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# HALP

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

Mar 13, 2021
#3
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oh ok how do i do it i its my fist time on here?

Valencia  Mar 13, 2021
#2
0 \([asy] size(4cm); pair o=(0,0); pair a=(-sqrt(2)/2,-sqrt(2)/2); pair b=(sqrt(2)/2,-sqrt(2)/2); path p=Arc(o,1,225,315)--b--a--cycle; fill(p,purple); dot(o); dot(a); dot(b); draw(o--b--a--o); draw(a/8--(a+b)/8--b/8); draw(Circle(o,1)); label("\$O\$",o,N); label("\$A\$",a,SW); label("\$B\$",b,SE); [/asy]\)

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Mar 13, 2021
edited by Valencia  Mar 13, 2021
edited by Valencia  Mar 13, 2021
#4
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This is a mess, I'm adding onto this mess

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

Mar 13, 2021
#5
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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)   Mar 13, 2021
#6
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ty!!!!!!!!!!!!!!!!!!!!!!!!!!!

Valencia  Mar 13, 2021
#7
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The "Twilight Zone"  has arrived......

Varxaax   beat me to my own answer    ......LOL!!!!!!   Mar 13, 2021