Question

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 ofthe bowl.

Answer 1

Hello!

• Inner radius, r = $\bf{5 \:cm}$
• Outer radius, R = 5 + 0.25 cm = $\bf{5.25\: cm}$

Then,

$\underline{\bf{Outer\:curved\:surface\:Area}}$ :

$\sf 4πr^{2} \\ \\ \implies 4 \times \dfrac{22}{7} \times (5.25)^{2} \\ \\ \implies \dfrac{2425.5}{7} \implies \boxed{\red{\bf{346.5\:cm^{2}}}}$

Cheers!

Answer 2

Answer:

$173.25\ cm^{2}$

Step-by-step explanation:

According to the question, Inner radius of the ball is 5 cm.

Thick of ball = 0.25 cm

we are to find the outer curved surface area.

In this case, we need to find outer radius.

Hence outer radius, r = 5 + 0.25

                                    = 5.25 cm

Now

we know that,

Curved surface area of hemisphere = (1/2) surface area of a sphere

                                                            = $\frac{1}{2}\times2\pi\ r^{2}$

                                                            = $2\pi\ r^{2}$

∴ Outer curved surface area of a hemispherical ball = $2\pi\ r^{2}$

                                                                                      = $2\times\frac{22}{7}\times5.25\times5.25$

                                                                                      = $173.25\ cm^{2}$

Hence the outer curved surface area of the ball is $173.25\ cm^{2}$

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