Question

A hemispherical bowl is made up of stone whose thickness is 5 cm. If the inner radius is 35 cm, find the total surface area of the bowl.
d'r' he inner radius Thickness of ring=5 cm​

Answer 1

↬Let R1 be outer radius and R2 be inner radius Thickness of ring =5cm.

∴R=(r+5)cm

=(35+5)cm

=40cm

⇒Total Surface Area = CSA of outer hemisphere + CSA of inner hemisphere + area of the ring.

$= 2\pi \: r {1}^{2} + 2\pi \: r2 {}^{2} + \pi(r1 {}^{2} - r2 {}^{2} )$

$= \pi \: (2r1 {}^{2} + 2r2 {}^{2} + r1 {}^{2} - r2 {}^{2} )$

$= \frac{22}{7} (3r {1}^{2} + r2 {}^{2} )$

$= \frac{22}{7} (3 \times {40}^{2} + {35}^{2}) cm {}^{2}$

$= \frac{6025 \times 22}{7} cm {}^{2}$

$= 18935.71cm {}^{2}$

Hope it's helpful to you

@ Dhruvshi23 ✨

Answer 2

Answer:

Let R1 be outer radius and R2 be inner radius Thickness of ring =5cm.

∴R=(r+5)cm

=(35+5)cm

=40cm

⇒Total Surface Area = CSA of outer hemisphere + CSA of inner hemisphere + area of the ring.

= 2\pi \: r {1}^{2} + 2\pi \: r2 {}^{2} + \pi(r1 {}^{2} - r2 {}^{2} )=2πr1

2

+2πr2

2

+π(r1

2

−r2

2

)

= \pi \: (2r1 {}^{2} + 2r2 {}^{2} + r1 {}^{2} - r2 {}^{2} )=π(2r1

2

+2r2

2

+r1

2

−r2

2

)

= \frac{22}{7} (3r {1}^{2} + r2 {}^{2} )=

7

22

(3r1

2

+r2

2

)

= \frac{22}{7} (3 \times {40}^{2} + {35}^{2}) cm {}^{2} =

7

22

(3×40

2

+35

2

)cm

2

= \frac{6025 \times 22}{7} cm {}^{2} =

7

6025×22

cm

2

= 18935.71cm {}^{2} =18935.71cm

2

Step-by-step explanation:

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