# Four Stacked Squares in a Circle in a Square +-- {.image} [[FourStackedSquaresinaCircleinaSquare.png:pic]] > What fraction of the big square is shaded? Shaded squares are identical. =-- ## Solution by [[Angle in a Semi-Circle]] and [[Pythagoras' Theorem]] +-- {.image} [[FourStackedSquaresinaCircleinaSquareLabelled.png:pic]] =-- The line segments $A D$ and $B C$ are [[parallel]], so their [[perpendicular bisectors]] are also parallel. As both are [[chords]], the centre lies on both of them, and so these perpendicular bisectors are actually the same line. As the line segments are of equal length, the line segment $D C$ is then parallel to this perpendicular bisector. Triangle $A D C$ is then a [[right-angled triangle]] and so $A C$ is a diameter since the [[angle in a semi-circle]] is a right-angle. Let $x$ be the length of a side of a small square and let $y$ be the side length of the large square. Then $D C$ has length $3 x$ and the diameter of the circle is $y$. Applying [[Pythagoras' theorem]] shows that: $$ y^2 = x^2 + (3 x)^2 = 10 x^2 $$ Therefore one small square is one tenth of the large square, so $\frac{2}{5}$ths of the large square is shaded. ## Solution by [[Intersecting Chords Theorem]] With $x$ and $y$ as above, $A G$ and $G D$ have length $\frac{x}{2}$, $E G$ has length $\frac{y - 3 x}{2}$ and $G F$ has length $\frac{y + 3 x}{2}$. Applying the [[intersecting chords theorem]] gives $$ \begin{aligned} \frac{x^2}{4} &= \frac{y - 3 x}{2} \frac{y + 3 x}{2} \\ &= \frac{y^2 - 9 x^2}{2} \\ 10 x^2 &= y^2 \end{aligned} $$ Which gives the fraction shaded as $\frac{2}{5}$ as before. and the same length, and are both [[chords]] of the circle. The [[perpendicular bisectors]] of both pass through the centre of the circle, and since the chords are parallel the perpendicular bisectors are also parallel.