question_answer

In Fig. 4, from the top of a solid cone of height 12 cm base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. \[\left[ \text{Use}\,\pi =\frac{22}{7}\text{and}\,\sqrt{5}=2.236 \right]\]

Answer:

Height of the given cone \[=12\text{ }cm\]

and the radius of the base \[=6\text{ }cm\]

Let the radius of the base of the smaller cone be x cm and height is 4 cm.

Now,                 \[\Delta \,ARQ\sim \Delta \,ACD\]

\[\therefore \frac{DC}{QR}=\frac{AD}{AQ}\]

\[\frac{6}{x}=\frac{12}{4}\Rightarrow x=2\,cm\]

\[\therefore l=RC=\sqrt{{{h}^{2}}+{{(R-r)}^{2}}}\]

\[=\sqrt{{{8}^{2}}+{{(6-2)}^{2}}}\]

\[=\sqrt{64+16}\]

\[=4\sqrt{5}\]

Total surface area of frustum PRCB

\[=[\pi l(R+r)+\pi {{r}^{2}}+\pi {{R}^{2}}]\]

\[=\frac{22}{7}\times 4\sqrt{5}(6+2)+\frac{22}{7}\times {{(2)}^{2}}+\frac{22}{7}+{{(6)}^{2}}\]

\[=\frac{22}{7}[32\times 2.236+4+36]\]

\[=\frac{22}{7}(111.552)\]

\[=350.592\,\,c{{m}^{2}}\]