# Find the blue area

• September 14th 2010, 05:25 AM
razemsoft21
Find the blue area
• September 14th 2010, 06:25 AM
Soroban
Hello, razemsoft21!

Quote:

Code:

              * * *           *    |r/2  *         *      A|      *       *        o        *               .|.       *      .:|:.      *       *      .::O::.      *       *    .::*|*::.    *           .::*:|:*::.       *  .::*::|::*::.  *         * ::*:::|:::*:: *         Bo - - - - - oC                 r

$\,O$ is the center of the circle.
Draw $AB$ and $AC.$

Note that $\Delta OBC$ is equilateral.
Its height is: . $\frac{\sqrt{3}}{2}r$

The height of $\Delta ABC$ is: . $\frac{r}{2} + \frac{\sqrt{3}}{2}r \:=\:\left(\frac{1+\sqrt{3}}{2}\right)r$

The area of $\Delta ABC$ is: . $\frac{1}{2}bh \;=\;\frac{1}{2}(r)\left(\frac{1+\sqrt{3}}{2}\righ t)r \;=\;\dfrac{1+\sqrt{3}}{4}\,r^2$

• September 14th 2010, 06:38 AM
razemsoft21
good job Soroban but I need the blue area
and as you see it is a part of circle not a full
circle to substact the area of triangle !!
• September 14th 2010, 09:07 AM
Soroban

Ha! . . . That was stupid!
I found the wrong area . . . *blush*

Let's see, the shaded area is the area of the circle,
. . minus the area of $\Delta ABC$, minus that missing segment.

The area of the segment is the area of the $60^o$ sector
. . minus equilateral triangle $OBC.$

The area of the sector is: . $\frac{1}{2}r^2(\frac{\pi}{3}) \:=\:\frac{\pi}{6}r^2$

The area of the equilateral triangle is: . $\frac{\sqrt{3}}{4}r^2$

So the area of the segment is: . $\frac{\pi}{6}r^2 - \frac{\sqrt{3}}{2}r^2 \:=\:\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2$

Therefore, the desired area is:

. . $\displaystyle \pi r^2 - \left(\frac{1+\sqrt{3}}{4}\right)r^2 - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2 \;=\;\left(\frac{5\pi}{6} - \frac{1}{4}\right)r^2$

• September 14th 2010, 10:31 AM
Opalg
Here is a slightly different method. In the attached picture (where the top of the white equilateral triangle is the centre of the circle), the blue and red regions together form 5/6ths of the area of the circle, so their combined area is $\tfrac56r^2$. From that we must subtract the red area. Each red triangle has area half the product of two sides times the sine of the included angle. In this case the sides are $\tfrac12r$ and $r$, and the included angle is $150^\circ$. So the area of each triangle is $\tfrac14r^2\sin(150^\circ) = \tfrac18r^2$. Therefore the blue area is $\bigl(\tfrac{5\pi}6-\tfrac14\bigr)r^2$, as Soroban already showed.

Sometimes it's worth doing a problem in more than one way, just to check that you have the correct answer.