Question

HELP!!!

Find the area of the portion of the circle that is outside triangle RST if the radius of the circle is 8 cm

Answer 1

heres the diagram

Answer 2

@denizen .. just think a little over it... to find.. the angles.. in the center... first....

Answer 3

the angles are 60, 70 and 50 right

Answer 4

first ol all... tell me the way u want to solve it...

Answer 5

cuz.. it appears.. that we are thinking over it in different ways..

Answer 6

@surjithayer .. what are u typing bro...? :p

Answer 7

i am thinking getting the area of the circle minus the area of the triangle

Answer 8

@denizen take a look on the following links
im sure that they are gonna help u alot in solving the problem
http://www.mathsisfun.com/geometry/circle-sector-segment.html
http://mathschallenge.net/library/geometry/circle_properties

Answer 9

enjoy the question...!.. just give it a little time.. n am sure that u would get the answer.. n its... not that damn tough as it looks like..

Answer 10

\[\cos 120=\frac{ 8^{2}+8^{2}-(ST)^{2} }{ 2*8*8 }\]
\[\cos \left( 180-60 \right)*128=128-(ST)^{2}\]
\[(-\cos 60)*128-128=-(ST)^{2}\]
\[-\frac{ 1 }{ 2 }*128-128=-(ST)^{2},(ST)^{2}=192=3*64\]
\[ST=8\sqrt{3}\]
Use sine formula to find RT
\[\frac{ RT }{ \sin 70 }=\frac{ ST }{ \sin 60 }\]
\[RT=\frac{ ST }{ \frac{ \sqrt{3} }{ 2 } }\sin 70=8\sqrt{3}*\frac{ 2 }{ \sqrt{3} }\sin 70=16 \sin 70\]=15.04
\[\frac{ RS }{ \sin 50 }=\frac{ ST }{ \sin 60 }=\frac{ 8\sqrt{3} }{ \frac{ \sqrt{3} }{ 2 } }=16\]
Rs=16 sin50=12.26
\[\frac{ SP }{ RS }=\sin 60=\frac{ \sqrt{3} }{ 2 },SP=\frac{ \sqrt{3} }{ 2 }*RS=\frac{ \sqrt{3} }{ 2 }*12.26=\sqrt{3}*6.13\]
\[\sqrt{3}*6.13=10.62\]
SP=10.62
Area of triangle\[=\frac{ 1 }{ 2 }*RT*SP=\frac{ 1 }{ 2 }*15.04*10.62=79.86\]
\[area of circle=\pi r ^{2}=\pi 8^{2}=64 \pi \]
required area\[=64\pi-79.86=201.06-79.86=121.2 sq. cm\]