Question

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

Answer 1

Answer:

Given

Inner radius = 5 cm

Hence outer radius = 5 +0.25=5.25

Therefore , outer curved surface area =2πr^2=2×22÷7×5.25×5.25

=173.25 cm^2

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

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