# Cups A and B are cone shaped and have heights of 25 cm and 15 cm and openings with radii of 7 cm and 3 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 22, 2018

Cup A will be filled $12 \text{cm}$ high.

#### Explanation:

First, we calculate the volume of cup B.

V_B=1/3pir_b^2*h=1/3pi*(3"cm")^2*15"cm"=45pi"cm"^3

Both the height and radius of cup A are greater than those of cup B, so ${V}_{a} > {V}_{b}$. Hence, cup A will not overflow.

The image below represents cup A with cup B's full content. ${V}_{\text{water}}$ is the volume of the water and ${h}_{\text{water}}$ is the height of the water.

${V}_{\text{water")=1/3pi*r_("water")^2*h_("water}}$

The 'water triangle' is similar to cup A, so $h \propto r$. This gives:

${V}_{\text{water")=1/3pi*(r_A*h_("water")/h_A)^2*h_("water}}$
${V}_{\text{water")=1/3pi*r_A^2/h_A^2*h_("water}}^{3}$
${V}_{\text{water")=(r_A^2pi)/(3h_A^2)*h_("water}}^{3}$

${V}_{\text{water}} = {V}_{B}$
(r_A^2pi)/(3h_A^2)*h_("water")^3=45pi"cm"^3
h_("water")^3=(45pi"cm"^3)/((r_A^2pi)/(3h_A^2))=(135h_A^2)/(r_A^2)"cm"^3
h_("water")=root(3)((135"cm"^3*h_A^2)/(r_A^2))=root(3)((135"cm"^3*(25"cm")^2)/((7"cm")^2))~~12"cm"

Cup A will therefore be filled $12 \text{cm}$ high.