# Thread: Find the area of the shaded region

1. ## Find the area of the shaded region

What is the area of the shaded region in this shape ؟
رفع الصور

2. ## Re: Find the area of the shaded region

Originally Posted by Mhmh96
What is the area of the shaded region in this shape ؟
رفع الصور
Are the shaded areas made from intersecting quarter circles?

3. ## Re: Find the area of the shaded region

Hello, Mhmh96!

Even if your diagram were symmetric, we're not sure what the figure is.

I'll assume that "Prove It" was correct . . . These are quarter-circles.

Code:
              * * *           * * *
*           *   *           *
*               *               *
*               *:*               *
:::
*               *:::*               *
*               *:::*               *
*               *:::*               *
.*.*.*.   :::   .*.*.*.
*  *:::::::::::**:**:::::::::::*  *
*:::::::::::::::*:::::::::::::::*
*  *:::::::::::**:**:::::::::::*  *
* * *    :::    * * *
*               *:::*               *
*               *:::*               *
*               *:::*               *
:::
*               *:*               *
*               *               *
*           *   *           *
* * *           * * *
Your diagram says the length of two "petals" is 4 units.
Then the radius of the circles is $\displaystyle \sqrt{2}.$

I'll let you work out the details.

I get an answer of $\displaystyle 4(\pi - 2)$

4. ## Re: Find the area of the shaded region

Yes quarter-circles,but how did you find the radius ?

5. ## Re: Find the area of the shaded region

Hello, Mhmh96!

How did you find the radius?

Code:
              * * *
*           *
*               *
*                 *

*         C         *
*         o         *
*    r  *   *  r    *
*.*.*.*.*
*  *:::::::::::*  *
A o - - - - - - - o B
*  *:::: 2 ::::*
* * *
*                   *
*                   *
*                   *

*                 *
*               *
*           *
* * *
Chord $\displaystyle AB$ is 2 units long.

Triangle $\displaystyle ABC$ is an isosceles right triangle.

Hence, its legs (the radius) is: $\displaystyle \frac{2}{\sqrt{2}} \,=\,\sqrt{2}$