Question

8. A hemispherical bowl is made of steel,
0.25 cm thick. The inner radius of the bowl
is 5 cm. Find the outer curved surface area
of the bowl.
​

Answer 1

Answer:

pls mark as brainlist answer

Step-by-step explanation:

r=5cm, thickness of steel sheet =0.25cm

⇒ R=5cm+0.25cm=5.25cm

outer curved surface area of the bowl =2πR

2

=2× 22/7× 525/100×525/100cm 2 =173.25cm 2

Answer 2

$given$

$inner \: radius \: = 5 \: cm$

$hence \: outer \: radius = 5 + 0.25 = 5.25$

Therefore, outer curved surface area :

$= 2 {\pi}^{2} = 2 \times 22 \div 7 \times 5.25 \times 5.25$

$= 173.25 \: {cm}^{2}$