Reginald Delacruz

2022-06-30

In triangle $\mathrm{\Delta}ABC$ $\measuredangle C={90}^{\circ}$. The angle bisectors of angles $\mathrm{\angle}A$ and $\mathrm{\angle}B$ cross at point $O$. The distance from point $d(O,\overline{AC})=3\mathrm{cm}$ and $d(O,\overline{AB})=15\mathrm{cm}$. Find the perimeter of triangle $\mathrm{\Delta}ABC$ in cm

Odin Jacobson

Beginner2022-07-01Added 17 answers

$O$ is the center of inscribed circle.

Let $M$, $N$ and $P$ be the tangent points of $AC$, $AB$ and $BC$ respectively, and

$AN=AM=x,NB=BP=y,x+y=15,CP=CN=3$ Then, perimeter $P=(x+y)+(y+3)+(3+x)=36$

Let $M$, $N$ and $P$ be the tangent points of $AC$, $AB$ and $BC$ respectively, and

$AN=AM=x,NB=BP=y,x+y=15,CP=CN=3$ Then, perimeter $P=(x+y)+(y+3)+(3+x)=36$

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