Mixing
Perimeter of a shaded part in a circle
$begingroup$
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
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
$endgroup$
add a comment |
$begingroup$
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
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
$endgroup$
$begingroup$
I personally think its including the 2 sides
$endgroup$
– Ahmed Elsayed
Jan 20 at 5:14
add a comment |
$begingroup$
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
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
$endgroup$
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
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
asked Jan 20 at 5:05
Ahmed ElsayedAhmed Elsayed
82
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $8pi+12$.
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $8pi+12$.
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
$endgroup$
add a comment |
$begingroup$
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 $8pi+12$.
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
$endgroup$
add a comment |
$begingroup$
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 $8pi+12$.
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
$endgroup$
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 $8pi+12$.
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
answered Jan 20 at 5:14
BadAtGeometryBadAtGeometry
188215
188215
add a comment |
add a comment |
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$begingroup$
I personally think its including the 2 sides
$endgroup$
– Ahmed Elsayed
Jan 20 at 5:14