Rs Aggarwal Solutions for Class 6 Math Chapter 16 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 6 students for Math Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal Book of Class 6 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal Solutions. All Rs Aggarwal Solutions for class Class 6 Math are prepared by experts and are 100% accurate.

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 180o​.
     + 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:

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 = 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:

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 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:

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 = 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:

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 = 180o
    ⇒ ∠B +∠C + ∠B + ∠C = 180o
​    ⇒ 2∠B + 2∠C = 180o
​    ⇒ 2(∠B +∠C) = 180o
​    ⇒ ∠B + ∠C = 180/2
    ⇒ ​∠B + ∠C = 90o
 ∠A = 90o
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:
 A = x3, B = x4, C = x6But, A + B + C = 180° x3 + x4 + x6 = 180°or 4x + 3x + 2x12 = 180°or 9x = 180° × 12 = 2160°or x = 240°  A = 2403 = 80°, B = 2404 = 60°, C = 2406 = 40°

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 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:

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 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:

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 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:

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 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:

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° = 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:

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) 80o
Let the measures of the given angles be (2x)o, (3x)o​ and (4x)o.
(2x)o + (3x)+ (4x)o = 180o
  ⇒ (9x)o = 180o
​  ⇒ x = 180 / 9
  ⇒ x = 20o
​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:

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 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:

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 = 70o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
 ​∠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:

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
∠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:

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 + 30o + 25o = 180o
​                    x = 180o - 55o
                     x = 125o

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 60o.

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.



View NCERT Solutions for all chapters of Class 6