<s>[INST] Area of triangle having an inscribed circle 
The radius of an inscribed circle in a triangle is $2 cm$. A point of tangency divides a side into $3 cm$ and $4 cm$. Find the area of the triangle.

We know that a side is $7 cm$, the others are $4+x$, $3+x$. I tried finding $2$ equations of the area, $S=pr \Rightarrow S=2x+14$ and by herone and equalizing, but that didn't give me a good answer.
 [/INST] 
$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$,   $OP=OQ=OR=2$.
$AO=\sqrt{3^2+2^2}=\sqrt{13}$;
$CO=\sqrt{4^2+2^2}=\sqrt{20}$.
$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$
Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get
$$
S=AB\cdot AC\cdot \sin \angle OAR \cdot \cos \angle OAR = CA\cdot CB \cdot \sin \angle OCR \cdot \cos \angle OCR
$$
$$
S=(3+x)\cdot 7 \cdot \dfrac{2}{\sqrt{13}}\cdot \dfrac{3}{\sqrt{13}} = (4+x)\cdot 7 \cdot \dfrac{2}{\sqrt{20}}\cdot \dfrac{4}{\sqrt{20}};  
$$
$$
S=\dfrac{42}{13}(3+x) = \dfrac{56}{20}(4+x);
$$
$$
840(3+x)=728(4+x);
$$
$$
112 x = 392;
$$
$$
x=\dfrac{7}{2};
$$
$$
S=21.
$$
</s>