Name the center of the circle J
Draw from vertex C a perpendicular line to AB ( The drawn line is the height and a diameter of the circle and bisects angle C and side AB)
Find the length of the height
(1)^2+(CD)^2=(2)^2
CD=√3
CJ=√3/2
Because CD is a diameter of the circle CJ is a radius
Name the point of intersection between line AC and the circle E
Name the point of the intersection between line BC and the circle F
Connect from the center J to point E and point F
We form the triangle EJC and triangle FJC
Because JE and JC are both radii of the circle then this triangle is an isosceles and the area of this triangle is (0.5×√3/2×√3/2×sin(120))
Because JF and JC are radii the triangle is isosceles and the area of this triangle is (0.5×√3/2×√3/2×sin(120))
To find the shaded area we need to find the area of the sector EJC then subtract the area of the triangle EJC then multiply the answer by 2 (the 2 shaded regions are equal in area)
(120/360×π×(√3/2)^2)-(0.5×√3/2×√3/2×sin 120)
Multiply the result by 2
Area of the shaded region is approximately 0.92
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h=2R=2√3/2=√3,R=(√3/2) S1=(1/2)(2Π/3)(3/4)=(1/4)*Π S2=(1/2)(3/4)*√3/2=3√3/16 S=2(Π/4-3√3/16)=Π/2-3√3/8= 0.92units^2🍀❤🌹🌷🌼🌼🌺
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