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avatar+2766 

\(ABCD\)  is a square with \(AB = 8\) cm. Arcs \(CD\)  and \(BC\) are semicircles. Express the area of the shaded region, in square centimeters, and in terms of \(\pi\). (As always, do not include units in your submitted answer.)

 

tertre  Apr 2, 2018
 #1
avatar+87629 
+3

We can use some symmetry to solve this

 

Draw a segment from the center of the square to the top right vertex of the square...this will divide the shaded area into to 2 equal parts...let this segment  be EC

 

If we let the center of the the circle on the right be the midpoint of BC  ....call this point, F....and the radius of this circle will  = 4

 

Then.....  the sector FEC of the circle on the right   will be the quarter area of this circle  = 

(1/4)pi (4)^2  =  4pi  units^2

 

And.....  the triangle formed by FEC  wll have the area  (1/2)(4)^2sin 90  = 8 units^2

 

So the top half of the shaded area is just the difference of these two areas  =

 

[ 4pi  -  8]    units^2

 

So....the whole shaded area is just twice this  =

 

[ 8pi - 16 ] cm^2  ≈   9.13 cm^2

 

Here's a pic, tertre :

 

 

 

 

cool cool cool

CPhill  Apr 2, 2018
 #2
avatar+2766 
+1

An amazing solution, CPhill! I understand much better now! Thanks for the diagram!

tertre  Apr 2, 2018

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