Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
The radius and height of a right solid circular cone are $r$ and $h$ respectively. A conical cavity of radius $\frac{r}{2}$ and height $\frac{h}{2}$ is cut out of this cone. What is the whole surface area of the rest of the portion?
Answer
Answer (as printed):
Explanation
Clearly, required surface area = Total surface area of bigger cone + Curved surface area of smaller cone - Area of base of smaller cone $$\begin{aligned} & =\left[\left(\pi r \sqrt{r^{2}+h^{2}}+\pi r^{2}\right)+\pi\left(\frac{r}{2}\right) \sqrt{\left(\frac{r}{2}\right)^{2}+\left(\frac{h}{2}\right)^{2}}-\pi\left(\frac{r}{2}\right)^{2}\right] \\ = & \pi r \sqrt{r^{2}+h^{2}}+\pi r^{2}+\frac{\pi r}{2} \sqrt{\frac{r^{2}+h^{2}}{4}}-\frac{\pi r^{2}}{4} \\ = & \pi r \sqrt{r^{2}+h^{2}}+\pi r^{2}+\frac{\pi r}{4} \sqrt{r^{2}+h^{2}}-\frac{\pi r^{2}}{4} \\ = & \frac{4 \pi r \sqrt{r^{2}+h^{2}}+4 \pi r^{2}+\pi r \sqrt{r^{2}+h^{2}}-\pi r^{2}}{4} \\ = & \frac{5 \pi r \sqrt{r^{2}+h^{2}}+3 \pi r^{2}}{4}=\frac{\pi r}{4}\left(5 \sqrt{r^{2}+h^{2}}+3 r\right) . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.