geometry
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.
Let $O$ be the centre of the incircle. Join $O$ to the three vertices.
It is not absolutely clear what ratio $12:5$ means. Let $M$ be the middle of the base. Maybe (i) $BO:OM=12:5$, or maybe (ii) $BO:OM=5:12$. (It will turn out that (ii) can't happen.)
We examine the consequences of (i). Let the base be $2y$ (I don't like fractions). Let $BO=12t$ and let $OM=5t$.
We calculate the area of triangle $BAC$ in two different ways. The area is half of base times height, so it is $(1/2)(2y)(17t)$, that is, $17yt$.
The area is also the sum of the areas of $\triangle BOA$, $\triangle BOC$, and $\triangle OAC$. This sum is $(1/2)(30)(5t)+(1/2)(30)(5t)+(1/2)(2y)(5t)$, which is $150t +5yt$. We conclude that $$17yt=150t+5yt.$$ Thus $12y=150$, and therefore $2y=25$.
Next we examine possibility (ii). The analysis is similar. We find pretty quickly that $y$ is "too big" (the Triangle Inequality is violated).
Make a construction like so 
Here, $OC = r$, $BC = \frac{l}{2}$, $AB = l$. Since $ABC \sim BOC$, taking ratios, we get $AC = \frac{l^2}{4r}$.
By the Pythagorean theorem, $AB^2 = AC^2 + BC^2$,
Therefore, $$l = \sqrt{\frac{l^2}{4} + \frac{l^4}{16r^2}}$$
Simplifying, we get $l = r\sqrt{12}$
The area would be $\frac{\sqrt{3}}{4}l^2$, which would be
$$3\sqrt{3}r^2$$
Best Answer
$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$, $OP=OQ=OR=2$.
$AO=\sqrt{3^2+2^2}=\sqrt{13}$;
$CO=\sqrt{4^2+2^2}=\sqrt{20}$.
$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$
Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get
$$ S=AB\cdot AC\cdot \sin \angle OAR \cdot \cos \angle OAR = CA\cdot CB \cdot \sin \angle OCR \cdot \cos \angle OCR $$
$$ S=(3+x)\cdot 7 \cdot \dfrac{2}{\sqrt{13}}\cdot \dfrac{3}{\sqrt{13}} = (4+x)\cdot 7 \cdot \dfrac{2}{\sqrt{20}}\cdot \dfrac{4}{\sqrt{20}}; $$
$$ S=\dfrac{42}{13}(3+x) = \dfrac{56}{20}(4+x); $$
$$ 840(3+x)=728(4+x); $$
$$ 112 x = 392; $$
$$ x=\dfrac{7}{2}; $$
$$ S=21. $$