# Thread: [SOLVED] I Have a Circle inside a Square, and I am asked to find a length

1. ## [SOLVED] I Have a Circle inside a Square, and I am asked to find a length

I have a square ABCD, with lengths 8cm
I have a circle inside the square, as big as it can possibly be, touching all sides. Radius 4cm
There is a diagonal of the square, passing through the circle.
Where the diagonal touches the circle, it is called P.

What I need to do, is prove that AP is equal to 1.66cm

How do I do this?
Thanks

2. $4\sqrt 2 - 4 \approx 1.66$

3. let point O be the circle's center.

$OP$ is a radius = 4

$OA$ is a leg of the right isosceles triangle $\Delta AOB$.

since $|AB| = 8$ , $|OA| = 4\sqrt{2}$

$|AP| = |OA| - |OP|$

4. The diagonal $d$ of a square with length $s$ is given by the equation:

$d=s\sqrt{2}$

5. Thanks a lot guys, that was really helpful

One more question, if it asks me to find DP, how would I get started?
Would I draw a triangle?
If so, how where do i go from there?
Thanks a lot, as you can see, circles are not my strong point

6. Originally Posted by Ben77mc
Thanks a lot guys, that was really helpful

One more question, if it asks me to find DP, how would I get started?
Would I draw a triangle?
If so, how where do i go from there?
Thanks a lot, as you can see, circles are not my strong point
Hint: use the law of cosines. angle DAP = 45 degrees