Question

Joanne has a cylindrical, above ground pool. the depth (height) of the pool is 1/2 of its radius, and the volume is 1570 cubic feet. What is the area of its bottom floor? Include equations or inequalities related.

Answer 1

Answer:

$\sqrt[3]{3140^2\pi}\approx 146.41\ ft^2$

Step-by-step explanation:

The volume of the cylinder is

$V_{cylinder}=\pi r^2\cdot H,$

where r is the base radius and H is the height.

Since $H=\dfrac{1}{2}r$ and V=1570 cubic feet, then

$1570=\pi r^2\cdot \dfrac{r}{2},\\ \\1570=\dfrac{\pi r^3}{2},\\ \\r^3=\dfrac{3140}{\pi},\\ \\r=\sqrt[3]{\dfrac{3140}{\pi}}\ ft.$

The area of its bottom floor is

$A_{floor}=\pi r^2=\pi\cdot \left(\sqrt[3]{\dfrac{3140}{\pi}}\right)^2= \sqrt[3]{3140^2\pi}\approx 146.41\ ft^2.$