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Let ABCD be a square of side length 4. Let M be on side line BC such that CM = 1, and let N be on side line AD such that DN = 1. We draw the quarter-circle centered at A.

Let x and y denote the areas of the shaded regions, as shown. Find x - y.

 

Thanks!

AnonymousConfusedGuy  Apr 13, 2018
 #1
avatar+92674 
+2

This one gets ugly, ACG....but....it is "do-able"

 

If we position the bottom left vertex of the square at (0,0)  the circle has a center of (0,0)  and a radius of 4.....so.....the equation is  x^2 + y^2  = 16

Note....since AN  = 3...the intersection  of  NM  and the circle occurs at  x  = 3

So....  3^2 + y^2  = 16

9+ y^2  = 16

y  = √[16 - 9 ]   = √7

 

Now...if you can picture this.....let this intersection be the point,  Q

Draw  AQ  and  QD

 

Now....we want to find the area of the sector AQD

Note that the  sine of the central  angle  QAD  is NQ / AQ =   √7/4

So...the measure [in degrees]  of this angle can be expressed as  arcsin (√7/4)

So....the area of the sector is  (1/2)r^2 *arcsin (√7/4) (/360° }=

(1/2) 4^2* [arcsin (√7/4) / 360°]   =  8 [arcsin (√7/4) / 360° ]     (1)

Don't worry....this looks ugly, but it will eventually disappear !!!!

 

And the area of the triangle AQD  is  given by

(1/2) r^2 sin QAD   =

(1/2) 4^2 (√7/4)  =

2√7    (2)

 

So....the area of the sector - the area of the triangle  =  the area between the chord QD and the edge of the circle   =  (1) - (2)

 

And this is part of the white area NQD

And the other part of this white area is the triangle with base ND  and height  NQ  =

(1/2)ND * NQ   =  (1/2)(1)*√7  =  √7/2      (3)

 

So..... the "white"  area NQD  is given by  [ (1) - (2) ] + (3)  =

 8 [arcsin (√7/4) / 360° ]  - 2√7  +  √7/2  =

8 [arcsin (√7/4) / 360° ] -  3√7/2      (4)

 

So.....Area  "y"   =  Area of rectangle  MCDN  - (4)  =

MC * CD  - (4)  =

1 * 4  -  (4) =

4  -  [8 arcsin (√7/4) / 360° ] -  3√7/2 ] =

[  4 + 3√7/2   - 8 [arcsin (√7/4) / 360°  ]   =  (5)

 

Now  ...area  "x"   is   just the area  of the quater circle  - the white area,   (4)

The area of the quarater circle  is  just  (1/4)pi (4)^2)  = 4pi

So "x"   =

4pi - [8 [arcsin (√7/4) / 360° ] -  3√7/2]   =

4pi + 3√7/2 -  [8 arcsin (√7/4) / 360°]   (6)

 

So..... "x"  - "y"  =   (6) - (5)  =

 

  (4pi + 3√7/2 -  [8 arcsin (√7/4) / 360°] )   -  ( [  4 + 3√7/2   - 8 [arcsin (√7/4) / 360°  ] )  =

 

( 4pi  -  4)     units^2

 

See???...I told you it was ugly.... LOL!!!!

 

 

 

cool cool cool

CPhill  Apr 13, 2018
 #2
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+1

How about this method:

Area of the 1/4 circle =4Pi...........................................................(1)

Area of the segment={[pi/2 - sin(pi/2)] /2 x 4^2} / 2 =2Pi - 4........(2)

Area of rectangle MCND =4 x 1 =4

Shaded area in this rectangle =4 - (2Pi - 4) =8 - 2Pi....................(3)

(1) - (2) - (3) =4Pi - (2Pi - 4) - (8 - 2Pi) =4Pi - 4

Guest Apr 13, 2018
 #3
avatar+92674 
0

Thanks, guest....that's easier than my approach  !!!!

 

cool cool cool

CPhill  Apr 13, 2018
 #4
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0

By the way, it just occurred to me that:

1/4 of the area of the circle which includes the unshaded segment =4Pi

The area of the rectangle which includes the unshaded area = 4

Then the shaded area of the 1/4 circle - shaded area of the rectangle =4Pi - 4

Guest Apr 13, 2018
 #5
avatar+1442 
+1

Thanks so much everyone this was very helpful!

AnonymousConfusedGuy  Apr 14, 2018

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