What is the radius of the circle? 
Please help with this grade nine math problem.  How does one calculate the radius if the two sides of the right angle triangle are 85cm.  The sides of the triangle are tangent to the circle.
 A: 
Hint. $AD = DB = AF$ and $FC = OE = r$.
A: 
It's useful to realize that the "left" and "right" radia, as drawn in the above picture, will be parallel to the respective cathetae.
Then you get:
$$C=\sqrt{A^2+A^2}=\sqrt{2}A$$
The height of the triangle is then:
$$h=\sqrt{A^2-\left(\frac{C}{2}\right)^2}=\sqrt{A^2-\frac{A^2}{2}}=\frac{1}{\sqrt{2}}A$$
Define x-axis along the base of the triangle and y-axis along the height.
Unit vectors at a 45° angle to the x-axis are given by:
$$\vec{u}_1=\frac{1}{\sqrt{2}}\left({1}\atop{1}\right)~~~~~,~~~~~\vec{u}_2=\frac{1}{\sqrt{2}}\left({1}\atop{-1}\right)$$
You can check $\vec{u}\cdot\vec{u}=1$.
Now use that the distance from any of the two 45° angles to the two nearest spots where the circle touches the triangle is the same, namely $C/2=A/\sqrt{2}$.
With this you can establish a vectorial relation between the following vectors:
$$h\vec{e}_y+\left(A-\frac{C}{2}\right)\vec{u}_2=R\vec{e}_y+R\vec{u}_1$$
Where $\vec{e}_y=(0,1)$ is the unit vector along the y-axis. This gives you two equations.
The y-axis equation is:
$$h-\frac{1}{\sqrt{2}}\left(A-\frac{C}{2}\right)=R+\frac{1}{\sqrt{2}}R\\\frac{A}{\sqrt{2}}-\frac{1}{\sqrt{2}}\left(A-\frac{A}{\sqrt{2}}\right)=\left(1+\frac{1}{\sqrt{2}}\right)R\\\frac{A}{2}=\left(1+\frac{1}{\sqrt{2}}\right)R\\R=\frac{A}{2+\sqrt{2}}$$
The x-axis equation is:
$$\frac{1}{\sqrt{2}}\left(A-\frac{C}{2}\right)=\frac{1}{\sqrt{2}}R\\\left(1-\frac{1}{\sqrt{2}}\right)A=R\\R=\frac{\left(1-\frac{1}{\sqrt{2}}\right)\left(2+\sqrt{2}\right)}{2+\sqrt{2}}A\\R=\frac{2-\sqrt{2}+\sqrt{2}-1}{2+\sqrt{2}}A\\R=\frac{A}{2+\sqrt{2}}$$
Both answers properly agree, so that the world is a happy and sunny place.
A: Hint : $A=rs$
where $A=\text{area},r=\text{inradius},s=\text{semiperimeter}$
A: Let $a\ (=85{\rm cm})$ be the length of the legs and $r$ the radius of the circle. Then
$${1\over2}\bigl(\sqrt{2}a\bigr)+r=a\ ,$$
which implies
$$r=\left(1-{\sqrt{2}\over2}\right)\>a\ .$$
