Area of Blue Shaded Region

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

See attachment.

[Math Tutor]

The blue part is just what's left of the circle after you cut out the square, so:

[tex]A_{\text{blue}} = A_{\text{circle}} - A_{\text{square}}[/tex]

From [tex]x^2 + y^2 = 36[/tex] the radius is [tex]r = 6[/tex], so

[tex]A_{\text{circle}} = \pi r^2 = 36\pi[/tex]

For the square, the handy fact is that its diagonal is a diameter of the circle, since the square is inscribed and centered at the origin. So [tex]d = 12[/tex] and

[tex]A_{\text{square}} = \frac{d^2}{2} = \frac{144}{2} = 72[/tex]

(If you'd rather do it with coordinates: the corner in the first quadrant is [tex](a,a)[/tex], so [tex]2a^2 = 36[/tex], [tex]a = 3\sqrt{2}[/tex], side [tex]= 2a = 6\sqrt{2}[/tex], and [tex]\left(6\sqrt{2}\right)^2 = 72[/tex]. Same thing.)

Putting it together:

[tex]A_{\text{blue}} = 36\pi - 72 = 36(\pi - 2) \approx 41.10[/tex] square units

Quick sanity check: the square takes up about 64% of the circle, and [tex]41.1[/tex] out of [tex]36\pi \approx 113.1[/tex] is about 36%, so that lines up.

[Eigenvalue]

Generalized area of shaded region for n-sided regular polygon:
a: apothem
l: 1/2 length of one of the polygon's sides
r: radius of the circle
n: number of sides the polygon has
a=r*cos(90(n-2)/n)
l=r*sin(90(n-2)/n)
Area(circle)-Area(polygon)=π[tex]r^{2 }[/tex]-([tex]\frac{r*cos(n-2)}{n}[/tex]*[tex]\frac{r*sin(90(n-2)}{n}[/tex]*0.5*2n)=[tex]r^{2 }\frac{(π-(n/2)sin(180(n-2))}{2}[/tex]

[nycmath]

Math Tutor,

A job well-done!

[nycmath]

Eigenvalue,

Thank you for sharing the formula. Nicely-done.