# Solution to the Rectangle Overlapping a Circle Puzzle +-- {.image} [[RectangleOverlappingaCircle.png:pic]] > The four marked angles are equal. If the rectangle’s width is $3$, what’s the area of the circle? =-- ## Solution by [[Angle in a Semi-Circle]] and [[Angles in the Same Segment]] +-- {.image} [[RectangleOverlappingaCircleLabelled.png:pic]] =-- With the points labelled as in the diagram, angle $A \hat{D} B$ is a right-angle so using the result that the [[angle in a semi-circle]] is a right-angle, $A B$ is a diameter of the circle. Then by the same result, angle $A \hat{C} B$ is also a right-angle. This means that the marked angle is $30^\circ$. Angle $D \hat{C} A$ is then $30^\circ$ and so as [[angles in the same segment]] are equal, angle $D \hat{B} A$ is also $30^\circ$. So triangle $A B D$ is a $30^\circ-60^\circ-90^\circ$ [[right-angled triangle]], which means it is half an [[equilateral]] triangle. This means that the length of $B D$ is $\frac{\sqrt{3}}{2}$ times the length of $A B$, so $A B$ has length $\frac{2}{\sqrt{3}} \times 3 = 2 \sqrt{3}$ and the radius of the circle is therefore $\sqrt{3}$. Its area is then $3 \pi$.