Question

solve this will gave u medal

Answer 1

@ChristopherToni

Answer 2

If you draw the radius from the center to any corner of the square, that tells you the diagonal length of the smaller square is \(\sqrt{2}\). Furthermore, the right triangle created by this radial line is the special \(1-1-\sqrt{2}\) triangle. Hence the sides of the smaller squares are 1, which in return makes square ABCD a \(2\times 2\) square. The part of the circle outside of the square has area \(\pi(\sqrt{2})^2-2^2 = 2\pi-4\). However, the shaded part we need outside of the square is only one-eighth of this value. Adding that result to the shaded area in the square (i.e. one-fourth of the total area of ABCD), we get \(\frac{1}{8}(2\pi-4) + \frac{1}{4}(4) = \frac{1}{4}\pi-\frac{1}{2} + 1 = \frac{1}{4}\pi+\frac{1}{2}\), which is choice (D).

Does this kind of make sense? :-)

Answer 3

why u minus two from area of circle @ChristopherToni

Answer 4

\(2\pi - 4\) is the area of the part you get if you remove the square from the circle. We need this value in order to find the area of the shaded part above the square in the diagram.

Answer 5

got that:) @ChristopherToni

Answer 6

What is \(1-\frac{1}{2}\)? Try getting a common denominator. :-)