# Solution to the [[A Quarter Circle and a Semi-Circle in a Triangle]] Puzzle +-- {.image} [[AQuarterCircleandaSemiCircleinaTriangle.jpeg:pic]] > What’s the area of the quarter circle? =-- ## Solution by [[Pythagoras' Theorem]] and [[Angle Between a Radius and Tangent]] +-- {.image} [[AQCircleandaSCircleinaTriangleLabelled.jpeg:pic]] =-- _Note: the semi-circle seems redundant in this puzzle_ With the points labelled as above, let $r$ be the radius of the quarter circle, $a$ the length of line segment $O A$, and $b$ the length of line segment $O B$. Since the red region is a quarter circle, angle $A \hat{O} B$ is $90^\circ$. Therefore the outer triangle is [[right-angled triangle|right-angled]]. Applying [[Pythagoras' theorem]] shows that: $$ 6^2 = a^2 + b^2 $$ As the [[angle between a radius and tangent]] is also $90^\circ$, triangles $O B A$ and $O B C$ are also [[right-angled triangle|right-angled]]. Applying [[Pythagoras' theorem]] to those shows that: $$ \begin{aligned} a^2 &= r^2 + 4 \\ b^2 &= r^2 + 16 \end{aligned} $$ Adding these equations together shows that $a^2 + b^2 = 2r^2 + 20$ and so $2 r^2 + 20 = 36$, from which $r^2 = 8$. Therefore, the area of the quarter circle is: $$ \frac{1}{4} \pi r^2 = 2 \pi $$ ## Solution by [[Angle Between a Radius and Tangent]], [[Angles in a Triangle]], and [[Similar Triangles]] As above, triangles $A B O$ and $C B O$ are [[right-angled triangles|right-angled]]. Angles $A \hat{O} B$ and $B \hat{O} C$ add to $90^\circ$, but then so also do angles $A \hat{O} B$ and $B \hat{A} O$. So angles $B \hat{A} O$ and $B \hat{O} C$ are the same, so triangles $O B C$ and $A B O$ are [[similar]]. Therefore, the ratios $4 : r$ and $r : 2$ are the same, leading to $r^2 = 8$ as before.