Notes
three circles in a rectangle ii solution

Solution to the Three Circles in a Rectangle II Puzzle

Solution by Similar Triangles

All three triangles are right-angled triangles, angles $E \hat{D} C$ and $A \hat{D} B$ add up to $90^\circ$, and angles $C \hat{B} E$ and $D \hat{B} A$ also add up to $90^\circ$, so all three triangles are similar. The length scale factors from the largest to the other two are $\frac{4}{5}$ and $\frac{3}{5}$. Let $x$ be the length of the diagonal, $B D$, so then $A D$ has length $\frac{4 x}{5}$ and $A B$ has length $\frac{3 x}{5}$.

To compute $x$, consider the area of the largest triangle. Using the formula for the area of a triangle, it is $\frac{1}{2} \times \frac{4 x}{5} \times \frac{3 x}{5} = \frac{6 x^2}{25}$. Using the centre of the circle, the triangle can be decomposed into three triangles each with height the radius of the in-circle and bases the sides of the original triangle. The area calculated this way is $\frac{1}{2} \times \frac{5}{2} \times \left(\frac{3 x}{5} + \frac{4 x }{5} + x\right) = 3 x$. Equating these gives $x = \frac{25}{2}$.

The sides of the rectangle are then $10$ and $7.5$ so its area is $75$.