Intersection of two Squares | AMC 8, 2004 | Problem 25

Two \(4\times 4\) 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?

Area of the square is \(\pi (r)^2\),where \(r\)=radius of the circle

Can you now finish the problem ..........

Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.

can you finish the problem........

Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.

The area of this region =Total area of larger two squares - the area of the intersection, the smaller square i.e \(4^2+4^2 -2^2=28 \)

Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle

Using the Pythagorean th. diameter of the circle be \(\sqrt{2^2 +2^2}=2\sqrt 2\)

Radius=\(\sqrt 2\)

area of the square=\(\pi (\sqrt2)^2\)=\(2\pi\)

Area of the shaded region= 28-2\(\pi\)

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