Question

Answer the Question in Attachment​

Answer 1

$\huge{\underline{\mathfrak{Solution:}}}$



$\sf{\large{\underline{\underline{Given \colon}}}}$

Inner Radius of Bawl = 5 cm

Thickness of Bawl = 0.25 cm



$\sf{\large{\underline{\underline{Explanation \colon}}}}$

Outer Radius of the Bawl
$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf r + t \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf (5 + 0.25) \: cm \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf 5.25 \: cm$

Outer Curved Surface Area of Bowl
$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf 2\pi R {}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf 2 \times \frac{22}{7} \times 5.25 \times 5.25 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf 2 \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf \frac{693}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf 173.25 \: cm {}^{2}$

Answer 2

Answer:

173.25 cm²

Step-by-step explanation:

We have,

Inner radius = 5 cm.

Thickness = 0.25 cm.

∴ Outer radius of the bowl = 5 + 0.25 = 5.25 cm.

Outer curved surface area = 2πr²

= 2 * (22/7) * (5.25)²

= 44 * 3.9375

= 173.25 cm²

∴ Outer curved surface area = 173.25 cm²

Hope it helps!