Finding the Area (parameter)

• Jul 31st 2013, 07:04 AM
Telo
Finding the Area (parameter)
Given is a square, ABCD, the lenght of a side is equal to "a". A and D are centers of circles, BD and AC are arcs of circles and they intersect each other in K. Find the area of brightened figure on the picture

Attachment 28933
• Jul 31st 2013, 07:56 AM
Soroban
Re: Finding the Area (parameter)
Hello, Telo!

Quote:

Given square ABCD with side $\displaystyle a.$
A and D are centers of quarter-circles.
The arcs intersect at K.
Find the area of shaded region.

Code:

    B *-*-------*-* C       |    K*:::::|       |  */..*:::|       | *./.....*:|       |*./.......*| a       |./.........|       */..........*     A *-----------* D             a

Draw chord $\displaystyle AK.$

The area of quarter-circle $\displaystyle AKCD$ is:-$\displaystyle \tfrac{1}{4}\pi a^2$

The area of sector $\displaystyle AKD$ is:-$\displaystyle \tfrac{1}{6}\pi a^2$

The area of segment $\displaystyle AK$ is:-$\displaystyle \tfrac{1}{6}\pi a^2 - \tfrac{\sqrt{3}}{4}a^2$

$\displaystyle \text{Shaded region} \:=\:\text{(quarter-circle)} - \text{(sector + segment)}$

. . . . . . . . . . $\displaystyle =\;\tfrac{1}{4}\pi a^2 - \left(\tfrac{1}{6}\pi a^2 + \left[\tfrac{1}{6}\pi a^2 - \tfrac{\sqrt{3}}{4}a^2\right]\right)$

. . . . . . . . . . $\displaystyle =\;\left(\tfrac{\sqrt{3}}{4} - \tfrac{\pi}{12}\right)a^2$
• Jul 31st 2013, 08:00 PM
ibdutt
Re: Finding the Area (parameter)
Quote:

Originally Posted by Telo
Given is a square, ABCD, the lenght of a side is equal to "a". A and D are centers of circles, BD and AC are arcs of circles and they intersect each other in K. Find the area of brightened figure on the picture