Question

10. The outer and inner diameters of a hemispherical bowl are 17cm and 15cm find the cost of polishing it over at 25 paise
percm^2

Answer 1

Answer:Curved surface area of hemisphere.

For outer CAS

= 2π R^2

=2π 17*17

=289*2*π

=578π

For inner CSA

=2π r^2

=2*π*15*15

=225*2*π

=450π

Now area of ring....

=2 π(R^2 - r^2)

=2 π(17^2 - 15^2)

=2 π (289 - 225)

=2 π 64

=138π

Total area to be polise

=578pi+450pi+138pi

=1166 pi

= 1166*22/7

=3664.6 cm^2

Cost of polishing

= 3664.4*24

=87950.4paisa

=87950.4/100

=₹879.5

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