# Cups A and B are cone shaped and have heights of 28 cm and 23 cm and openings with radii of 11 cm and 9 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?

Dec 26, 2017

Cup A will be filled upto 15.3967 cm

#### Explanation:

Volume of cone B ${V}_{b} = \left(\frac{1}{3}\right) \pi {r}_{2}^{2} {h}_{2}$

Given ${r}_{2} = 9 c m , {h}_{2} = 23 c m$

${V}_{b} = \left(\frac{1}{3}\right) \cdot \pi \cdot {9}^{2} \cdot 23 = 1950.929 \pi$ $c {m}^{3}$

Similarly Volume of cone A ${V}_{a} = \left(\frac{1}{3}\right) \pi {r}_{1}^{2} {h}_{1}$

Given $r - 1 = 11 c m , {h}_{1} = 28 c m$

${V}_{a} = \left(\frac{1}{3}\right) \cdot \pi \cdot {11}^{2} \cdot 37 = 3547.9053 \pi$ $c {m}^{3}$

As volume of cone A is greater than the volume of cone B, cup A will not overflow.

Volume of partly filled cone A ${V}_{p} = \left(\frac{1}{3}\right) \pi {r}_{1}^{2} {h}_{3}$

${V}_{p} = = {V}_{b}$

$\left(\frac{1}{3}\right) \pi \cdot {9}^{2} \cdot 23 = \left(\frac{1}{3}\right) \cdot \pi \cdot {11}^{2} \cdot {h}_{3}$

${h}_{3} = \frac{\cancel{\left(\frac{1}{3}\right) \cdot \pi} \cdot {9}^{2} \cdot 23}{\cancel{\left(\frac{1}{3}\right) \cdot \pi} \cdot {11}^{2}}$

${h}_{3} = \frac{1863}{121} = 15.3967$ $c m$