Notes
two circles in a triangle solution

Solution to the Two Circles in a Triangle Puzzle

Solution by Angle Between a Radius and Tangent and Pythagoras' Theorem

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:

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:

Thus the length of $A E$ is $y + 1.5 + 1 + x = 10.5$ and so the area of the triangle is $\frac{1}{2} \times 4 \times 10.5 = 21$.