Mathematics Question and Solution
Let a be the radius of the inscribed circle.
b = ½(2a+1+4)
b = ½(2a+5) units.
b is the radius of the ascribed semi circle.
c = b-4
c = ½(2a+5)-4
c = ½(2a-3) units.
d = a-c
d = a-½(2a-3)
d = (3/2) units.
e = b-a
e = ½(2a+5)-a
e = (5/2) units.
It implies;
Calculating a.
a²+(3/2)² = (5/2)²
a² = (25/4)-(9/4)
a² = 16/4
a= 2 units.
Again, a is the radius of the inscribed circle.
Recall.
b = ½(2a+5) units.
And a = 2 units.
b = ½(4+5)
b = ½(9) units.
Again, b is the radius of the ascribed semi circle.
Therefore, area red is;
Area semi circle with radius ½(9) units - Are circle with radius 2 units.
= ½(½(9))²π - 2²π
= ⅛(81π)-4π
= ⅛(81π-32π)
= ⅛(49π) square units.
= 19.2422550032 square units.
