Mathematics Question and Solution
½*a*b = 10
a = (20/b) cm.
a is CD.
b is OD.
b² - 30
(20/b)² - 10
Cross Multiply.
10b² = 12000/b²
10b⁴ = 12000
b⁴ = 1200
b⁴ = 1200
b² = √(1200)
b² = 20√(3)
b = √(20√(3))
b = 5.8856619128 cm.
Again, b is OD.
Recall.
a = (20/b)
And b = 5.8856619128 cm.
a = 20/5.8856619128
a = 3.3980884897 cm.
Again, a is CD.
Therefore, the required red angle is;
Let it be c.
tanc = b/a
c = atan(5.8856619128/3.3980884897)
c = atan(1.7320508076)
c = atan(√(3))°
c = 60°
Again, c is the required red angle.
Calculating area BCFB.
sin60 = b/d
d = 5.8856619128/sin60
d = 6.7961769794 cm.
d is OC.
e = 0.5*6.7961769794*5.8856619128sin(60+90)
e = 10 cm²
e is area triangle OCF.
f² = 6.7961769794²+5.8856619128²-2*6.7961769794*5.8856619128cos(60+90)
f = 12.2519822882 cm.
f is the radius of the quarter circle.
sing = a/f
g = asin(3.3980884897/12.2519822882)
g = 16.102113752°
g is angle AOF.
h = 90-g
h = 73.897886248°
h is angle BOF.
Area BCFB is;
Area sector with radius 12.2519822882 units and angle 73.897886248° - 40 cm² - 10 cm²
= (73.897886248π*12.2519822882²/360)-50
= 96.8037337915-50
= 46.8037337915 cm²
