#### To determine

**To find:**

The area of the given region

#### Answer

4342-5

#### Explanation

**1) Concept:**

Without loss of generality we may assume that the centre of square is at origin. Then the square is symmetric about both axes. Divide the square into four right angled triangles of equal area using the diagonals. Owing to symmetry to find the total area it suffices to find the area of the shaded region R consisting of all points in the triangle that are closer to the center than to that *side. We* can then multiply this area by 4 to find the total area of the region.

2) **Calculation:**

Find the equation of the curve whose points are equidistant from the center and side. This curve(purple) shall be the top-boundary of the area. Consider an arbitrary point (x, y) on the curve. Then the distance of the point from the side is 1-y, and its distance from the center is x2+y2.

So the distances are equal if x2+y2=1-y

Take square of both sides.

x2+y22=1-y2

By simplifying,

x2+y2=1-2y+y2

Subtract x2+y2 from both sides.

x2+y2-x2+y2=1-2y+y2-x2-y2

By simplifying,

0=1-2y-x2

Add 2y to both sides.

0+2y=1-2y-x2+2y

By simplifying,

2y=1-x2

Divide by 2, and simplify.

y=12(1-x2)

Thus the crescent upper boundary of the region belongs to a parabola that opens down.

The area R is equal to the area of the black triangle plus the blue crescent shaped area

R=Ac+AT

*To find these areas, find *y- coordinate h of the horizontal line separating them.Consider the top right corner point of R. We may note that this point is of the type (h,h) for some h. This can be seen in two ways. Notice that the diagonal of the square is part of the line y=x, and the point lies on the line. Or notice that the top right corner point is part of top boundary curve as well as right boundary curve. Suppose the point is (h,h') and its distance from origin is d. Then by construction the distance from top side of square to point is same as its distance from origin. Similarly the distance from right side of square to point is also same as its distance from origin d. Thus 1-h'=d=1-h. So we have h=h'

Let (h,h) be the corner point. Then,

1-h=2h

Add h on both sides.

1-h+h=2h+h

By simplifying,

1=(2+1)h

Divide both sides by 2+1, and simplify.

h=11+2=2-1 Notice that the last answer is obtained by multiplying 2-1 to both denominator and numerator.Due to symmetry the top left corner shall be (-h,h)

Thus area of the crescent shaped section (blue) is

Ac=∫-hh121-x2-hdx

By simplifying(the function is symmetric),

=2∫0h12-h-12x2dx

Take integration.

=212-hx-16x30h

Put limits.

=212-hh-16h3-[12-h0-1603]

By simplifying,

=h-2h2-13h3 The area of black triangle is 1/2(base)(height)= (1/2)(2h)h =h2

So the area of the region, R is

A=h-2h2-13h3+h2

Thus total area is 4A=4[h-2h2-13h3+h2]

Factor out h, and simplify.

4A=4h(1-h-13h2)

But h=2-1

So, put h=2-1 in the above equation.

=42-11-2-1-132-12

By simplifying,

=42-11-132

By simplifying,

=4342-5

Therefore, the area of the given region is 4342-5

**Conclusion:**

The area of the given region is 4342-5