Ex 12.1 NCERT Solutions Class 9 Chapter 12

The solutions to exercise 12.1 (from class 9th's maths NCERT book) are given below. The exercise has 6 questions which have been explained thoroughly with proper reasoning. 

class 9 Exercise 12.1 question 1


The traffic signal board is an equilateral triangle with a perimeter 180 cm

Perimeter = sum of all sides

          180 = 3a

     60 cm  = a

Semi-perimeter = 90 cm

The area of the triangle according to Heron’s formula is 

Area = √s(s-a)(s-a)(s-a)  (since all sides are equal)

        = √90 (90-60)(90-60)(90-60)

        =    √90 (30)(30)(30)

        =  3 x 3 x 100√ 3

        = 900√3 cm2


class 9 Exercise 12.1 question 2

Let a = 122 m, b = 22 m, c = 120 m; the three sides of the triangular-shaped wall Area of the triangle = 1/2 x base x height = 1/2 x 120 x 22 = 60 x 22 = 1320 m2 I m2 of the wall is rented for = Rs 5000/year 1320 m2 of the wall will fetch rent = 5000 x 1320 => Rs 6600000/year Rent for 12 months = Rs 6600000 Rent for 3 months will be = 6600000 x 3/12 = Rs 1650000


class 9 Exercise 12.1 question 3

Let a= 11 m, b= 6 m and c = 15 m be the sides of the side walls of the triangular slide Semi perimeter, s = (a+b+c)/2 s= (11+6+15)/2 s= 32/2 s = 16 m Area of the painted side wall = √(s(s-a)(s-b)(s-c)) = √(16(16-11)(16-6)(16-15)) = √(16(5)(10)(1)) =20√2 m2


class 9 Exercise 12.1 question 4

Two sides of a triangle = a= 18 cm and b = 10 cm; let the third side be c

Perimeter of the triangle = 42 cm

18 cm + 10 cm + c = 42 cm  {i.e. sum of all the sides of the triangle = 42 cm}

28 + c = 42

c=42-28

c=14

Semi perimeter: (a+b+c)/2

s= (18+10+14)/2

s=42/2

s= 21 cm

Area of the triangle: √s(s-a)(s-b)(s-c) = √(21(21-18)(21-10)(21-14)) = √(21)(3)(11)(7)

= 21√11 cm2 


class 9 Exercise 12.1 question 5

Let the sides of the triangle be a=12x, b=17x, and c=25 x

Perimeter of the triangle = 540 cm

12x + 17x + 25x = 540

54x = 540

x= 10 cm


So the sides are 120 cm, 170 cm and 250 cm

Semi perimeter = Perimeter/2 =  540/2 = 270 cm

Area of the triangle: √(s(s-a)(s-b)(s-c) ) = √(270(270-120)(270-170)(270-250) ) = √(270(150)(100)(20) ) = √((9)(30)(30)(5)(5)(20)(20) ) = 3 x 30 x 5 x 20 = 9000 m2 


class 9 Exercise 12.1 question 6

Let the sides of the isosceles triangle be a= 12 cm, b= 12 cm and c

Perimeter = a+ b+ c

30 = 12+ 12 + c

30 – 24 = c

6 cm = c

Semi perimeter = Perimeter/2 = 30/2 = 15 cm

Area of the triangle: √(s(s-a)(s-b)(s-c) ) = √(15(15-12)(15-12)(15-6) ) = √(15(3)(3)(9) ) = 9√15 cm2