# Cups A and B are cone shaped and have heights of 25 cm and 26 cm and openings with radii of 9 cm and 7 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?

Jun 14, 2018

See below for explanation

#### Explanation:

Cone Volume is given by $V = \frac{1}{3} \pi {r}^{2} h$

We have cone $A$ which volume is V_A=1/3pi9^2·25=675pi

For cone $B$: V_B=1/3pi7^2·26=1274/3pi

Comparing both volume it is obvius that $\frac{1274}{3} < 675$ then the cup B do not overfilled cup A

In this case, what will be the height of cup A filled?. Lets see.

With volume B into cone A, we have a situation like this

By Thales theorem we know that $\frac{9}{r} = \frac{25}{h}$ then $h = \frac{25 r}{9}$ (1)

Revolving DEC around CA axis, we have the cone produced by volume B in cup A, so his volume is known. Then

1274/3pi=1/3pir^2·25r/9 we can remove $\pi$ and $\frac{1}{3}$

$1274 = 25 {r}^{3} / 9$ from here we have r^3=1274/25·9=458.64

Then $r = \sqrt[3]{458.64} = 7.71182$ cm

In (1), we have h=25·7.71182/9=21.42 cm