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In the figure below, $O$ is the center of the circle, and the radius of the circle is $8$. Angle $\angle BOA$ is right. What is the area of the filled-in (purple) region? Give your answer in exact form (which may involve pi). In other words, don't round off (but do simplify if you can).

 Mar 13, 2021
 #1
avatar+30829 
+1

Need to see the PURPLE area ...is it in the 90 degree area ...or the 270 degree area ...or other?

 Mar 13, 2021
 #3
avatar+42 
-1

oh ok how do i do it i its my fist time on here?

Valencia  Mar 13, 2021
 #2
avatar+42 
0

\([asy] size(4cm); pair o=(0,0); pair a=(-sqrt(2)/2,-sqrt(2)/2); pair b=(sqrt(2)/2,-sqrt(2)/2); path p=Arc(o,1,225,315)--b--a--cycle; fill(p,purple); dot(o); dot(a); dot(b); draw(o--b--a--o); draw(a/8--(a+b)/8--b/8); draw(Circle(o,1)); label("$O$",o,N); label("$A$",a,SW); label("$B$",b,SE); [/asy]\)

.
 Mar 13, 2021
edited by Valencia  Mar 13, 2021
edited by Valencia  Mar 13, 2021
 #4
avatar+311 
+1

This is a mess, I'm adding onto this mess

 

https://web2.0calc.com/questions/help-me-out-plz

 Mar 13, 2021
 #5
avatar+117637 
+1

I remember  this  one from the  other  day.....it  just  wants  the area  between  sector BOA  and  triangle BOA

 

Area of sector  =  pi * 8^2 * (90/360)  =   16 pi

 

Area of triangle  BOA   = (1/2) (8)^2  =  32

 

Purple area =    16 pi -32  =   16  ( pi  - 2)

 

 

cool cool cool

 Mar 13, 2021
 #6
avatar+42 
+1

ty!!!!!!!!!!!!!!!!!!!!!!!!!!!

Valencia  Mar 13, 2021
 #7
avatar+117637 
0

The "Twilight Zone"  has arrived......

 

Varxaax   beat me to my own answer    ......LOL!!!!!!

 

 

 

cool cool cool

 Mar 13, 2021

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