Mathematics Question and Solution
a²+8² = 10²
a = 6 units.
b = ½(10)
b = 5 units.
b is the radius of the ascribed semi circle.
6² = 5²+5²-2*5*5cosc
50cosc = 50-36
cosc = 14/50
c = acos(7/25)°
c = 73.7397952917°
d = 2c
d = 147.4795905834°
e² = 5²+5²-2*5*5cos147.4795905834
e = 9.6 units.
f = ½(e)
f = 4.8 units.
cosg = 4.8/6
g = acos((24/5)/6)
g = acos(⅘)°
g = 36.8698976458°
h = ½(g)
h = ½(acos(⅘))°
h = 18.4349488229°
tan(½(acos(⅘))) = j/(4.8)
j = ⅓*(24/5)
j = ⅕(8) units.
Where j is the radius of the inscribed shaded area (circle).
It implies;
Shaded Inscribed Area Circle is;
π(⅕(8))²
= (64/25)π square units.
= 2.56π square units.
