Find the area of the triangular region KHQ,

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?

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My progress $$\triangle ADB \sim \triangle CDB \\ \frac{R}{r}=\frac{DC}{DB}\\ \tanβ=\frac{DC}{DB}=\frac{4}{3} \implies \frac{\sinβ}{\cosβ}=\frac{4}{3}\\ 1+(\frac{\sinβ}{\cosβ})^2=1+(\frac{4}{3})^2 \implies\\ \cosβ=\sinα=\frac{3}{5}\\ \sinβ=\cosα=\frac{4}{5}\\ \cosα=\cos^2\frac{α}{2}−\sin^2\frac{α}{2} \implies\\ \cos\frac{α}{2}=\frac{3}{\sqrt{10}}\\ \therefore \sin\frac{α}{2}=\frac{1}{\sqrt{10}}\\ \cosβ=\cos^2\frac{β}{2}−\sin^2\frac{β}{2} \implies\\ \cos\frac{β}{2}=\frac{2}{\sqrt5}\\ \sin\frac{β}{2}=\frac{1}{\sqrt5}\\ KH\cdot\sin(90^{\circ}−\frac{α}{2})=QH\cdot\sin(90^{\circ}−\frac{β}{2})\\ KH\cdot\cos(\frac{α}{2})=QH\cdot\cos(\frac{β}{2})\\ KH\cdot\frac{3}{\sqrt{10}}=QH\cdot\frac{2}{\sqrt5}\\ KH=\frac{2\sqrt2}{3}\cdot QH\\ KH\cdot\cos(90^{\circ}−\frac{α}{2})+QH\cdot\cos(90^{\circ}−\frac{β}{2})=QK=R+r=7\\ KH\cdot\sin(\frac{α}{2})+QH\cdot\sin(\frac{β}{2})=7\\ KH\cdot\frac{1}{\sqrt{10}}+QH\cdot\frac{1}{\sqrt5}=7\\ \frac{2\sqrt2}{3}\cdot QH\cdot\frac{1}{\sqrt{10}}+QH\cdot\frac{1}{\sqrt5}=7\\ \therefore QH=\frac{21}{\sqrt5}\\ \mathrm{Area}=QH\cdot\frac{\sin(90^{\circ}−\frac{β}{2})\cdot7}{2}\\ \mathrm{Area}=KH\cdot\frac{\cos(\frac{β}{2})\cdot7}{2}\\ \mathrm{Area}=\frac{\frac{21}{\sqrt5}\cdot\frac{2}{\sqrt5}\cdot7}{2}\\ \therefore \boxed{\color{red}S_{KHQ}=\frac{147}{5}}$$ enter image description here

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 $\frac12 (7)(8.4) = 29.4$.

HINT.-By property of incenter we have $$(x+3)^2+h^2=(x+h-3)^2\text{ in triangle ABD }\\(y+4)^2+h^2=(y+h-4)^2\text{ in triangle CBD }\\(x+h-3)^3+(y+h-4)^2=(x+y+7)^2\text{ in triangle ABC }$$ where $AK=x,BD=h$ and $QC=y$. This system determines $x,y$ and $h$ (we need just $x$ and $y$).

On the other hand we have (taking as origine of coordinates the point $A$) pente of line $TK=-\tan\left(\dfrac{\alpha}{2}\right)=-\dfrac 3x$ and pente of line $SQ=\tan\left(\dfrac{\beta}{2}\right)=\dfrac 4y$ so one has the equations of lines $TK$ and $SQ$: $$Y=-\frac{3(X-x)}{x}\\Y=\frac{4(X-7x)}{y}$$ from which the absolute value of $Y$ is the height of the triangle $KQH$ of base $KQ=7$. We are done.

Property: Consider a $3:4:5$ right triangle $ABC$ with its angle bisectors as in the figure below.

It is well-known (or can be easily proven via angle bisector theorem) that $\triangle ABD$'s perpendicular sides are in the ratio $1:2$ and in $\triangle CBE$ that is $1:3$.

In this problem, given inradii of two (similar) right triangles, the sides $AB$ and $BC$ are in the ratio $r:R$, implying $\triangle ABC$ is a $3:4:5$ triangle.

Then a quick angle chasing shows that $\triangle KGH$ and $\triangle QGH$ correspond to aforementioned special right triangles.

Therefore $GH=2KG=3GQ\implies GH=\dfrac65KQ.$

And $KQ=7$.

Hence, $$[\triangle KHQ]=\frac12\cdot KQ\cdot GH=\frac12\cdot7\cdot\frac65\cdot7=\frac{147}5$$

After posting the answer, I realized it was very similar to previous answer of Misha Lavrov. So here is another solution from me -

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I will start from the points once we have established the side lengths of given triangles using similar triangles and inradius formula of right triangle (please see details in my earlier solution, below this solution).

$U$ is the point of tangency of bigger circle with $BG$. A bit of angle chasing shows that $\triangle USQ \sim \triangle QHK$

We will first find area of $\triangle USQ$

$S_{\triangle BGC} = \frac 12 \cdot 16 \cdot 12 = 96$

$S_{\triangle SQC} = \frac 12 \cdot QC \cdot SJ = \frac{216}{5}$

$S_{\triangle SBU} = \frac 12 \cdot BU \cdot SM = \frac{128}{5}$

$S_{\triangle GQU} = 8$

$S_{\triangle USQ} = 96 - \left(\frac{216}{5} + \frac{128}{5} + 8\right) = \frac{96}{5}$

Now ratio of sides of $\triangle USQ$ and $\triangle KHQ$ is $KQ:UQ = 7:4\sqrt2$.

$ \therefore S_{\triangle KHQ} = \left(\frac{7}{4\sqrt2}\right)^2 \cdot \frac{96}{5} = \frac{147}{5}$

Earlier solution (please refer to this to see how to find side lengths):

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