# Solution to the [[A Circle in a Rectangle]] Puzzle +-- {.image} [[ACircleinaRectangle.jpeg:pic]] > What’s the area of this rectangle? =-- ## Solution by [[Pythagoras' Theorem]], [[Angle Between a Radius and Tangent]], [[Vertically Opposite Angles]], and [[Similar Triangles]] +-- {.image} [[ACircleinaRectangleLabelled.jpeg:pic]] =-- 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 triangle|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: $$ (r+1)^2 = r^2 + 2^2 $$ 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 $6 \times \frac{9}{2} = 27$.