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.

  • $\begingroup$ Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? $\endgroup$
    – 5xum
    Commented Jun 3, 2014 at 9:08
  • 1
    $\begingroup$ I edited and I could add that i tried to find what type of triangle it is, but with no luck $\endgroup$
    – user130101
    Commented Jun 3, 2014 at 9:12
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    $\begingroup$ You know two sides (one side is radius) of two right angled triangles, find the third sides of both. Then you repeat the process and get the lengths of all sides. $\endgroup$
    – tpb261
    Commented Jun 3, 2014 at 9:15

2 Answers 2


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$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$,   $OP=OQ=OR=2$.


$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$

Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get

$$ S=AB\cdot AC\cdot \sin \angle OAR \cdot \cos \angle OAR = CA\cdot CB \cdot \sin \angle OCR \cdot \cos \angle OCR $$

$$ S=(3+x)\cdot 7 \cdot \dfrac{2}{\sqrt{13}}\cdot \dfrac{3}{\sqrt{13}} = (4+x)\cdot 7 \cdot \dfrac{2}{\sqrt{20}}\cdot \dfrac{4}{\sqrt{20}}; $$

$$ S=\dfrac{42}{13}(3+x) = \dfrac{56}{20}(4+x); $$

$$ 840(3+x)=728(4+x); $$

$$ 112 x = 392; $$

$$ x=\dfrac{7}{2}; $$

$$ S=21. $$


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.

  • $\begingroup$ so we have tg(OBP)=2/3, what is the angle? $\endgroup$
    – user130101
    Commented Jun 3, 2014 at 9:30
  • $\begingroup$ you can find sin, cos if you like. Go tan^{-1} to get back the angle. $\endgroup$
    – DeepSea
    Commented Jun 3, 2014 at 9:36
  • $\begingroup$ I actually tried this before asking, but I didn't get an exact angle. As this is an exam question, i thought i should get a standard angle $\endgroup$
    – user130101
    Commented Jun 3, 2014 at 9:46

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