# Cups A and B are cone shaped and have heights of 39 cm and 25 cm and openings with radii of 17 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?

Sep 7, 2016

Cup A will not overflow. Cup A will be filled to a height of 28.12cm

#### Explanation:

cone's volume formula = $\pi {r}^{2} \times \frac{h}{3}$

Cup A, height 39, radius 17, $v o l = 11803.0 c {m}^{3}$
Cup B, height 25, radius 13, $v o l = 4424.4 c {m}^{3}$

As Cup A's volume is larger than Cup B's, Cup A will not overflow.

Now if $4424.4 c {m}^{3}$ of water is poured into Cup A, then Cup A will be filled to a height of h1, at this height, the radius is, say, r1.

$\frac{r}{h} = \frac{r 1}{h 1} \implies r 1 = \frac{r h 1}{h}$
$V = \pi {\left(r 1\right)}^{2} \frac{h 1}{3}$
$= \pi {\left(r \frac{h 1}{h}\right)}^{2} \frac{h 1}{3}$
$= \pi \left({r}^{2} {\left(h 1\right)}^{2} / {h}^{2}\right) \frac{h 1}{3}$
$= \left(\frac{\pi}{3}\right) \left(\frac{{r}^{2} {\left(h 1\right)}^{3}}{{h}^{2}}\right)$
${\left(h 1\right)}^{3} = \frac{3 V {h}^{2}}{\pi {r}^{2}}$
${\left(h 1\right)}^{3} = \frac{3 \times 4424.4 \times {39}^{2}}{\pi \times {17}^{2}} = 22236.0$

$h 1 = 28.12 c m$

$r 1 = 17 \times \frac{28.12}{39} = 12.25 c m$