Question

A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which expresses the ratio of the volume of the candle to its surface area(including the base)? for cone, v=1/3pir^2h and sa=pir^2 pir sqrt r^2 h^2.

Answer 1

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 $\frac{r(1-\sqrt{2})}{-3}$

Answer 2

Rational Expressions QC

1.B

2.C

3.D

4.D

5.B