2 quadrants within a square of length 10cm.

how to find the area of the shaded area in yellow?

thanks!

http://img.photobucket.com/albums/v3...untitled-7.jpg

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- Sep 2nd 2007, 08:23 AMbellsfind area of shaded area (yellow)
2 quadrants within a square of length 10cm.

how to find the area of the shaded area in yellow?

thanks!

http://img.photobucket.com/albums/v3...untitled-7.jpg - Sep 2nd 2007, 08:29 AMKrizalid
Find an equilateral triangle. There's a lot of hidden information there.

- Sep 2nd 2007, 08:52 AMred_dog
You can split the area in two sectors with the same area and an equilateral triangle.

Each sector has an angle of $\displaystyle 30^{\circ}$. - Sep 2nd 2007, 09:25 AMSoroban
Hello, bells!

Did you catch red_dog's excellent hint?

Quote:

Two quadrants within a square of length 10cm.

. . Find the area of the shaded region.

http://img.photobucket.com/albums/v3...untitled-7.jpg

Label the square $\displaystyle A\!-\!B\!-\!C\!-\!D$, starting at the upper-left and labelling clockwise.

Label the intersection of the two arcs with $\displaystyle O$.

Draw radii $\displaystyle OC$ and $\displaystyle OD$.

Triangle $\displaystyle OCD$ is equilateral with side 10cm.

. . I'm sure you can find its area.

Sectors $\displaystyle DAO$ and $\displaystyle CBO$ have radius 10 and central angles 30°.

. . and you can find their areas, too, right?

- Sep 2nd 2007, 03:06 PMbells
why is it an equilateral triangle?

thanks. - Sep 2nd 2007, 09:37 PMearboth
- Sep 2nd 2007, 09:49 PMJhevon
earboth gave you detailed information on how to find the answer, but allow me to fill in a bit of the background.

recall that the yellow regions are outlined by circular arcs with radius 10. consider the left arc. any line drawn from the bottom left corner to any point along the arc will have the length of the radius, which is 10. the same thing holds true for the arc on the right. since the base of the triangle drawn is one side of the square, it will have a length of 10 as well. thus we have a triangle with sides 10 units of length each, thus it is an equilateral triangle - Sep 4th 2007, 06:11 AMjanvdl
Wow, this was an excellent problem. I like it. :)

- Sep 4th 2007, 08:34 AMbells
got it!

thanks! :)