From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm , a concial cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm.

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Solution

$\begin{matrix} G i v e n, H e i g h t o f c y l i n d e r = H e i g h t o f c o n i v a l c a v i t y, h = 2.4 c m D i a m e t e r, d = 1.4 c m ∴ r = \frac{1.4}{2} = 0.7 c m S l a n t h e i g h t, l = \sqrt{h^{2} + r^{2}} = \sqrt{{(2.4)}^{2} + {(0.7)}^{2}} = \sqrt{5.76 + 0.49} = \sqrt{6.25} = 2.5 c m N o w, T S A o f t h e r e m a i n g i n g s o l i d = C S A o f c y l i n d r i c a l p a r t + C S A o f c o n i c a l p a r t + A r e a o f c y l i n d r i c a l b a s e T S A = 2 π r h + π r l + π r^{2} = π r [2 (h) + l + r] = \frac{22}{7} \times 0.7 [2 \times 2.4 + 2.5 + 0.7] = 17.6 c m^{2} H e n c e, T S A o f r e m a i n g i n g s o l i d i s = 17.6 c m^{2} . \end{matrix}$

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