# Arcs Inside a Triangle +-- {.image} [[ArcsInsideaTriangle.png:pic]] > Two arcs are drawn from corners of the green equilateral triangle, which has area $3$. What's the area of the blue equilateral triangle? =-- ## Solution by [[Lengths in an Equilateral Triangle]] and [[Pythagoras' Theorem]] +-- {.image} [[ArcsInsideaTriangleLabelled.png:pic]] =-- Let $x$ and $y$ be the side lengths of the green and blue triangles, respectively. Using the relationships between the [[lengths in an equilateral triangle]], the height of the green triangle is $\frac{\sqrt{3}}{2} x$. As the right-hand edge of the green triangle is tangent to the arc centred at $A$, angle $A \hat{D} C$ is the [[angle between a radius and tangent]] so is $90^\circ$. This establishes $D$ as the [[midpoint]] of that side and so the length of $A D$ is $\frac{\sqrt{3}}{2} x$. Then $A E$ has the same length. The point marked $B$ is the [[midpoint]] of $A C$, making $A B E$ a [[right-angled triangle]]. The length of $E B$ is the height of the green equilateral triangle, so is $\frac{\sqrt{3}}{2} y$. The length of $A B$ is $\frac{1}{2} x$. Applying [[Pythagoras' theorem]] then shows that: $$ \begin{aligned} \left( \frac{\sqrt{3}}{2} x\right)^2 &= \left(\frac{1}{2} x\right)^2 + \left( \frac{\sqrt{3}}{2} y\right)^2 \\ \frac{3}{4} x^2 &= \frac{1}{4} x^2 + \frac{3}{4} y^2 \\ y^2 &= \frac{2}{3} x^2 \end{aligned} $$ The [[area scale factor]] from the green to the blue triangles is therefore $\frac{2}{3}$. Since the green triangle has area $3$, the blue triangle has area $2$.