Mathematics Question and Solution
Let the inscribed bigger quarter circle radius be a.
Let the inscribed big quarter circle radius be b.
c = 5 cm.
c is the radius of the inscribed semi circle.
It implies;
a+b = 10 --- (1).
a+5 =√(10²+5²)
a = √(125)-5
a = 5(√(5)-1) cm.
Again, a is the inscribed bigger quarter circle radius.
And;
a+b = 10
b = 10-a
b = 10-5(√(5)-1)
b = (15-5√(5)) cm.
Again, b is the inscribed big quarter circle radius.
Area Blue is;
= (10*10)-½(5²)π-¼(5√(5)-5)²π-¼(15-5√(5))²π
= 100-½(25π)-¼(125-50√(5)+25)π-¼(225-150√(5)+125)π
= 100-½(25π)-¼(150-50√(5))π-¼(350-150√(5))π
= ¼(400-50π-150π+50√(5)π-350π+150√(5)π)
= ¼(400-550π+200√(5)π)
= ½(200-275π+100√(5)π)
= 19.2717466834 cm²
