# Find the radius in dependency of a (the side length)

I really need help with this task, because it´s supposed to be in my exam...I added a picture of the geomtric figure. The task is to find out about radius r in dependency of a. So at the end it should look something like r=k*a +z+a (for example). I was trying with the Pathagoras making up these formula:

1)y²= (a-r)² + r²

2)r²=y² – (a-r)²

But it looks very complicated, like this at the end. Has anyone please got an idea how to make it easier?? and to get r just on the left side of the equation.Thank you!

all together into 2) : r²= [(a-r)² +r²]² – (a-r)²

r²= [((a-r)²)² + 2(a-r)² +r²+(r²)²] – (a-r)²

r²= [(a-r)²(a-r)² +2(a-r)²+r²+ (r²)²] -(a-r)²

r²=[((a-r)(a-r))²+ 2(a-r)² + r²-(r²)²] -(a-r)²

r²=[(a²-2ar+r²)² + 2(a-r)²+r²+(r²)²]- (a-r)²

r²=[(a²-2ar+r²)(a²-2ar+r²)) + [2(a-r)²+r²+(r²)²]- (a-r)²]

r²=[(a²)²-2a³r+a²r² +4a²r²-2ar³-2a³r+r²a²-2ar³+(r²)²] +[ 2(a-r)²+r²+(r²)²]- (a-r)²]

r²=[(a²)²-4a³r+6a²r²-4ar³+(r²)² +2(a-r)²+r² +(r²)²)] -(a-r)²

r²=[(a²)²-4a³r+6a²r²-4ar³+(r²)² ] +2[(a-r)(a-r)]+2r² +(r²)² -[(a-r)(a-r)]

r²=[(a²)²-4a³r+6a²r²-4ar³+(r²)² ] +[2(a²-2ar+r²) +2r² +(r²)² ] -[a²-2ar+(r²)²]

r²=[(a²)²-4a³r+6a²r²-4ar³+(r²)² ] +[a²-2ar+r²+2r²+(r²)²]

r²= (a²)² -4a³r+6a²r² -4ar²-4ar³+2(r²)² +a² -2ar+r²+2r²

-
Looking at picture, $r$ is not dependent of $a$. –  gev Mar 22 '13 at 14:51
not but should find an equation, how its compared to a. for aexample. if a was 5 centimeters, how big is r. but we dont have numbers.so it should be just an exquation. bith, radius and a are in the same square with length of sides a. –  Sophia Mar 22 '13 at 14:53
When I said not dependent I meant $r$ cannot be expressed in terms of $a$. You can make $a$ as big as you want without affecting $r$. –  gev Mar 22 '13 at 14:55
mh but i am sure there must be a different answer.. otherwise the professor wouldnt have said "your exam task will look like this.." –  Sophia Mar 22 '13 at 14:57
sorry I dont really know what to do either..seems hard the task –  Sophia Mar 22 '13 at 14:58