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Let $AD$ be an altitude in right angled $\triangle{ABC}$ with $\angle{A}=90^{o}$ and $D$ on $BC$. Suppose that the radii of incircles of triangles $ABD$ and $ACD$ are $33$ $(r_1)$ and $56$ $(r_2)$ respectively. Let $r$ be the radius of incircle of the $\triangle{ABC}$. Find the value of $3(r+7)$.

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(Figure is rough)

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Both the triangles $DBA$ and $DAC$ are similar to the original triangle $ABC$.
It follows that the ratio of their inradii, $\frac{33}{56}$, equals $\frac{AB}{AC}$. By the Pythagorean theorem, since $\sqrt{33^2+56^2}=65$, we have that $\frac{BC}{AB}=\frac{65}{33}$, hence $r_{ABC}=\color{red}{65}$.

In order to clarify: if we have two similar right triangles, the ratio of their inradii equals the ratio of the lengths of two corresponding legs. That follows from the fact that we may obtain one triangle from the other through a suitable dilation. The dilation factor is the same for both the inradii or two corresponding leg lengths.

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  • $\begingroup$ please provide links for "the ratio of their inradii, $\frac{33}{56}$, equals $\frac{AB}{AC}$" or prove here if possible. $\endgroup$ – mnulb Sep 12 '16 at 17:19
  • $\begingroup$ @Ayushakj: if you have two similar right triangles, the ratio of their inradii equals the ratio of the lengths of two corresponding legs. That follows from the fact that we may obtain one triangle from the other through a suitable dilation. The dilation factor is the same for both the inradii or two corresponding leg lengths. Added to the answer. $\endgroup$ – Jack D'Aurizio Sep 12 '16 at 17:21

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