# Area Problem #9

A square overlaps a circle as shown.

Two vertices lie on the edge of the circle.

Write a suitable formula for the area of the remainder of the circle.

Use r for radius, s for a side of the square, and θ.

## 7 thoughts on “Area Problem #9”

General Approach: Area of Circle – Portion of Square within Circle – Curved Sliver made by two vertices
Area of Circle = pi * r^2
Portion of Square is half of square = 1/2 * s^2
Approach to include curved sliver: Area of sector – Isosceles Triangle made by joining the two vertices and drawing in two radii
Area of Sector = (theta * pi * r^2)/360
Area of Isosceles Triangle = 0.5 * s * sqrt2 * r * cos(0.5theta)
Therefore, final formula:
pi * r^2 – 1/2 * s^2 – [(theta * pi * r^2)/360 – 0.5 * s * sqrt2 * r * cos(0.5theta)]

• I’m confused over your area of isosceles. Can you explain it? I did it with the same method but used 1/2abSinC for area of the isosceles… Which if we’re talking about the same triangle gives (r^2. Sin θ) /2. Someone else has highlighted a neater way by taking the ‘kept’ portion of the sector and dividing it into two similar triangles, then finding their area.

The isosceles triangle is made with a base of s * sqrt 2 (diagonal of square), and two legs made with the radius of the circle. To find the height of the triangle, it is the longer leg of a right triangle where r is the hypotenuse and 0.5theta would be the angle of reference. Using the cosine ratio, I get cos(0.5theta) = height/r so the height = r cos(0.5theta).

• Got it. You’re absolutely right. Messy formula at the end, but a correct one! Thanks for explaining.

3. James says:

I don’t see why you need theta here.

The area of the circle is pi*r^2 and the portion of the square that overlaps the circle is (pi*s^2)/4.

So remaining circle area is pi(r^2 – s^2/4)

• James says:

Never mind, I see where I screwed it up.

• Actually you can do it without theta, but theta makes it accessible for younger students