Chapter 11: Surface Areas and Volumes

Volume of cone = (1/3) πr²h cube units

Where r be radius and h be the height of the cone

(i) Radius of the cone, r = 6 cm

Height of the cone, h = 7cm

Ley V be the volume of the cone, so we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm³.

(ii) Radius of the cone, r = 3.5cm

Height of the cone, h = 12cm

Volume of the cone = (1/3)×(22/7)×3.5²×7 = 154

Hence,

The volume of the cone is 154 cm³.

(i) Radius of the cone, r =7 cm

Slant height of the cone, l = 25 cm

or h = 24

Height of the cone is 24 cm

Now,

Volume of the 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, the capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm³)

= 1.232 Liters.

(ii) Height of the cone, h = 12 cm

Slant height of the cone, l = 13 cm

r = 5

Hence, the radius of the cone is 5 cm.

Now, Volume of the cone, V = (1/3)πr²h

V = (1/3)×(22/7)×52×12 cm³

= 2200/7

Volume of the cone is 2200/7 cm³

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm³)

= 11/35 litres

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 the 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 the cone 10 cm.

Height of cone, h = 9cm

Volume of cone =48π cm³

Let r be the radius of the cone.

As we know, volume of the cone, V = (1/3) πr²h

So, 1/3 π r²(9) = 48 π

r² = 16

r = 4

Radius of the cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of the base is 8cm.

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 the cone is 38.5 m³

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

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.

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.

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 is the height of the cone.

= (1/3)×π×12×12×5

= 240 π

The volume of the cones formed is 240π cm³.

So, the required ratio = (the result of question 7) / (the result of question 8) = (100π)/(240π) = 5/12 = 5:12.

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.