Question

# From a solid cylinder whose height is $$2.4\ cm$$ and diameter $$1.4\ cm$$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $$cm^{2}$$.

Solution

## Given :radius $$(r) =\dfrac {diameter}{2}= \dfrac {1.4}{2}=0.7 \ cm$$height $$(h)= 2.4 \ cm$$Total surface area$$=$$ surface area of cylinder$$+$$ surface area of cone$$-2$$(surface area of above circle).                              $$=2\pi rh+2\pi r^2+\pi r^2+\pi r\sqrt{h^2+r^2}-2\pi r^2$$                              $$=2\pi rh+\pi r\sqrt{h^2+r^2}+\pi r^2$$                              $$=\pi (2rh+r^2+r\sqrt{h^2+r^2})$$                              $$=\dfrac{22}{7}(2\times 0.7\times 2.4+(0.7)^2+0.7\sqrt{(0.7)^2+(24)^2})$$                              $$=\dfrac{22}{7}(3.36+0.49+1.75)$$                             $$=\dfrac{22}{7}(5.6)$$Total surface area$$=17.6cm^2$$.Hence, this is the answer.Mathematics

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