Math, asked by honeyhansuja, 3 months ago

The internal and external radii of a hollow hemispherical vessel are
12 cm and 12.5 cm respectively. If the cost for painting 1 cmof the
surface area is 30.05, then the total approximate cost of painting the vessel all over is?

use pi =22/7​

Answers

Answered by Anonymous
12

Answer:given,

internal diameter=24cm

internal radius=24/2=12cm

external diameter=25 cm

external radius =25/2=12.5cm

surface area of internal bowl=2πr²

=2(22/7)(12²)

=2×22×12×12/7

=905.14cm²

surface area of external bowl=2πR²

=2(22/7)(12.5²)

=2×22×12.5×12.5/7

=982.14cm²

surface area of ring=π(R²-r²)

=22/7(12.5²-12²)

=22/7(12.25)

=22×12.25/7

=38.5cm²

cost of internal bowl=905.14×0.05=Rs.45.25

cost of external bowl=982.14×0.05=Rs.49.1

cost of ring=38.5×0.05=Rs.1.925

cost of bowl=45.25+49.1+1.9=96.25 rupees approx

Step-by-step explanation:

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