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

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

$\displaystyle OP$ is a radius = 4

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

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

$\displaystyle |AP| = |OA| - |OP|$

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

$\displaystyle d=s\sqrt{2}$

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