Hello, realintegerz!
Find the radius of a circle that is inscribed in a right triangle
whose short sides are 27 and 36 units long. Code:
A *
| \
| \
| \
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| * * * \ 45
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| * *
27 |* o *\
| o r \
* P o * \
* o o o o * * \
* r o * \
| o \
|* o r * \
| * o * \
| * o * \
B * - - - * * * - - - - - - - - - - - - * C
36
We have right triangle $\displaystyle ABC\!:\;\;AB = 27,\;BC = 36$
. . Using Pythagorus, we have: .$\displaystyle AC = 45$
The area is: .$\displaystyle A \:=\:\tfrac{1}{2}(36)(27) \:=\:486$ .[1]
The inscribed circle has center $\displaystyle P$ and radius $\displaystyle r$.
Draw line segments $\displaystyle PA, PB\text{ and }PC,$
. . and we have three smaller triangles.
. . Area of $\displaystyle \Delta PAB \:=\:\tfrac{1}{2}(27)r \:=\:\tfrac{27}{2}r$
. . Area of $\displaystyle \Delta PBC \:=\:\tfrac{1}{2}(36)r \:=\:18r$
. . Area of $\displaystyle \Delta PAC \:=\:\tfrac{1}{2}(45)r \:=\:\tfrac{45}{2}r$
The total area is: .$\displaystyle A \;=\;\tfrac{27}{2}r + 18r + \tfrac{45}{2}r \:=\:54r$ .[2]
We just described the area of $\displaystyle \Delta ABC$ in two ways.
Equate [1] and [2]: .$\displaystyle 54r \:=\:486 \quad\Rightarrow\quad\boxed{ r \:=\:9}$