Question

A semicircular and thin sheet of metal of radius 14 cm is bent and an open conical Cup is made the capacity of the cup is

Answer 1

Answer:

depth of the conical cup = 12.124 cm and it's capacity is 622.365 cm³.

Step-by-step explanation:

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

  = 12.124 cm

Height or depth of the conical cup = 12.124 cm

So, 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 cm³.

i hope it will helps you friend

Answer 2

$\huge\bold {Hey!!!}$

Hey mate here is yur answer :

Radius, r = 14 cm

Circumference of semi circle = πr = (22/7) x 14 = 44 cm

This becomes circumference of base of cone 2πR = 44 R = 44 x (7/44) R = 7 cm

The radius of semi circular sheet = slant height of conical cup That is l = 7 cm

We know that r2 + h2 = l2 142 + h2 = 72 196 – 49 = h2 h2 = 147

Therefore, h = 7√3 cm That is depth of the conical cup is 7Ö3 cm Capacity of cup = (1/3) πr2h = (1/3) x (22/7) x 72 x 7√3 = 622.37 cu cm