Math Prompt

Two  squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

The question, in particular, prompts to find the area of the shaded region, given two intersecting squares and a circle inscribed within this intersection. Inscribed in the circle, is a square denoting the exact area that the squares intersect. The question provides information that the squares bisect their intersecting sides, which forms the inscribed square in particular. Since the squares bisect the sides, meaning divided in half, the side lengths of the inscribed square are 2, given that the side lengths for the 2 intersecting squares are 4 each. Given the square inscribed side lengths within the circle, finding the diagonal measure through the Pythagorean Theorem will help us find the diagonal measure, which will represent the diameter of the circle. We can find the diameter of the circle if we solve for the diagonal of the square since they are equivalent. Then if we divide the diameter in half we will get the radius. 

To find and calculate the length of the diagonal of the square, we need to set up a right triangle to find the diagonal length through the Pythagorean Theorem. Since the inscribed square’s side lengths were found to be 2, a right triangle can be drawn where the base and leg are 2 and the hypotenuse is unknown. Using Pythagorean Theorem (a²+b²=c²), we get the following equation: (2² + 2² = c²), where a = 2 and b = 2. Simplifying the equation further will result in: (c² = 8). Square rooting each term will result in the equation to: (c = sqrt(8)). Simplifying sqrt(8) using Prime Factorization will result in 2*sqrt(2), which is the hypotenuse of the diagram and the diagonal measure of the inscribed circle. 

The Diameter of the circle is the diagonal of the square. So the diagonal is 2*sqrt(2). Then we divide that in half to find the radius. The 2’s cross each other out and we end up with sqrt(2) as the radius. To find Area the formula is πr². So we want to square sqrt(2) and so we end up with just a 2. And since it does not specify what π is we leave it at 2π for the area of the circle. To find the area of the square inside the circle we just multiply 2*2 to get 4 since each side of that square is 2.

The bigger squares’ sides are each 4 so each triangle has an area of 16, and since there are 2 big triangles, the combined equivalence is 32. But since they are overlapped with the smaller, inscribed square in the circle, we have to subtract a 4 from the 32 to end up with a 28 in the shaded area. So to find the total area of the shaded region we subtract 2π, the area of the circle, from the 28. Since we cannot simplify any further we end up with 28 – 2π, leaving the answer in terms of π.