Ex 13.4 NCERT Solutions Class 9 Chapter 13

The solutions to exercise 13.4 (from class 9th's maths NCERT book) are given below. The exercise has 9 questions which have been explained thoroughly with proper reasoning. 



(i) TSA of the sphere

= 4 x πr 2

=4 x (22/7) x 10.5 x 10.5

= 4 x 22 x 1.5 x 10.5

= 22 x 6 x 10.5

= 132 x 10.5

= 1386 cm2

 

(ii)TSA of the sphere

= 4 x πr 2

= 4 x (22/7) x 5.6 x 5.6

= 4 x 22 x 0.8 x 5.6

= 88 x 4.48

= 394.24 cm2

 

(iii)TSA of the sphere

= 4 x πr 2

= 4 x (22/7) x 14 x 14

= 4 x 22 x 2 x 14

= 2464 cm2

 

(i) TSA of the sphere

= 4 x πr 2

= 4 x (22/7) x 7 x 7

= 4 x 22 x 7

= 616 cm2

 

(ii) TSA of the sphere

= 4 x πr 2

= 4 x (22/7) x (21/2) x (21/2)

= 22 x 3 x 21

= 66 x 21

=1386 cm2

 

(iii) TSA of the sphere

= 4 x πr 2

= 4 x (22/7) x (3.5/2) x (3.5/2)

= 22 x 0.5 x 3.5

= 38.50 cm2

TSA of the hemisphere

= 3 x πr 2

= 3 x (3.14) x 10 x 10

= 3 x 314

= 942 cm2





Finding the TSA of the balloon when r = 7 cm

= 4 x πr 2

TSA1 = 4 x (22/7) x 7 x 7 …….(1)

Finding the TSA of the balloon when r = 14 cm

= 4 x πr 2

TSA2= 4 x (22/7) x 14 x 14 ………..(2)


TSA1/TSA2= { 4 x (22/7) x 7 x 7}/ { 4 x (22/7) x 14 x 14}

TSA1/TSA2=  ¼





The inside of the hemispherical bowl is to be tin painted,

so we will find CSA of the hemisphere

Inner CSA of the hemispherical bowl

= 2 x πr 2

= 2 x (22/7) x (10.5/2) x (10.5/2)

= 11 x 1.5 x 10.5

= 173.25  cm2


Cost of tin painting 100 = Rs 16

Cost of tin painting 173.25 =

= {(16 x 173.25) / 100}

= 16 x 1.73 = Rs 27.68


TSA of the sphere = 4 x πr 2

154 = 4 x (22/7) x r 2

(154 x7)/ 88 = r 2

(14 x 7)/8 = r 2

(7 x 7)/ 4 = r 2

7 2/2 2 = r 2

r= 3.5 cm





Let the diameter of the earth is n, and the moon's diameter be n/4

TSA of Earth

= 4 x πr 2

= 4 x π x (n/2) 2


TSA of Moon

= 4 x πr 2

= 4 x π x (n/8) 2



Ratio of TSA of Earth to that of Moon

= {4 x π x (n/2) 2}/ {4 x π x (n/8) 2}

= 64/4

= 16:1



Inner radius = 5 cm



Outer radius = inner radius + the thickness of the bowl = 5 cm + 0.25 cm = 5.25 cm



Outer CSA of the hemispherical bowl

= 2 x πr 2

= 2 x (22/7) x (5.25) 2

= 2 x 22 x 0.75 x 5.25

= 173.25  cm2




Height of the cylinder equals the diameter of the sphere

 (i) TSA of the sphere

= 4 x πr 2

(ii) CSA of cylinder

= 2 x πrh

= 2 x 2 x πr 2

(iii) Ratio of TSA of sphere to CSA of cylinder

= (4x πr 2)/(4 x πr 2)

= 1:1