A cylindrical pail containing water drains into a cylindrical tub $40 \mathrm{~cm}$ across and $50 \mathrm{~cm}$ deep, while resting at an angle of $45^{\circ}$ to the horizontal, as shown. How deep is the water in the tub when its level reaches the pail?
(A) $10 \mathrm{~cm}$
(D) $35 \mathrm{~cm}$
(B) $20 \mathrm{~cm}$
(C) $30 \mathrm{~cm}$
(E) $40 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-145.jpg?height=466&width=358&top_left_y=743&top_left_x=1339)
A cylindrical pail containing water drains into a cylindrical tub $40 \mathrm{~cm}$ across and $50 \mathrm{~cm}$ deep, while resting at an angle of $45^{\circ}$ to the horizontal, as shown. How deep is the water in the tub when its level reaches the pail?
(A) $10 \mathrm{~cm}$
(B) $20 \mathrm{~cm}$
(C) $30 \mathrm{~cm}$
(D) $35 \mathrm{~cm}$
(E) $40 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-270.jpg?height=540&width=404&top_left_y=1790&top_left_x=1424)

## Solution

We label the points $A, B$ and $C$ as shown in the diagram. From symmetry, we note that $\triangle B A C$ is an isosceles rightangled triangle.

From $A$ we draw a line perpendicular to $B C$ meeting the line at point $D$. This construction allows us to conclude that $\triangle A B D$ is also a right-angled isosceles triangle and specifically that $B D=D A$. Since $B D=D A$ and $B D=D C=20$, we find $D A=20$.

This makes the depth of the water $50-20$ or 30 .

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-271.jpg?height=393&width=395&top_left_y=259&top_left_x=1426)

ANSWER: (C)