Ex 12.2 NCERT Solutions Class 9 Chapter 12

The solutions to exercise 12.2 (from class 9th's maths NCERT book) are given below. The exercise has 9 questions which have been explained thoroughly with proper reasoning.


class 9 Exercise 12.2 question 1

Draw a diagonal BD, such that triangle BCD is a right-angled triangle

Using Pythagoras theorem

BD2 = BC2 + CD2 

       = 122 +52 

       = 144 + 25

      = 169

BD2 = 132 

BD= 13 cm

Area of the right triangle BCD = 1/2 x base x height = 1/2 x 12 x 5 = 30 cm2 

In Triangle ABD, AB= 9 cm, BD = 13 cm and AD = 8 cm

Semi perimeter = (a+ b+ c)/2  = (9+13+ 8)/2  = 30/2  = 15 cm

Area of triangle ABD = √(s(s-a)(s-b)(s-c) ) = √(15(15-9)(15-13)(15-8) ) = √(15(6)(2)(7) )

= 6√35 cm2 = 6 x 5.916 = 35.49 cm2 

Area of the quadrilateral = Area of triangle ABD + Area of triangle BCD

                                             = 35.49 +30 = 65.49 cm2 


class 9 Exercise 12.2 question 2

Area of triangle ABC

a= 3cm, b= 4cm and c= 5cm

Semi perimeter = (a+ b+ c)/2  = (3+4+ 5)/2  = 12/2  = 6 cm

Area of triangle ABC = √(s(s-a)(s-b)(s-c) ) = √(6(6-3)(6-4)(6-5) ) =√(6(3)(2)(1) )

= 6 cm2 

Area of triangle ACD

a=5 cm, b=4 cm, c=5 cm

Semi perimeter = (a+ b+ c)/2  = (5+4+ 5)/2  = 14/2  = 7 cm

Area of triangle ACD = √(s(s-a)(s-b)(s-c) ) = √(7(7-5)(7-4)(7-5) ) =√(7(2)(3)(2) )

= 2√21 cm2 = 2 x 4.5825 = 9.16 cm2 

Area of the Quadrilateral= Area of triangle ABC + Area of triangle ACD

                                              = 6 + 9.16 => 15.16 cm2


class 9 Exercise 12.2 question 3


Area of IV & V = 2 x area of right triangle

 Area of IV & V =2 x (  1/2 x base x height)

                                    = 1.5 x 6 => 9 cm2

Area of II = area of the rectangle => length x breadth

                 = 6.5 x 1 => 6.5 cm2 


Area of I

a=5 cm, b=5 cm, c=1 cm

Semi perimeter = (a+ b+ c)/2  = (5+1+ 5)/2  = 11/2  = 5.5 cm

Area of triangle ACD = √(s(s-a)(s-b)(s-c) ) = √(5.5(5.5-5)(5.5-1)(5.5-5) ) =√(5.5(0.5)(4.5)(0.5) ) => 2.48 cm2 

Area of III

Square of side 1 cm

Area = 1 cm2

Two right angled triangles with base 0.5 cm and height = 0.866 cm

Area of 2 such triangles = 2 x (  12 x base x height)

                                           = base x height => 0.5 x 0.866 = 0.433 cm2 


Area III = 1 + 0.433 = 1.433 cm2

Area of paper used= 9 + 6.5 +2.48 + 1.433 = 19.42 cm2


class 9 Exercise 12.2 question 4


Area of triangle 

a=26 cm, b=28 cm, c=30 cm

Semi perimeter = (a+ b+ c)/2  = (26+28+ 30)/2  = 84/2  = 42 cm

Area of triangle = √(s(s-a)(s-b)(s-c) ) = √(42(42-26)(42-28)(42-30) ) =√(42(16)(14)(12) ) =√((7)(6)(16)((7)(2))((2)(6)) ) = 7 x 6 x 4 x 2 = 336 cm2


Parallelogram’s base = 28 cm

Area of Parallelogram = base x height

336 = 28 x height

Height = 12 cm


class 9 Exercise 12.2 question 5

Smaller triangle 

OA2 +  OD2 =AD2

242 +  OD2 = 302

OD2 = 900 – 576

OD2 = 324

OD = 18 cm


So the smaller diagonal is 36 cm (d2) long

Area of the rhombus = 12 x d1 x d2 = 12 x 48 x 36 = 24 x 36 = 864 cm2

Each cow will get grazing area = Area of the rhombus/18 = 864/18 = 48 cm2

class 9 Exercise 12.2 question 6

Area of 1 triangle 

a=20 cm, b=50 cm, c=50 cm

Semi perimeter = (a+ b+ c)/2  = (50+50+20)/2  = 120/2  = 60 cm

Area of triangle = √(s(s-a)(s-b)(s-c) ) = √(60(60-50)(60-20)(60-50) ) =√(60(10)(40)(10) ) =100√((6)(4)) =200√6

There are 5 blue and 5 white triangles

Cloth required for each color 

= 200√6 x 5 = 1000√6 cm2


class 9 Exercise 12.2 question 7

Let the side of the square be ‘a’

Diagonals bisect each other at 90 degrees, so consider this figure 

Area of region I = 1/2 X base x height

                 = 1/2 X 32 x 16 = 256 cm2 = Area of region II


Area of region III 

It’s an isosceles triangle with sides 6 cm each and base 8 cm

a=6 cm, b=6 cm, c=8 cm

Semi perimeter = (a+ b+ c)/2  = (6+6+8)/2  = 20/2  = 10 cm


Area of triangle = √(s(s-a)(s-b)(s-c) ) = √(10(10-6)(10-6)(10-8) ) =√(10(4)(4)(2) ) = √((2)(5)(4)(4)(2) ) =8√5 cm2 = 17.88 cm2


class 9 Exercise 12.2 question 8



Area of 1 triangular tile 

a=9 cm, b=28 cm, c=35 cm

Semi perimeter = (a+ b+ c)/2  = (9+28+35)/2  = 72/2  = 36 cm


Area of triangle = √(s(s-a)(s-b)(s-c) ) = √(36(36-9)(36-28)(36-35) ) =√(36(27)(8)(1) ) = 36√6 = 88.18 cm2


Area of 16 such tiles = 88.16 cm2 x 16 = 1410.906 cm2

Cost of polishing 1 cm2 of a tile = Rs 0.50

Cost of polishing 1410.906 cm2 = 1410.906 x 0.50 => Rs 705.45


class 9 Exercise 12.2 question 8

Area of triangle BCE

a=13 cm, b=14 cm, c=15 cm

Semi perimeter = (a+ b+ c)/2  = (13+14+15)/2  = 42/2  = 21 cm


Area of triangle = √(s(s-a)(s-b)(s-c) ) = √(21(21-13)(21-14)(21-15) ) =√(21(8)(7)(6) ) = √((3)(7)(4)(2)(7)(3)(2)) = 3 X 7 X 2 X 2 = 84 cm2

Also, Area of triangle = 1/2 x base x height

84 = 1/2 x 15 x height

Height = 11.2 cm


Area of trapezium = 1/2 x (sum of parallel sides) x height

                                  = 1/2 x (10+25) x 11.2

                                  = 35 x 5.6 = 196 cm2