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

  1. 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.

  2. 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).

  3. 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)

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