# Cups A and B are cone shaped and have heights of 35 cm and 21 cm and openings with radii of 12 cm and 11 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?

$27.856$ cm from apex of cup A

#### Explanation:

The volume (${V}_{A}$) of cone cup-A with vertical height $35$ cm & radius $12$ cm is

${V}_{A} = \frac{1}{3} \setminus \pi {r}^{2} h = \frac{1}{3} \setminus \pi \left({12}^{2}\right) \left(35\right) = 1680 \setminus \pi \setminus c {m}^{3}$

The volume (${V}_{B}$) of cone cup-B with vertical height $21$ cm & radius $11$ cm is

${V}_{B} = \frac{1}{3} \setminus \pi {r}^{2} h = \frac{1}{3} \setminus \pi \left({11}^{2}\right) \left(21\right) = 847 \setminus \pi \setminus c {m}^{3}$

Since, the volume of cone cup A is more than that of cone cup B hence when content of full cup B is poured into cup A, cup A wouldn't overflow.

Let $h$ be the vertical height from apex up to which cup A is filled when content of full cup B is poured. If $r$ is the radius of circular surface of content of cup A then using properties of similar triangles

$\setminus \frac{r}{h} = \setminus \frac{12}{35}$

$r = \frac{12}{35} h$

Now, the volume filled in cone cup A will be equal to the volume of full cone cup B hence we have

$\frac{1}{3} \setminus \pi {r}^{2} h = 847 \setminus \pi$

${r}^{2} h = 2541$

${\left(\frac{12}{35} h\right)}^{2} h = 2541$

${h}^{3} = \setminus \frac{2541 \setminus \times {35}^{2}}{{12}^{2}}$

$h = \setminus \sqrt[3]{\setminus \frac{2541 \setminus \times {35}^{2}}{{12}^{2}}}$

$h = 27.856 \setminus c m$

Thus, the cone cup A will be filled to a vertical height $27.856$ cm from apex.