No 196:

Let 3∠A = 4 ∠B = 6 ∠C = x
Then, we have:

#### Page No 196:

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

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

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

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

#### Page No 197:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60o.
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 60o.
(vi) Isosceles
Two angles are equal in measure.
(vii) Scalene
All the angles are unequal.

#### Page No 197:

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

#### Page No 197:

(i) No
If the two angles are 90o each, then the sum of two angles of a triangle will be 180o​, which is not possible.
(ii) No
For example, let the two angles be 120o and 150o. Then, their sum will be 270o​, which cannot form a triangle.
(iii) Yes
For example, let the two angles be 50o and 60o​, which on adding, gives 110o. They can easily form a triangle whose third angle is 180o $-$ 110o = 70o​.
(iv) No
For example, let the two angles be 70o​ and 80o, which on adding, gives 150o. They cannot form a triangle whose third angle is 180o$-$ 150= 30o, which is less than 60o.
(v) No
For example, let the two angles be 50o and 40o, which on adding, gives 90o . Thus, they cannot form a triangle whose third angle is 180o $-$ 90o = 90o​, which is greater than 60o.
(vi) Yes
Sum of all angles = 60o + 60o + 60o​ = 180o

#### Page No 197:

(i) A triangle has 3 sides 3 angles and 3 vertices.
(ii) The sum of the angles of a triangle is 180o
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60o.
(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:

Correct option: (c)
A triangle has 6 parts: three sides and three angles.

#### Page No 197:

Correct option: (b)
(a) Sum = 30° + 60° + 70° = 160o
This is not equal to the sum of all the angles of a triangle.
(b) Sum = 50° + 70° + 60° = 180o
Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185o
This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190o
This is not equal to the sum of all the angles of a triangle.

#### Page No 197:

(b) 80o
Let the measures of the given angles be (2x)o, (3x)o​ and (4x)o.
$\therefore$ (2x)o + (3x)+ (4x)o = 180o
⇒ (9x)o = 180o
​  ⇒ x = 180 / 9
⇒ x = 20o
$\therefore$ ​2x =  40o, 3x = 60o, 4x = 80o

Hence, the measures of the angles of the triangle are 40o​, 60o, 80o.
Thus, the largest angle is 80o.

#### Page No 198:

Correct option: (d)
The measure of two angles are complimentary if their sum is 90o degrees.
Let the two angles be x and y, such that x + y = 90o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180o​.
x + y + z​ = 180o
⇒ 90o + z = 180o
⇒ = 180o $-$ 90
= 90o
​The third angle is 90o.

#### Page No 198:

Correct option: (c)
Let ∠A = 70o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
$\therefore$ ​∠B = 70o
​We need to find the vertical angle ​∠C.
Now, sum of all the angles of a triangle is 180o.
∠A + ∠B + ∠C = 180o
⇒​ 70o + 70o​ + ∠C = 180o
⇒ 140+ ​∠C = 180o
⇒ ∠C = 180o $-$ 140o
⇒ ∠C = 40o

#### Page No 198:

Correct option: (c)
A triangle having sides of different lengths is called a scalene triangle.

#### Page No 198:

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 = 180o
⇒  40o + 2∠B = 180o
⇒ 2∠B = 140o
⇒ ∠B = 70o
Bisectors of an angle divide the angle into two equal angles.
So, in  ∆BOC:
∠OBC = 35o and ∠OCB = 35o
∠BOC + ∠OBC + ∠OCB = 180​o
⇒ ∠BOC + 35o + 35o = 180o
⇒ ∠BOC = 180o​ - 70o
⇒ ∠BOC = 110o

#### Page No 198:

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:

Correct option: (d)
Two angles of a triangle measure 30° and 25°, respectively.
Let the third angle be x.
x + 30o + 25o = 180o
​                    x = 180o $-$ 55o
x = 125o

#### Page No 198:

Correct option: (c)
Each angle of an equilateral triangle measures 60o.