Radius of circle inside right triangle

realintegerz · #1 $Old$ November 27th, 2008, 05:40 PM

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...

Soroban · #2 $Old$ November 27th, 2008, 10:03 PM

Hello, realintegerz!

Quote:

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 $ABC\!:\;\;AB = 27,\;BC = 36$
. . Using Pythagorus, we have: . $AC = 45$
The area is: . $A \:=\:\tfrac{1}{2}(36)(27) \:=\:486$ .[1]

The inscribed circle has center $P$ and radius $r$ .

Draw line segments $PA, PB\text{ and }PC,$
. . and we have three smaller triangles.

. . Area of $\Delta PAB \:=\:\tfrac{1}{2}(27)r \:=\:\tfrac{27}{2}r$

. . Area of $\Delta PBC \:=\:\tfrac{1}{2}(36)r \:=\:18r$

. . Area of $\Delta PAC \:=\:\tfrac{1}{2}(45)r \:=\:\tfrac{45}{2}r$

The total area is: . $A \;=\;\tfrac{27}{2}r + 18r + \tfrac{45}{2}r \:=\:54r$ .[2]

We just described the area of $\Delta ABC$ in two ways.

Equate [1] and [2]: . $54r \:=\:486 \quad\Rightarrow\quad\boxed{ r \:=\:9}$

dr.g0nz0 · #3 $Old$ November 7th, 2009, 06:04 AM

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

ukorov · #4 $Old$ November 8th, 2009, 01:53 AM

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