Answer:
\(\frac{r(1-\sqrt{2})}{-3}\)
Step-by-step explanation:
Volume of cone = \(\frac{1}{3} \pi r^{2} h\)
Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius
= \(\frac{1}{3} \pi r^{2} \times r\)
= \(\frac{1}{3} \pi r^{3}\)
Surface area of cone including 1 base = \(\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}\)
Since r = h
So, area = \(\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}\)
= \(\pi r^{2} +\pi\times r \times \sqrt{2r^2}\)
= \(\pi r^{2} +\pi\times r^2 \times \sqrt{2}\)
Ratio of volume of cone to its surface area including base :
\(\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}\)
\(\frac{\frac{1}{3}r}{1+\sqrt{2}}\)
\(\frac{r}{3(1+\sqrt{2})}\)
Rationalizing
\(\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\)
\(\frac{r(1-\sqrt{2})}{-3}\)
Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is \(\frac{r(1-\sqrt{2})}{-3}\)
