# Thread: Given Radius of incricle and Hypotenuse solve for the other 2 sides?

1. ## Given Radius of incricle and Hypotenuse solve for the other 2 sides?

The right angle triangle has a hypotenuse of 403 and a incricle with a radius of 74 and i need to find the lengths of the other 2 sides....... i honestly have no idea how to do this everything i try does not work.

2. ## Re: Given Radius of incricle and Hypotenuse solve for the other 2 sides?

Sum of areas of triangles COB, COA, and AOB = area of triangle ABC

$\displaystyle \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc = \frac{1}{2}ab$

$\displaystyle r(a+b+c) = ab$

you are given $\displaystyle r$ and $\displaystyle c$ ... and you know $\displaystyle a = \sqrt{c^2 - b^2}$

finish.

3. ## Re: Given Radius of incricle and Hypotenuse solve for the other 2 sides?

Hi,
an other solution:
The tangent points beetwen sides und incircle are D (on BC), E (on AC), F (on AB).
DC=CE=r=74;
BD=BF=x
EA=AF=403-x
=>
a=x+74
b=74+403-x= 477-x
c=403
You have to use the Pithagoras Formula and have a ecuation second grades in x
Find the adequat solution for x and introduce in a and b