Perimeter of a shaded part in a circle












1












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https://cdn.discordapp.com/attachments/334723040099434498/536393147538997250/Screenshot_20190120-055431.jpg This is a question that i want to know will the solution of it be 8 pi or 8 pi + 12 , will the 2 sides of the triangle be counted in the "perimeter of the shaded part" or just the circumference






circle







share|cite|improve this question









$endgroup$












  • $begingroup$
    I personally think its including the 2 sides
    $endgroup$
    – Ahmed Elsayed
    Jan 20 at 5:14
















1












$begingroup$


https://cdn.discordapp.com/attachments/334723040099434498/536393147538997250/Screenshot_20190120-055431.jpg This is a question that i want to know will the solution of it be 8 pi or 8 pi + 12 , will the 2 sides of the triangle be counted in the "perimeter of the shaded part" or just the circumference






circle







share|cite|improve this question









$endgroup$












  • $begingroup$
    I personally think its including the 2 sides
    $endgroup$
    – Ahmed Elsayed
    Jan 20 at 5:14














1












1








1





$begingroup$


https://cdn.discordapp.com/attachments/334723040099434498/536393147538997250/Screenshot_20190120-055431.jpg This is a question that i want to know will the solution of it be 8 pi or 8 pi + 12 , will the 2 sides of the triangle be counted in the "perimeter of the shaded part" or just the circumference






circle







share|cite|improve this question









$endgroup$




https://cdn.discordapp.com/attachments/334723040099434498/536393147538997250/Screenshot_20190120-055431.jpg This is a question that i want to know will the solution of it be 8 pi or 8 pi + 12 , will the 2 sides of the triangle be counted in the "perimeter of the shaded part" or just the circumference






circle




circle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 20 at 5:05









Ahmed ElsayedAhmed Elsayed

82




82












  • $begingroup$
    I personally think its including the 2 sides
    $endgroup$
    – Ahmed Elsayed
    Jan 20 at 5:14


















  • $begingroup$
    I personally think its including the 2 sides
    $endgroup$
    – Ahmed Elsayed
    Jan 20 at 5:14
















$begingroup$
I personally think its including the 2 sides
$endgroup$
– Ahmed Elsayed
Jan 20 at 5:14




$begingroup$
I personally think its including the 2 sides
$endgroup$
– Ahmed Elsayed
Jan 20 at 5:14










1 Answer
1






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Notice that because the area is $36pi$, the radius is 6. Also, $sin angle{ABM}=frac {1}{2}$, $angle{ABM}=30°$. Since $ABM$ is an isosceles triangle, $angle{BAM}=30°$ and $angle{AMB}=120°$. Then the larger arc is $360°-120°=240°$. Applying the circumference of a sector gives $frac{240°}{360°}2pi r=frac{2}{3}*2*pi*6=8pi$. Adding the 2 radii from the cut-off part gives $8pi+12$.






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    $begingroup$

    Notice that because the area is $36pi$, the radius is 6. Also, $sin angle{ABM}=frac {1}{2}$, $angle{ABM}=30°$. Since $ABM$ is an isosceles triangle, $angle{BAM}=30°$ and $angle{AMB}=120°$. Then the larger arc is $360°-120°=240°$. Applying the circumference of a sector gives $frac{240°}{360°}2pi r=frac{2}{3}*2*pi*6=8pi$. Adding the 2 radii from the cut-off part gives $8pi+12$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Notice that because the area is $36pi$, the radius is 6. Also, $sin angle{ABM}=frac {1}{2}$, $angle{ABM}=30°$. Since $ABM$ is an isosceles triangle, $angle{BAM}=30°$ and $angle{AMB}=120°$. Then the larger arc is $360°-120°=240°$. Applying the circumference of a sector gives $frac{240°}{360°}2pi r=frac{2}{3}*2*pi*6=8pi$. Adding the 2 radii from the cut-off part gives $8pi+12$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Notice that because the area is $36pi$, the radius is 6. Also, $sin angle{ABM}=frac {1}{2}$, $angle{ABM}=30°$. Since $ABM$ is an isosceles triangle, $angle{BAM}=30°$ and $angle{AMB}=120°$. Then the larger arc is $360°-120°=240°$. Applying the circumference of a sector gives $frac{240°}{360°}2pi r=frac{2}{3}*2*pi*6=8pi$. Adding the 2 radii from the cut-off part gives $8pi+12$.






        share|cite|improve this answer









        $endgroup$



        Notice that because the area is $36pi$, the radius is 6. Also, $sin angle{ABM}=frac {1}{2}$, $angle{ABM}=30°$. Since $ABM$ is an isosceles triangle, $angle{BAM}=30°$ and $angle{AMB}=120°$. Then the larger arc is $360°-120°=240°$. Applying the circumference of a sector gives $frac{240°}{360°}2pi r=frac{2}{3}*2*pi*6=8pi$. Adding the 2 radii from the cut-off part gives $8pi+12$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 5:14









        BadAtGeometryBadAtGeometry

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