As the image above shows, a right triangle $ABC$ has $O_1$ as its inscribed circle, and $O_2,~O_3,~O_4$ as its escribed circle. The radius of a circle $O_i$ is $r_i.~(i\in \{1,~2,~3,~4\})$

Given that $r_1=1$ and that the area of triangle $ABC$ is $\dfrac{15}{2},$ find the value of $r_2+r_3+r_4.$

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This problem is a part of
<Grade 10 CSAT Mock test> series
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The answer is 14.

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Let $\overline{AB},~\overline{BC},~\overline{CA}$ be $c,~a,~b$ respectively.

Then by above picture, we know that $(b-r_1)+(a-r_1)=c.$

$r_1=\dfrac{a+b-c}{2}.$

Also, by above picture, we know that $b+r_2=c+(a-r_2).$

$r_2=\dfrac{a-b+c}{2}.$

Using similar method with $r_3$ and $r_4,$ we get

$r_3=\dfrac{-a+b+c}{2};$

$r_4=\dfrac{a+b+c}{2}.$

Therefore,

$r_1+r_2+r_3+r_4=a+b+c=15.$

$\therefore~r_2+r_3+r_4=\boxed{14}.$