# Help with solving this probability problem

• March 18th 2013, 06:22 PM
arb123
Help with solving this probability problem
I know that you have to do

area of circle-area of inner square
Area of outer square

the area of the circle= (3.1413)r^2
the area of outer square= 4r^2

But i cant figure out the area of the inner circle?

can anyone help me?
• March 18th 2013, 07:14 PM
chiro
Re: Help with solving this probability problem
Hey arb123.

What do you mean for the inner circle? I only see one circle in your diagram and you have the area of that circle in your post.
• March 18th 2013, 07:18 PM
arb123
Re: Help with solving this probability problem
I meant the inner square. Sorry!
• March 18th 2013, 07:27 PM
Paze
Re: Help with solving this probability problem
Hi arb123.

Notice that if you split the inner square into two triangles, you will have two triangles with base 2r and height r.

We have:

A circle with area: $r^2\cdot \pi$

A square with area: $2\cdot \frac{(2r\cdot r)}{2}$

A larger square with area: $2r \cdot 2r$

The shaded area is thus: $\left(r^2\cdot \pi-2\cdot \frac{(2r\cdot r)}{2}\right)$

The probability of the dart hitting the shaded area and NOT the white area inside the larger square, between its corners and the circle is thus:

$\frac{\left(r^2\cdot \pi-2\cdot \frac{(2r\cdot r)}{2}\right)}{2r\cdot 2r}$
• March 18th 2013, 07:28 PM
Paze
Re: Help with solving this probability problem
Perhaps chiro can verify my answer?
• March 18th 2013, 08:52 PM
arb123
Re: Help with solving this probability problem
why wouldn't the height also be 2r? when you draw in out the height of each side of the triangle is equal to the base.
• March 18th 2013, 09:06 PM
Paze
Re: Help with solving this probability problem
Does this image help you understand why the height is r and the base is 2r? http://s9.postimage.org/5zb7sblnj/expl.png
• March 18th 2013, 09:11 PM
arb123
Re: Help with solving this probability problem
Yes, i see now! Thank you.