# Find the area of the shaded region

• Apr 10th 2012, 11:28 AM
Mhmh96
Find the area of the shaded region
What is the area of the shaded region in this shape ؟
http://im26.gulfup.com/2012-04-10/1334082502991.jpgرفع الصور
• Apr 10th 2012, 11:56 AM
Prove It
Re: Find the area of the shaded region
Quote:

Originally Posted by Mhmh96
What is the area of the shaded region in this shape ؟
http://im26.gulfup.com/2012-04-10/1334082502991.jpgرفع الصور

• Apr 10th 2012, 07:40 PM
Soroban
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 $\sqrt{2}.$

I'll let you work out the details.

I get an answer of $4(\pi - 2)$
• Apr 12th 2012, 12:04 PM
Mhmh96
Re: Find the area of the shaded region
Yes quarter-circles,but how did you find the radius ?
• Apr 12th 2012, 02:18 PM
Soroban
Re: Find the area of the shaded region
Hello, Mhmh96!

Quote:

How did you find the radius?

Code:

              * * *           *          *         *              *       *                *                                 *        C        *       *        o        *       *    r  *  *  r    *             *.*.*.*.*       *  *:::::::::::*  *       A o - - - - - - - o B       *  *:::: 2 ::::*               * * *                *                  *       *                  *       *                  *       *                *         *              *           *          *               * * *
Chord $AB$ is 2 units long.

Triangle $ABC$ is an isosceles right triangle.

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