NCERT Solutions Class 9th Maths Chapter – 11 Surface Areas and Volumes Exercise 11.3

(i) radius 6cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm

Solution – Volume of cone = (1/3) πr2h cube units
Where r be radius and h be the height of the cone

(i) Radius of cone, r = 6 cm
Height of cone, h = 7cm
Say, V be the volume of the cone, we have
V = (1/3)×(22/7)×36×7
= (12×22)
= 264
The volume of the cone is 264 cm3.

(ii) Radius of cone, r = 3.5cm
Height of cone, h = 12cm
Volume of cone = (1/3)×(22/7)×3.52×7 = 154
Hence,
The volume of the cone is 154 cm3.

(i) radius 7cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm

Solution
(i) Radius of cone, r =7 cm
Slant height of cone, l = 25 cm

or h = 24
Height of the cone is 24 cm
Now,
Volume of cone, V = (1/3) πr2h (formula)
V = (1/3)×(22/7) ×72×24
= (154×8)
= 1232
So, the volume of the vessel is 1232 cm3
Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3)
= 1.232 Liters.

(ii) Height of cone, h = 12 cm
Slant height of cone, l = 13 cm

r = 5
Hence, the radius of cone is 5 cm.
Now, Volume of cone, V = (1/3)πr2h
V = (1/3)×(22/7)×52×12 cm3
= 2200/7
Volume of cone is 2200/7 cm3
Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm3)
= 11/35 litres

Solution – Height of the cone, h = 15 cm
Volume of cone =1570 cm3
Let r be the radius of the cone
As we know: Volume of cone, V = (1/3) πr2h
So, (1/3) πr2h = 1570
(1/3)×3.14×r2 ×15 = 1570
r2 = 100
r = 10
Radius of the base of cone 10 cm.

Solution – Height of cone, h = 9cm
Volume of cone =48π cm3
Let r be the radius of the cone.
As we know: Volume of cone, V = (1/3) πr2h
So, 1/3 π r2(9) = 48 π
r2 = 16
r = 4
Radius of cone is 4 cm.
So, diameter = 2×Radius = 8
Thus, diameter of base is 8cm.

Solution – Diameter of conical pit = 3.5 m
Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m
Height of pit, h = Depth of pit = 12m
Volume of cone, V = (1/3) πr2h
V = (1/3)×(22/7) ×(1.75)2×12 = 38.5
Volume of cone is 38.5 m3
Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone

Solution – Volume of a right circular cone = 9856 cm3
Diameter of the base = 28 cm
(i) Radius of cone, r = (28/2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V = (1/3) πr2h
(1/3) πr2h = 9856
(1/3)×(22/7) ×14×14×h = 9856
h = 48
The height of the cone is 48 cm.

Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl
= (22/7)×14×50
= 2200
curved surface area of the cone is 2200 cm2.

Solution – Height (h)= 12 cm
Radius (r) = 5 cm, and
Slant height (l) = 13 cm

Volume of cone, V = (1/3) πr2h
V = (1/3)×π×52×12
= 100π
Volume of the cone so formed is 100π cm3.

Solution

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.
Volume of cone = (1/3) πr2h; where r is the radius and h be the height of cone
= (1/3)×π×12×12×5
= 240 π
The volume of the cones of formed is 240π cm3.
So, required ratio = (result of question 7) / (result of question 8) = (100π)/(240π) = 5/12 = 5:12.

Solution – Radius (r) of heap = (10.5/2) m = 5.25
Height (h) of heap = 3m
Volume of heap = (1/3)πr2h
= (1/3)×(22/7)×5.25×5.25×3
= 86.625
The volume of the heap of wheat is 86.625 m3.
Again,

= (22/7)×5.25×6.05
= 99.825
Therefore, the area of the canvas is 99.825 m2.