Question

The thickness of hemispherical bowl is 0.25cm. The inner radius of the bowl is 5cm. Find the outer curved surface area of the bowl.​

Answer 1

Answer:

Answer

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

⇒ R=5cm+0.25cm=5.25cm

outer curved surface area of the bowl =2πR

2

=2×

7

22

×

100

525

×

100

525

cm

2

=173.25cm

2

Step-by-step explanation:

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Answer 2

Answer:

$\large \bold \pink{answer:}$

Inner radius of the bowl (r) = 5cm

Thickness of steel = 0.25cm

Therefore, outer radius of bowl(r) = (5 to 25)cm = 5.25

Outer Curved surface area of bowl

$2\pi\: {r}^{2} = 2 \times \frac{22}{7} \times ({5.25})^{2} \: {cm}^{2} \\ = 173.25 \: {cm}^{2}$