1. ## Radius of circle inside right triangle

Find the radius of a circle that is inscribed in a right triangle whose short sides are 27 and 36 units long.

I have no clue how to do this

This is mind boggling...

2. 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 *
| \
|   \
|     \
|       \
|         \
|           \
|       * * * \   45
|   *           *
| *               *
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]: .$\displaystyle 54r \:=\:486 \quad\Rightarrow\quad\boxed{ r \:=\:9}$

3. ## simple solution

This can be done really simply. Area of the triangle is equal to s*r.
A = s*r , r is the radius and s = (a+b+c)/2 is called semiperimeter or half of the triangles perimeter. In your case you can easily find the Area, a*b/2 and using the pythagorean theorem find the third side. And then just compare the ttwo equations a*b/2 = (a+b+c)*r/2. Hope this helps

4. Let's forget the area stuff. Just use the geometric theorems on circles.
Since the circle is inscribed into triangle, AB, AC, BC are its tangents with the points of contact at X, Y, Z respectively (diagram).
Produce AO, BO, CO and so we have 3 pairs of congruent triangles, such as AXO and AYO.
therefore AY = AX = 27 - r, also
YC = ZC = 36 - r
Since AC = AY + YC = 45, we get
(27 - r) + (36 - r) = 45
63 - 45 = 2r
r = 9

,

,

,

,

# circle inside triangle

Click on a term to search for related topics.