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

Please Help!

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- Jul 31st 2013, 07:04 AMTeloFinding 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

Please Help!

Attachment 28933 - Jul 31st 2013, 07:56 AMSorobanRe: 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

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 PMibduttRe: Finding the Area (parameter)
- Jul 31st 2013, 08:45 PMjohngRe: Finding the Area (parameter)
Hi,

Soroban beat me to the same solution. I was busy constructing the attached drawing; I enjoy drawing pictures. I'll post my version of his solution anyway.

Attachment 28934