# Cups A and B are cone shaped and have heights of 33 cm and 37 cm and openings with radii of 10 cm and 13 cm, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Aug 23, 2016

$29.9 c m$

#### Explanation:

We need to find and compare the Volumes of A and B.
Check first whether they are similar in shape - this would make some of the calculations easier.

Are the sides in the same ratio?
$\frac{13}{10} = 1.3 \mathmr{and} \frac{37}{33} = 1.12 \text{ "rArr " A and B are not similar}$

$V o {l}_{\text{cone}} = \frac{\pi {r}^{2} h}{3}$

$V o {l}_{A} = \frac{{13}^{2} \times 37 \times \pi}{3} \text{ "and " } V o {l}_{B} = \frac{{10}^{2} \times 33 \times \pi}{3}$

By inspection we can see that $V o l A > V o l B$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} {13}^{2} \times 37 > {10}^{2} \times 33$

Therefore A will not overflow but we need the height.

The cone formed by the water in A and the whole cone of A are similar in shape.

The ratio of the cubes of the heights is equal to the ratio of the volumes.

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} {h}^{3} / {H}^{3} = \frac{v}{V}$

${h}^{3} / {37}^{3} = \frac{{10}^{2} \times 33}{{13}^{2} \times 37}$

${h}^{3} = \frac{{37}^{3} \times {10}^{2} \times 33}{{13}^{2} \times 37} = 26 , 731.95$

$h = \sqrt[3]{26 , 731.95}$

$29.9 c m$