# Solution to the [[Two Triangles in a Semi-Circle]] Puzzle +-- {.image} [[TwoTrianglesinaSemiCircle.jpg:pic]] > These triangles are congruent and isosceles. What’s the area of the semicircle? =-- ## Solution by [[Perpendicular Bisector of a Chord]], Properties of [[Isosceles Triangles]], and [[Congruent Triangles]] +-- {.image} [[TwoTrianglesinaSemiCircleLabelled.jpg:pic]] =-- In the diagram above, $O$ is the centre of the semi-circle and $D$ is the [[midpoint]] of [[chord]] $A C$. As such, the [[perpendicular bisector]] of $A C$, which passes through $D$, also passes through $O$. Since $A C$ is the base of the [[isosceles triangle]] $A B C$, that perpendicular bisector also passes through $B$. So $B D O$ is a straight line. Triangles $B D A$ and $O D A$ are both [[right-angled triangle|right-angled]] at $D$. Since the blue and purple triangles are [[congruent]], angles $B \hat{A} D$ and $D \hat{A} O$ are equal. Therefore, triangles $A D B$ and $A D O$ have the same interior angles and share side $A D$ so are [[congruent]]. This means that line segments $A B$ and $A O$ have the same length. Then since the triangles are [[congruent]] and [[isosceles]], $A B$ has length $4$, so the radius of the semi-circle is also $4$ and its [[circle area|area]] is then: $$ \frac{1}{2} \pi 4^2 = 8 \pi $$