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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?

 

 May 3, 2020
 #1
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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 

 

 

cool cool cool

 May 3, 2020
edited by CPhill  May 3, 2020
 #2
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+1

Thank you so much!!

Guest May 3, 2020

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