A square of side length 1 inch is drawn with its center A on a circle O of radius 1 inch such that a side of the square is perpendicular to segment OA. What is the area of the shaded region?

Guest May 3, 2020

#1**+1 **

We can form a chord connecting the two points where the circle and the square intersect

Call these points B and C

Now triangle BOC is equilateral becuse OB = OC = BC = 1

This means that angle AOB = 60°

And we can find the shaded area between the chord and the edge of the circle thusly :

Area of sector BOC - area of the equilateral triangle =

pi (1)^2 (60/360) - (1/2)*(1)^2 * √ 3 / 2 =

[ pi/6 - √ 3/4 ]

Now the height of the equlteral triangle can be found as

area = (1/2)base * height

√ 3/4= (1/2)(1) * height

2√ 3/4 = height = √ 3/2

If we connect AO we can call the intersection of this segment and the side of the square Q

And note that QO = 1/2

And the remaining rectagular shaded area has a width of 1

And its height = height of the equlateral triangle - OQ = √ 3/2 - 1/2 = [ √ 3 - 1] / 2

So the remaining rectagular shaded area = (1) [ √ 3 - 1 ] / 2 = [ √ 3 - 1 ] / 2

So.....the total shaded area = [ pi/6 - √ 3/4] + [√ 3 - 1] / 2 =

pi/6 + √ 3/2 - √ 3/4 - 1/2 =

[ pi/6 + √ 3/4 - 1/2] units^2 ≈ .46 units^2

CPhill May 3, 2020