A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length `3,4 and 5` units. Then area of the triangleis equal to:

Options

A. `(9sqrt3(sqrt1+sqrt3))/pi^2` sq unit
B. `(9sqrt3(sqrt3-1))/pi^2` sq unit
C. `(9sqrt3(sqrt3+1))/(2pi^2)`
D. `(9sqrt3(sqrt3-1))/(2pi^2)` sq unit

Answer

A

Explanation

Let arc(BC)=3 , arc(CA)=4 , arc(AB)=5
Let r be the radius of the circle , then
`3=ralpha , 4=rbeta , 5 =rgamma ( because "Angle"="arc"/"radius")`
Now , 3+4+5=r `(alpha + beta+ gamma)=r.2pi rArr r =6/pi`
`triangleABC=triangleOBC+triangleOCA+triangleOAB`
`=1/2r^2 sin alpha + 1/2 r^2 sin beta + 1/2 r^2 sin gamma = 1/2r^2 { sin (3/r) + sin (4/r) + sin (5/r) }`
`=1/2xx36/pi^2["sin"pi/2 + "sin" (2pi)/3 + "sin " (5pi)/6]=18/pi^2{1+sqrt3/2+1/2}=(9sqrt3(1+sqrt3))/pi^2` sq unit