Question

A semi circular sheet of paper of diameter 42cm is bent into an open conical cup. Find the depth and volume of the cup.

Answer 1

Solution:-
Slant height of the conical cup, 'l' = radius of the semi-circular sheet, R = 14 cm
Let radius and height of the conical cup be 'r' and 'h' respectively.
Circumference of the base of the cone = Length of arc of the semi-circle
Or, 2πr = (1/2)2πR
Or, 2πr = (1/2)(2π)(14)
Or, r = 7 cm
Now, we know that l² = h² + r²
(14)² = (h)² + (7)²
h² = 196 - 49
h = √147
Height or depth of the conical cup = 12.124 cm
Now, capacity of the conical cup = 1/3πr²h
= 1/3*22/7*7*7*12.124
= 13069.672/21
Capacity of the conical cup = 622.365 cu cm
Answer.

Answer 2

Answer:

Height or depth of the conical cup = 12.124 cm

Capacity or volume of the conical cup = 622.365 cu cm

Step-by-step explanation:

Given:

A semicircular sheet of paper of diameter 42cm is bent into an open conical cup.

To find:

Depth and volume of the cup

Solution:

'l' = radius of the semi-circular sheet, R = 14 cm Slant height of the conical cup

Let's call the radius and height of the conical cup 'r' and 'h', respectively.

Circumference of the cone's base = arc length of the semi-circle

2$\pi$r = (1/2)2$\pi$R

Alternatively, 2$\pi$r = (1/2)(2$\pi$)(14)

And, r = 7 cm

We now know that $l^{2}$= $h^{2}$ + $r^{2}$

$14^{2}$ = $h^{2}$ + (7)²

h² = 196 - 49

h = √147

12.124 cm is the height or depth of the conical cup.

Now, the conical cup's capacity = 1/3$\pi$$r^{2}$h 

= 1/3*22/7*7*7*12.124

= 13069.672/21.

Conical cup capacity = 622.365 cu cm

Hence, 12.124 cm is the height or depth of the conical cup and Conical cup capacity = 622.365 cu cm.

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