Maths NCERT Solutions Class 9 Exercise 13.7

 

Ex 13.7 NCERT Solutions Class 9 Chapter 13

The solutions to exercise 13.7 (from class 9th's maths NCERT book) are given below. The exercise has 9 questions which have been explained thoroughly with proper reasoning.

(i)   Volume of a right circular cone

= ⅓ x π r 2 h

= ⅓ x 22/7 x 6 x 6 x 7

= 22 x 2 x 6

= 264 cm3

(ii) Volume of a right circular cone

= ⅓ x π r 2 h

= ⅓ x 22/7 x 3.5 x 3.5 x 12

= 22 x 0.5 x 3.5 x 4

= 11 x 14

= 154 cm3

(i)   l 2 = h 2 + r 2

25 2 = h 2 + 7 2

625 - 49 = h 2

576 = h 2

h= 24 cm

Volume of a right circular cone

= ⅓ x π r 2 h

= ⅓ x 22/7 x 7 x 7 x 24

= 22 x 7 x 8

= 1232 cm3

= 1.232 litres

(ii)    l 2 = h 2 + r 2

13 2 = 12 2 + r 2

169 - 144= r 2

25 = r 2

h= 5 cm

Volume of a right circular cone

= ⅓ x π r 2 h

= ⅓ x 3.14 x 5 x 5  x 12

= 3.14 x 25 x 4

= 314 cm3

= 0.314 litres

Volume of a right circular cone = ⅓ x π r 2 h

1570 = ⅓ x 3.14 x r 2 x 15 

1570 = 3.14 x r 2 x  5

1570 /(3.14 x 5)  = r 2

157000 /(314 x 5)  = r 2

500 /( 5)  = r 2

r = 10 cm

Volume of a right circular cone = ⅓ x π r 2 h

48 π = ⅓ x π x r 2 x 9

48 = r 2 x 3

16= r 2 

r= 4 cm

Diameter = 2 r = 8 cm

r= 3.5/2 = 1. 75 cm

Volume of a right circular cone 

= ⅓ x π r 2 h

= ⅓ x 22/7 x 1.75 x 1.75 x 12

= 22 x 0.25 x 1.75 x 4

= 22 x 1 x 1.75

= 38.5 m3

1 m3= 1000 litres, or

1 m3= 1 kilolitres

Therefore, V= 38.5 kiloliters

(i)Volume of a right circular cone =  ⅓ x π r 2 h

r= diameter/2 = 28/2= 14 cm

9856 = ⅓ x π r 2 h

9856 = ⅓ x 22/7 x  14 x 14 x h

9856 = ⅓ x 22 x  2 x 14 x h

(9856 x 3)/(22 x  2 x 14)= h

h= 48 cm

(ii)l 2 = h 2 + r 2

l 2 = 48 2 + 14 2

l 2 = 2304  + 196

l 2 = 2500

Slant height = 50 cm

(iii) CSA of cone = π r l

= 22/7 x 14 x 50

= 22 x 2 x 50

= 2200 m2

It is revolved around the side 5 cm, so it acts as our radius, 12 cm is the height and slant height is 13 cm

Volume of a right circular cone 

=  ⅓ x π r 2 h

= ⅓ x π x 5 x 5 x 12

= π x 25 x 4

= 100 π cm2, or

= 314 cm2

Therefore, 5 cm now acts as the height, 12 is the radius and 13 is the slant height

Volume of a right circular cone 

=  ⅓ x π r 2 h

= ⅓ x π x 12 x 12 x 5

= 240 π cm2

Ratio 

 100 π : 240 π

5 : 12

Volume of the conical heap

=  ⅓ x π r 2 h

= ⅓ x 22/7 x 5.25 x 5.25 x 3

= 22 x 0.75 x 5.25

= 86.625 cm3

l 2 = h 2 + r 2

l 2 = 3 2 + 5.25 2

l 2 = 36.5625

Slant height = 6.046 cm

CSA of cone

= π r l

= 22/7 x 5.25 x 6.046

= 22 x 0.75 x 6.05

= 99.825 cm2