Mathematics Question and Solution
a = (x+17) cm.
a is the length of the the ascribed rectangle.
b = a-9
b = (x+17)-9
b = (x+8) cm.
b is the width of the ascribed rectangle and also, the diameter of the inscribed semi circle and the radius of the inscribed quarter circle.
c² = 2(8+x)²
c² = 2(64+16x+x²)
c = √(128+32x+2x²) cm.
d = ½(b)
d = ½(x+8) cm.
d is the radius of the inscribed semi circle.
Therefore, calculating x.
Observing the side length ratios of similar plane shapes (right-angled triangle).
x - ½(x+8)
½(x+8) - 9
Cross Multiply.
9x = ¼(x+8)²
36x = x²+16x+64
x²-20x+64 = 0
Resolving the above quadratic equation via completing the square approach to get x.
(x-10)² = -64+(-10)²
(x-10)² = -64+100
x-10 = ±√(36)
x = 10±6
It implies;
x ≠ 10+6
x = 10-6
x = 4 cm.
b = (x+8)
And x = 4 cm.
b = 4+8
b = 12 cm.
Again, b is the width of the ascribed rectangle and also, the diameter of the inscribed semi circle and the radius of the inscribed quarter circle.
d = ½(x+8)
And x = 4 cm.
d = ½(4+8)
d = 6 cm.
Again, d is the radius of the inscribed semi circle.
tane = 4/6
e = atan(⅔)°
Therefore, shaded area blue is;
Area rectangle with length 21 cm and width 12 cm - 2(area sector with radius 6 cm and angle atan(⅔)°) - 2(area triangle with height 9 cm and base 6 cm) - Area quarter circle with radius 12 cm.
= (12*21)-2(atan(2/3)π*6*6/360)-2(½*6*9)-(¼*12*12π)
= 252-⅕(atan(⅔)π)-54-36π
= 198-36π-⅕(atan(⅔)π)
= ⅕(990-180π-atan(⅔)) cm²
= 63.7345707431 cm²
