# Solution to the [[Two Circles in a Triangle]] Puzzle +-- {.image} [[TwoCirclesinaTriangle.png:pic]] > What's the area of this triangle? =-- ## Solution by [[Angle Between a Radius and Tangent]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoCirclesinaTriangleAnnotated.png:pic]] =-- Let the points be labelled as above. Consider the right-hand triangle, $G C E$, with its [[in-circle]] of radius $1$. The length of $I D$ is $1$, and $I J$ is [[perpendicular]] to $G C$ since the [[angle between a radius and tangent]] is $90^\circ$, so $G J$ has length $4 - 1 = 3$. By [[symmetry]], $G F$ also has length $3$. Let $F E$ have length $x$, so then $D E$ also has length $x$, and the sides of the [[right-angled triangle]] $G C E$ are therefore $4$, $1 + x$, and $3 + x$. Applying [[Pythagoras' theorem]] shows that: $$ \begin{aligned} (3 + x)^2 &= (1 + x)^2 + 4^2 \\ 9 + 6 x + x^2 &= 1 + 2 x + x^2 + 16 \\ 4 x &= 8 \\ x &= 2 \end{aligned} $$ Now consider the left-hand triangle, $G C A$. Let $A B$ have length $y$, then by a similar argument the lengths of $G C A$ are $y + 1.5$, $4$, and $y + 2.5$. Applying Pythagoras' theorem again shows that: $$ \begin{aligned} (y + 2.5)^2 &= (y + 1.5)^2 + 4^2 \\ y^2 + 5 y + 6.25 &= y^2 + 3 y + 2.25 + 16 \\ 2 y &= 12 \\ y &= 6 \end{aligned} $$ Thus the length of $A E$ is $y + 1.5 + 1 + x = 10.5$ and so the area of the triangle is $\frac{1}{2} \times 4 \times 10.5 = 21$.