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 \[=[\pi l(R+r)+\pi {{r}^{2}}+\pi {{R}^{2}}]\] \[=\frac{22}{7}\times 4\sqrt{5}(6+2)+\frac{22}{7}\times {{(2)}^{2}}+\frac{22}{7}+{{(6)}^{2}}\] \[=\frac{22}{7}[32\times 2.236+4+36]\] \[=\frac{22}{7}(111.552)\] \[=350.592\,\,c{{m}^{2}}\]
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