In the diagram, if the area of the isosceles triangle AOB is 77.52 sq in and the area of sector OACB is 125.45 sq in, what is the area of the segment ACB?
To find the area of the segment ACB, we need to subtract the area of the isosceles triangle AOB and the area of sector OACB from the area of the circle OACB.
Let's denote the area of the circle OACB as A, the area of the isosceles triangle AOB as B, and the area of sector OACB as C.
We know that A = πr^2, where r is the radius of the circle.
We also know that B = 77.52 and C = 125.45.
Since the angle of the sector, in radians, is equal to the central angle of the sector, we can find the central angle θ using the formula C = (θ/2) * r^2. Rearranging this equation, we get θ = (2C)/r^2.
Substituting the values 125.45 for C and r for r in this equation, we can find θ.
θ = (2*125.45)/r^2 = 250.9/r^2
Now we can use θ to find the area of the segment ACB.
The area of the segment ACB = (θ/2) * r^2 - B
Substituting the values 250.9/r^2 for θ and 77.52 for B in this equation, we can find the area of the segment ACB.
Area of segment ACB = (250.9/r^2)*(r^2) - 77.52 = 250.9 - 77.52 = 173.38 sq in.
So, the area of the segment ACB is 173.38 sq in.