The internal and external diameters of a hollow hemisph

SOLUTION

Given,
internal diameter=21cm  $\frac{22}{7}$
internal radius= $\frac{21}{2}$ =10.5cm
external diameter=25.2cm
external radius=$\frac{25.2}{2}$ =12.6cm
surface area of internal bowl=2πr²
=2 × $\frac{22}{7}$×(10.5²)
= $\frac{2\times 22\times 10.5\times 10.5}{7}$
=693cm²
surface area of external bowl=2πR²
=2 ×$\frac{22}{7}$ × (12.6²)
= $\frac{2\times 22\times 12.6\times 12.6}{7}$
=997.9cm²
Surface area of ring=π(R²-r²)
=$\frac{22}{7}$(12.6²-10.5²)
=$\frac{22}{7}$(48.51)
= $\frac{22\times 48.51}{7}$
=152.46cm²
Cost of internal bowl=693×0.1= Rs. 69.3
Cost of external bowl=997.9×0.1= Rs. 99.79
Cost of ring=152.46×0.1= Rs.15.246
Cost of bowl=69.3+99.7+15.2= Rs.184.2