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 (Headbang)

This is mind boggling...

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- Nov 27th 2008, 04:40 PMrealintegerzRadius 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 (Headbang)

This is mind boggling... - Nov 27th 2008, 09:03 PMSoroban
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

. . Using Pythagorus, we have: .

The area is: . .[1]

The inscribed circle has center and radius .

Draw line segments

. . and we have three smaller triangles.

. . Area of

. . Area of

. . Area of

The total area is: . .[2]

We just described the area of in two ways.

Equate [1] and [2]: .

- Nov 7th 2009, 05:04 AMdr.g0nz0simple 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 - Nov 8th 2009, 12:53 AMukorov
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