Question

a hemisphericsl bowl is made of steel ,0.25 cm thick .The inner r=5 cm .find it's outer c.s.a of the bowl​

Answer 1

Answer:

thickness = 0.25 cm

inner r = 5cm

outer radius = 5+0.25=5.25 cm

CSA of a hemispherical bowl = 2πr^2

= 2×22/7 × 5.25×5.25

= 173.25 cm^2

Answer 2

$\huge\mathfrak{Answer}$

Inner radius of bowl, r = 5 cm

Thickness of steel, t = 0.25 cm

Outer radius of bowl, R = r + t

= 5 + 0.25 = 5. 25 cm

Outer curved surface area of bowl = 2$\pie$R²

$= 2 \times \frac{22}{7} \times 5.25 \times 5.25$

$= 2\times \frac{22}{7} \times \frac{525}{100} \times \frac{525}{100}$

$= 2 \times \frac{227}{7} \times \frac{21}{4} \times \frac{21}{4}$

$= \frac{11 \times 3 \times 21}{4}$

$= \frac{693}{4} = 173.23 {cm}^{2}$

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