# Two Circles Inside a Square +-- {.image} [[TwoCirclesInsideaSquare.png:pic]] > Given the areas of the two circles inside this square, can you find the area of the rectangle? =-- ## Solution by [[Lengths in a Square]] +-- {.image} [[TwoCirclesInsideaSquareLabelled.png:pic]] =-- With the points labelled as above, let $r$ be the radius of the smaller circle, $R$ of the larger, and let $x$ be the side length of the square. The unknown rectangle has sides $x - 2 R$ and $x - 2 r$. The radii satisfy $\pi r^2 = 2 \pi$ and $\pi R^2 = 8 \pi$ so $r^2 = 2$ and $R^2 = 8$. Triangle $B C D$ is an [[isosceles]] [[right-angled triangle]] since the centres of the circles lie on the diagonal of the square. The hypotenuse has length $r + R$, so the length of $B C$ is $\frac{r + R}{\sqrt{2}}$ using the relationships between [[lengths in a square]]. The side length of the square is then $r + \frac{r + R}{\sqrt{2}} + R = \frac{1 + \sqrt{2}}{\sqrt{2}}(r + R)$. The sides of the red rectangle are therefore: $$ \begin{aligned} x - 2 R &= \frac{ (1 + \sqrt{2}) r + (1 - \sqrt{2})R}{\sqrt{2}} \\ x - 2 r &= \frac{ (1 - \sqrt{2}) r + (1 + \sqrt{2})R}{\sqrt{2}} \end{aligned} $$ The area of this rectangle is therefore: $$ \frac{ (1 + \sqrt{2}) r + (1 - \sqrt{2})R}{\sqrt{2}} \times \frac{ (1 - \sqrt{2}) r + (1 + \sqrt{2})R}{\sqrt{2}} = \frac{1}{2} \left( 6 rR -r^2 - R^2 \right) $$ Since $r^2 = 2$ and $R^2 = 8$, $(r R)^2 = 16$ and so $r R = 4$. This then simplifies to $7$.