Notes
a circle in a rectangle solution

Solution to the A Circle in a Rectangle Puzzle

Solution by Pythagoras' Theorem, Angle Between a Radius and Tangent, Vertically Opposite Angles, and Similar Triangles

In the diagram above, $O$ is the centre of the circle. Let $r$ be the radius of the circle.

As the angle between a radius and tangent is $90^\circ$, line segments $O B$ and $O E$ are vertical and horizontal, respectively. So $A E$ has length $r$. Since angles $O \hat{D} C$ and $E \hat{D} A$ are vertically opposite, they are equal. Then since triangles $A E D$ and $O C D$ are right-angled and $A E$ and $O C$ have the same length, they are congruent. Hence $C D$ has length $2$.

The other sides in triangle $O C D$ have lengths $r + 1$ and $r$, so applying Pythagoras' theorem shows that:

This simplifies to $2r + 1 = 4$ which means that $r = \frac{3}{2}$.

The width of the rectangle is then $2 + 1 + r + r = 6$. The sides of the rectangle are in the same ratio to the corresponding sides of triangle $A E D$, so the height of the rectangle is $3 r = \frac{9}{2}$.

The area of the rectangle is therefore $6 \times \frac{9}{2} = 27$.