Question

A HEMISPHERICAL BOWL IS MADE OF STEEL 0.25 CM THICK. THE INNER RADIUS OF THE BOWL IS 5CM. FIND THE OUTER CURVED SURFACE AREA OF THE BOWL (PIE=22/7)

Answer 1

Thickness = 0.25 cm
Inner radius = 5 cm
Now, the outer radius = 5.25 cm
Thus, the outer curved surface area of the bowl = 2/3×22/7×r×r×r
= 2/3×22/7×5.25×5.25
= 57.75 cm^2

.....hope you got your answer.

Answer 2

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$\small\bold{Inner \: radius \: of \: the \: bowl \: = }$ 5cm

$\small\bold{Thickness \: of \: the \: steel \: =}$ 0.25cm

$\small\bold{Outer \: radius \: of \: the \: bowl \: =}$ 5+0.25=5.25cm

$\small\bold{Outer \: Curved \: surface \: of \: the \: hemispherical \: bowl \: =}$ 2πr²

= 2× $\large\dfrac{22}{7}$ × (5.25)² = 172.25 cm²

${\huge{\red{\underline{\overline{\mathscr{\red{Hope \: This \: Helps \: You}}}}}}}$