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.
 A: 
$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$,   $OP=OQ=OR=2$.
$AO=\sqrt{3^2+2^2}=\sqrt{13}$;
$CO=\sqrt{4^2+2^2}=\sqrt{20}$.
$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$
Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get
$$
S=AB\cdot AC\cdot \sin \angle OAR \cdot \cos \angle OAR = CA\cdot CB \cdot \sin \angle OCR \cdot \cos \angle OCR
$$
$$
S=(3+x)\cdot 7 \cdot \dfrac{2}{\sqrt{13}}\cdot \dfrac{3}{\sqrt{13}} = (4+x)\cdot 7 \cdot \dfrac{2}{\sqrt{20}}\cdot \dfrac{4}{\sqrt{20}};  
$$
$$
S=\dfrac{42}{13}(3+x) = \dfrac{56}{20}(4+x);
$$
$$
840(3+x)=728(4+x);
$$
$$
112 x = 392;
$$
$$
x=\dfrac{7}{2};
$$
$$
S=21.
$$
A: Hint: Let $\triangle ABC$ be such that: $O$ be the incenter, and $P$ be the point on side $BC$ such that $OP\perp BC$, and $PB = 3$, and $PC = 4$, then you can find angles: $\text{<OBP}$, and $\text{<OCP}$, then double them to find angles $\text{<B}$, and $\text{<C}$, then you can find $\text{<A}$, and then half of $\text{<A}$, then you finally can find $x$, and you are done. Use tangent in your calculations get you to the answer quicker!
For example: to find $\text{<B}$, we use: $tan(B) = tan(2t) = \dfrac{2tant}{1 - tan^2t}$, with
$tant = tan(\text{<OBP}) = \dfrac{2}{3}$. Then take $tan^{-1}$ to get the angle.
