For reference: Calculate the area of the triangular region $KHQ$, if: $r=3$ and $R=4$.(Answer:$\frac{147}{5}$)
Would there be a simpler algebraic solution?
My progress
$$\triangle ADB \sim \triangle CDB \\
\frac{R}{r}=\frac{DC}{DB}\\
\tanβ=\frac{DC}{DB}=\frac{4}{3} \implies
\frac{\sinβ}{\cosβ}=\frac{4}{3}\\
1+(\frac{\sinβ}{\cosβ})^2=1+(\frac{4}{3})^2 \implies\\
\cosβ=\sinα=\frac{3}{5}\\
\sinβ=\cosα=\frac{4}{5}\\
\cosα=\cos^2\frac{α}{2}−\sin^2\frac{α}{2} \implies\\
\cos\frac{α}{2}=\frac{3}{\sqrt{10}}\\
\therefore \sin\frac{α}{2}=\frac{1}{\sqrt{10}}\\
\cosβ=\cos^2\frac{β}{2}−\sin^2\frac{β}{2} \implies\\
\cos\frac{β}{2}=\frac{2}{\sqrt5}\\
\sin\frac{β}{2}=\frac{1}{\sqrt5}\\
KH\cdot\sin(90^{\circ}−\frac{α}{2})=QH\cdot\sin(90^{\circ}−\frac{β}{2})\\
KH\cdot\cos(\frac{α}{2})=QH\cdot\cos(\frac{β}{2})\\
KH\cdot\frac{3}{\sqrt{10}}=QH\cdot\frac{2}{\sqrt5}\\
KH=\frac{2\sqrt2}{3}\cdot QH\\
KH\cdot\cos(90^{\circ}−\frac{α}{2})+QH\cdot\cos(90^{\circ}−\frac{β}{2})=QK=R+r=7\\
KH\cdot\sin(\frac{α}{2})+QH\cdot\sin(\frac{β}{2})=7\\
KH\cdot\frac{1}{\sqrt{10}}+QH\cdot\frac{1}{\sqrt5}=7\\
\frac{2\sqrt2}{3}\cdot QH\cdot\frac{1}{\sqrt{10}}+QH\cdot\frac{1}{\sqrt5}=7\\
\therefore QH=\frac{21}{\sqrt5}\\
\mathrm{Area}=QH\cdot\frac{\sin(90^{\circ}−\frac{β}{2})\cdot7}{2}\\
\mathrm{Area}=KH\cdot\frac{\cos(\frac{β}{2})\cdot7}{2}\\
\mathrm{Area}=\frac{\frac{21}{\sqrt5}\cdot\frac{2}{\sqrt5}\cdot7}{2}\\
\therefore \boxed{\color{red}S_{KHQ}=\frac{147}{5}}$$
Best Answer
Let $B'$ be the foot of the perpendicular from $B$ onto $AC$
When right triangle $\triangle ABC$ is divided into two triangles by $BB'$, they are similar. In this case, we also know that they have inscribed circles with radius $r=3$ and with radius $R=4$, so they are similar with a $3:4$ ratio. Therefore $AB : BC = 3 : 4$.
Therefore $AB : BC : AC = AB' : B'B : AB = BB' : B'C : BC = 3 : 4 : 5$.
In a triangle with sides $3,4,5$ the inradius is $1$. So here $\triangle AB'B$ has sides $9,12,15$ and $\triangle BB'C$ has sides $12,16,20$. We can figure out many side lengths from here.
Let $T', H', S'$ be the feet of the perpendiculars from $T, H, S$ onto $AB$. We know $\triangle AT'T$ is a $3:4:5$ right triangle, and $AT = AK = AB' - KB' = 9-3 =6$, so $TT' = 4.8$ and $AT' = 3.6$; also, $T'K = AK - AT' = 6-3.6 = 2.4$. Since $TT' : T'K = 2 : 1$ and $\triangle HH'K \sim \triangle TT'K$, we know $HH' : H'K = 2:1$ as well.
Let's do the same thing from the other side, looking at $\triangle CSS'$, $\triangle QSS'$, and $\triangle QHH'$. We get $SS' = 7.2$ and $QS'=2.4$, so $SS' : QS' = 3:1$ and therefore $HH' : QH' = 3:1$ as well.
It follows that $H'K : HH' : QH' = 3 : 6 : 2$ so $H'K : QH' = 3 : 2$. But we know $H'K + QH' = 7$, so $H'K = 4.2$ and $QH' = 2.8$, and therefore $HH' = 8.4$. The area of the triangle we want is $\frac12 (7)(8.4) = 29.4$.