Question

A square with sides of 3sqrt2 is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the square?

help meeee

Answer 1

A side of the square is 3*sqrt(2)

So, using pythagoras, we can find the diagonal of the square (since the diagonal forms a right angle with the square's sides):

C^2=A^2+B^2 - in our case A and B are 3*sqrt(2)
==> C^2 = 18 + 18 = 36
==> C = 6
The diagonal of the square is 6, and that is also the diameter of the circle.
The radius of the circle will be half the diameter.
==> r=3

The area of the entire circle will be pi * r^2 = 9 * pi

So, one sector of the circle (one slice) will be a fourth of the entire area:

The answer will be:
(9/4)*pi

Answer 2

For such an inscribed square, the radius "R" of the circle is given by the formula

a²
R² = ----- where a is the length of the side of the rectangle
2

In this case

(3√2)² 9*2
R² = ------ = ----- = 9
2 2

so R = √9 => R = 3

Answer 3

Now radii to the vertices form an angle of 90°

so we have to find the area of a sector of angle 90°

angle 90° 22
area of sector = -------- * πR² = ------ * ----- * 3²
360° 360° 7
1 22 22 * 9 198
= ----- * ----- * 9 = -------- = ------- = 7.07 sq units
4 7 4 * 7 28