Mathematics Question and Solution
Let a be the radius of the ascribed semi circle.
b = a-½(7)
b = (a-3.5) units.
15² = b²+c²
c² = 225+(a-3.5)²
c² = 225-(a²-7a+12.25)
c = √(225-a²+7a-12.25)
c = √(212.75+7a-a²) units.
Therefore;
a² = 3.5²+c²
a² = 12.25+212.75+7a-a²
2a² = 225+7a
2a²-7a-225 = 0
Resolving the above quadratic equation via completing the square approach to get a, radius of the ascribed semi circle.
(a-(7/4))² = (225/2)+(-7/4)²
(a-(7/4))² = (1800+49)/16
a = (7/4)±√(1849/16)
a = ¼(7±43)
It implies;
a ≠ ¼(7-43)
a = ¼(7+43)
a = ¼(50) units.
a = ½(25) units.
a = 12.5 units.
Again, a is the radius of the ascribed semi circle.
Therefore;
d = a+½(7)
d = 12.5+3.5
d = 16 units.
c = √(212.75+7a-a²)
And a is 12.5 units.
c = √(212.75+7(12.5)-12.5²)
c = 12 units.
c is the height of the inscribed green trapezoid.
Therefore, area green trapezoid is;
½(CD+AE)*CE
= ½(7+16)*12
= 23*6
= 138 square units.
