Answer:
[tex]\frac{r(1-\sqrt{2})}{-3}[/tex]
Step-by-step explanation:
Volume of cone = [tex]\frac{1}{3} \pi r^{2} h[/tex]
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
= [tex]\frac{1}{3} \pi r^{2} \times r[/tex]
= [tex]\frac{1}{3} \pi r^{3}[/tex]
Surface area of cone including 1 base = [tex]\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}[/tex]
Since r = h
So, area = [tex]\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}[/tex]
= [tex]\pi r^{2} +\pi\times r \times \sqrt{2r^2}[/tex]
= [tex]\pi r^{2} +\pi\times r^2 \times \sqrt{2}[/tex]
Ratio of volume of cone to its surface area including base :
[tex]\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}[/tex]
[tex]\frac{\frac{1}{3}r}{1+\sqrt{2}}[/tex]
[tex]\frac{r}{3(1+\sqrt{2})}[/tex]
Rationalizing
[tex]\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}[/tex]
[tex]\frac{r(1-\sqrt{2})}{-3}[/tex]
Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is [tex]\frac{r(1-\sqrt{2})}{-3}[/tex]
