?
The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl ?
Answer & Solution
Correct Answer:
Option
B
Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :
$$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$
External radius, R = 5 cm
Total surface area :
$$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$
Join the Discussion
Login to post a comment or share your explanation.
Login