# Solution to the Triangle Inside a Square Puzzle +-- {.image} [[TriangleInsideaSquare.png:pic]] > The right-angled triangle covers $\frac{1}{4}$ of the square. What fraction does the isosceles triangle cover? =-- ## Solution by [[Angle in a Semi-Circle]] and [[Similar Triangles]] +-- {.image} [[TriangleInsideaSquareLabelled.png:pic]] =-- In the above diagram, a second square has been placed on top of the original one. As $O A$, $O C$, and $O D$ are all the same length, the circle centred at $O$ that passes through $A$ also passes through $C$ and $D$. Since triangle $O A E$ is [[isosceles]], it also passes through $E$. Triangles $O E A$ and $O E D$ have the same area since their "bases" $O A$ and $O D$ have the same length and they have the same apex, $E$, above the line $A D$. So triangle $O A E$ has half the area of triangle $D A E$. Triangle $D A E$ is [[right-angled triangle|right-angled]] since the [[angle in a semi-circle]] is $90^\circ$ and is [[similar]] to triangle $D A B$. So also is triangle $A E B$. Since the length of $A D$ is twice that of $A B$, the length of $A E$ is twice that of $E B$, and that of $D E$ is twice that of $A E$. So $D E$ has length four times that of $E B$, meaning that triangle $D E A$ is four fifths of triangle $D A B$. Hence triangle $O A E$ is two fifths of triangle $D A B$. Since triangle $D A B$ has the same area as the original square, the yellow triangle has area $\frac{2}{5}$ths of that square.