Question

A hemispherical bowl is made of 0.25 cm thick steel. Internal radius of this bowl is 5 cm. Find the external curved surface of the bowl.

Answer 1

$\huge\mathfrak{Solution:}$

Internal radius r = 5 cm

So, external radius R = internal radius + thickness of steel = 5 cm + 0.25 cm = 5.25 cm

So, external curved surface area of the bowl = 2 × pi × R^2

= 2 × 22/7 × (5.25)^2 cm^2

= 2 × 22/7 × 525/100 × 525/100 cm^2

= 2 × 22/7 × 21/4 × 21/4 cm^2

= 693/4 cm^2

= 173.25 cm^2

__________________

HOPE IT HELPS ❤❤

Answer 2

HERE is ur Answer ✍️✍️

Internal radius

r = 5 cm

So,

external radius R

= internal radius + thickness of steel

= 5 cm + 0.25 cm

= 5.25 cm 

So,

external curved surface area of the bowl

= 2 × pi × R^2 

= 2 × 22/7 × (5.25)^2 cm^2

= 2 × 22/7 × 525/100 × 525/100 cm^2

= 2 × 22/7 × 21/4 × 21/4 cm^2

= 693/4 cm^2 

= 173.25 cm^2 

hope it helps u❤️❤️