Find the area of the given triangle

Question

Another mathematical puzzle:

Find the area of $\triangle FGH$, given that $FG=FH$ and the radii of the circles shown are $2$ and $1$

Answer 1

The answer is:

$16\sqrt2$

Because:

if you draw radii where the circles are tangent to the triangles with tangent points marked as $D$ (smaller circle) and $E$ (larger circle; this forms right angles), you get the following similar triangles: $\triangle FDA \sim \triangle FEB \sim \triangle FXG$. $A$ is the center of the small circle and $B$ is the center of the large one. $X$ is the midpoint of $HG$.

From this, we get this proportion:

$\frac{FA}{FB} = \frac{DA}{EB}$ OR $\frac{FA}{FA+3} = \frac{1}{2}$ because of the radii lengths.

So:

$FA=3$

Using Pythagorean Theorem:

$FD = \sqrt8$

Using our similar triangle relations from the beginning, we get:

$XG = 2\sqrt2$

Finally, we use the area of a triangle formula to get:

$16\sqrt2$

Answer 2

Let $h$ denote the height of the triangle from F onto GH. Then, by Pythagoras' theorem, $$FG^2=h^2+(GH/2)^2,$$ and by triangles' similarity $$h:(GH/2) = (FG-GH/2):2$$ and $$(h-4):h = 1:2.$$

Can you continue from this?

From the last we get $h-4=h/2$ which resolves to $h=8.$ Plug it to previous two to get

$$\begin{cases}FG^2=64+(GH/2)^2\\8:(GH/2) = (FG-GH/2):2\end{cases}$$ The last equation is equivalent to $$16:(GH/2) = (FG-GH/2)$$

hence

$$FG = (GH/2)+16:(GH/2)$$ and $$FG^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$

Compare it to the first equation: $$64+(GH/2)^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$ $$64 = 32 + 256:(GH/2)^2$$ $$32 = 256:(GH/2)^2$$ $$(GH/2)^2 = 256:32 = 8$$ $$GH/2 = \sqrt 8 = 2\sqrt 2$$

Finally the area sought is

$$S_{\triangle FGH} = h\cdot GH/2 = 16\sqrt 2.$$

Answer 3

Let me chip in with a streamlined answer.

$8\sqrt 8$

All the smaller circle is telling us is that the height of the triangle must be $8$, so let's jot that down and from here pretend the small circle never existed.

The points where the circle (there is only one circle!) touches the triangle divide the long sides into two segments of lengths $y>x$ and the base into two equal segments of length $x$. Let us now compute the area $A$ of the triangle from base and height $A=8x$, from incircle radius and circumference $A=2(2x+y)$ and from Heron's formula $A^2=x^2y(2x+y)$. Comparing the first two gives $2x=y$, together with the last we get $x=\sqrt 8$ and $A=8\sqrt 8$.