A Problem of Quarter Circles

For 31 answer question should be G.I.F. as nearest integer is 32. (31.51467 is the answer calculated value.)

First the area of the quarter circle is 78.5. To get the area of the shaded region and the two "arrows" attached to it we subtract 21.5 from 78.5, which is 57. Cut the square in half into to separate parts with a length of 5 and width of 10. Make a line from the bottom right vertex of the square to the top of the shaded region. This gives you a right triangle with a base length of 5 and a hypotenuse of 10 (We know this because the line is the radius of the quarter circle). The ratio of the two sides tells us that this triangle is a 30-60-90 triangle so the height of this triangle is 5 Sqrt(3). The area of this triangle is then 21.65. Afterwards you are left with part of the quarter circle(the part that is next to the triangle we just solved for). We know the angle of this "triangle" is 30 degrees(b/c of the 30-60-90 triangle). This means that the area of this "triangle" is 1/3 of the quarter circle it makes, which is 26.167. Now if you subtract the area of these two triangles from the half of the square you get the 1/2 the area of "curve" directly above the shaded region, which is 2.18. Now we will find the area of the arrow. We know the area of a arrow and two curves which is 21.5. So 21.5-4 (2.18), gives us the area of the area of the arrow which is 12.77. Finally we subtract 12.77*2 from the area of the shaded region and two areas to give use the area of the shaded region, which is 31.5.

Integrate[4 (-5 + Sqrt[100 - (5 + x)^2]), {x, 0, Sqrt[75] - 5}]

Answer~31.515=32

The area of the green region is given by $A=x^2\left(1-\sqrt{3}+\dfrac{\pi}{3}\right)$ . We have

$A=10^2\left(1-\sqrt{3}+\dfrac{\pi}{3}\right)\approx$ $\boxed{32}$

via integration very simple.....