Mathematics Question and Solution
a = 4+6
a = 20 cm.
a is the radius of the quarter circle.
tanb = 10/4
b = atan(5/2)°
c = b+60
c = (60+atan(5/2))°
c = 128.1985905136°
(10/sin128.1985905136) = (4/sind)
d = 18.3215084855°
e = 180-d-c
e = 180-18.3215084855-128.1985905136
e = 33.4799010009°
(10/sin128.1985905136) = (f/sin33.4799010009)
f = 7.0195191472 cm.
g = 180-atan(5/2)-60
g = (120-atan(5/2))°
g = 51.8014094864°
Therefore, are blue is;
Area quarter circle with radius 10 cm - Area triangle with height 10 cm and base 4 cm - Area triangle with height 6sin51.8014094864 cm and base 7.0195191472 cm - Area sector with radius 10 cm and angle 33.4799010009° + Area triangle with height 10cm and base 10sin33.4799010009 cm.
¼(10*10)π-½(10*4)-½(6*7.0195191472sin51.8014094864)-(33.4799010009π*10*10÷360)+½(10*10sin33.4799010009)
= 78.5398163397-20-16.5493328837-29.2167252854+27.5822214728
= 40.3559796434 cm²
Or
h = 90-33.4799010009
h = 56.5200989991°
j² = 10²+4²
j = 10.7703296143 cm.
Area blue is;
(0.5*10.7703296143*7.0195191472sin60)+(56.5200989991π*100/360)-(0.5*100sin56.5200989991).
= 32.7368579283+49.3230910544-41.7039693389
= 40.3559796438 cm²
