# Semi-Circle and Tangents in a Square +-- {.image} [[SemiCircleandTangentsinaSquare.png:pic]] > The yellow lines are tangents to the semicircle. What fraction of the square is shaded? =-- ## Solution by [[Angle Between a Radius and Tangent]] and [[Similarity]] +-- {.image} [[SemiCircleandTangentsinaSquareLabelled.png:pic]] =-- Since the [[angle between a radius and tangent]] is $90^\circ$ and the [[angles in a quadrilateral]] add up to $360^\circ$, angles $F \hat{C} B$ and $B \hat{O} F$ add up to $180^\circ$, therefore angle $F \hat{C} B$ is the same as angle $F \hat{O} A$. This shows that the kites $B C F O$ and $A O F E$ are [[similar]]. Since $A E$ is the same length as the diameter of the semi-circle, the scale factor is $2$. Therefore the length of $C B$ of half of that of $O B$, and so a quarter of the length of the side of the square. Setting $x$ to be the side length of the square, the area of the shaded region is then $2 \times \frac{1}{2} \times \frac{1}{2} x \times \frac{3}{4} x = \frac{3}{8} x^2$ and so $\frac{3}{8}$ths of the square is shaded. ## Solution by [[Pythagoras' Theorem]] The lengths of $C B$ and $C F$ are the same as $B C F O$ is a kite, similarly $F E$ and $A E$ have the same length. With $x$ the length of the side of a square and $y$ the length of $C B$, applying [[Pythagoras' theorem]] to triangle $E C D$ gives: $$ (x + y)^2 = x^2 + (x - y)^2 $$ This simplifies to $4 x y = x^2$ and so $x = 4 y$. This shows that $C B$ is a quarter of the length of the square. The calculation of the area continues from there as above.