Mathematics Question and Solution
Let a be the side length of the square and the diameter of the biggest inscribed semi circle.
Let b = diameter of the bigger inscribed semi circle.
Let c = diameter of the big inscribed semi circle.
It implies;
(½(a+c))² = (½(a))²+(8+(½(c)))²
¼(a²+2ac+c²) = ¼(a²)+64+8c+¼(c²)
½(ac) = 64+8c
ac = 128+16c --- (1).
(½(b+c))² = (½(c))²+(6+(½(b)))²
¼(b²+2bc+c²) = ¼(c²)+36+6b+¼(b²)
½(bc) = 36+6b
bc = 72+12b --- (2).
8+c = 6+b --- (3).
a = 8+c --- (4).
Substituting (4) in (1) to get c.
(8+c)c = 128+16c
c²+8c = 128+16c
c²-8c-128 = 0
Resolving the above quadratic equation via completing the square approach to get c.
(c-4)² = 128+(-4)²
(c-4)² = 144
c-4 = ±√(144)
c = 4±12
It implies;
c ≠ 4-12
c = 4+12
c = 16 cm.
Again, c is the diameter of the big inscribed semi circle.
At (4).
a = 8+c
And c is 16 cm.
a = 8+16
a = 24 cm.
Again, a is the side length of the square and the diameter of the biggest inscribed semi circle.
Therefore;
b = 24-6
b = 18 cm.
Again, b is the diameter of the bigger inscribed semi circle.
Calculating shaded area blue.
It is;
Area square with side length 24 cm - Area half circle with radius 12 cm - Area half circle with radius 8 cm - Area half circle with radius 9 cm.
= (24*24)-½(12²)π-½(8²)π-½(9²)π
= 576-72π-32π-½(81π)
= ½(1152-144π-64π-81π)
= ½(1152-289π) cm²
= 122.0398615563 cm²
