Find Area of Shaded Region

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

See attachment.

I know the circle is centered at the origin.

Let r = radius = 3.

The square can be divided into four right triangles.

Let h = height.

x^2 + y^2 = h^2

Let x = y = r = 3.

(3)^2 + (3)^2 = h^2

18 = h^2

Taking the square root on both sides of the equation, I get
h = 3 • [tex]\sqrt{2}[/tex]

Area = (side)^2

A = [3 • [tex]\sqrt{2}[/tex]]^2

A = 18 square units.

You say?

[Math Tutor]

Yep, [tex]18[/tex] square units is the right answer.

Your method works, but I'd relabel things a bit - the [tex]h[/tex] you solved for isn't a height, it's actually the side of the square. Drawing both diagonals cuts the square into 4 right isosceles triangles whose legs are radii, so legs of [tex]3[/tex] and [tex]3[/tex] give a hypotenuse of [tex]3\sqrt{2}[/tex], and that hypotenuse is a side:

[tex]s = 3\sqrt{2}, \qquad A = s^2 = \left(3\sqrt{2}\right)^2 = 18[/tex]

A cleaner way to see it: by symmetry the corner in the first quadrant is [tex](a,a)[/tex]. Plug that into the circle:

[tex]a^2 + a^2 = 9 \quad \Rightarrow \quad 2a^2 = 9 \quad \Rightarrow \quad a = \frac{3}{\sqrt{2}}[/tex]

The side is [tex]2a = 3\sqrt{2}[/tex], so [tex]A = 18[/tex] again.

Or skip the side completely. The diagonal of the square is just the diameter, [tex]d = 6[/tex], and for any square:

[tex]A = \frac{d^2}{2} = \frac{36}{2} = 18[/tex]

One small thing to watch: don't set [tex]x = y = 3[/tex] out of habit. Here [tex]3[/tex] is the distance from the center to a corner, not to the midpoint of a side. It lined up for you because you were really using the two radii as the legs.

[nycmath]

Math Tutor,

Your reply is right on point.