Mathematics Question and Solution
Calculating x, the diameter of the semi circle.
Notice, we have a cyclic quadrilateral in the complete circle.
Therefore;
x² =11²+y²
y² = x²-121 --- (1).
cosa = (11/x) --- (2).
y² = 7²+2²+2*2*7cosa --- (3).
Therefore, substituting (2) and (1) in (3) to get x, the diameter of the semi circle.
x²-121 = 7²+2²+2*2*7(11/x)
x²-121 = 53+28(11/x)
x³-121x = 53x+308
x³-174x-308 = 0
Resolving the above polynomial equation to get x, the diameter of the half circle.
x³-14x²+14x²-174x-308 = 0
It implies;
x³-14x²+14x²-196x+22x-308 = 0
Taking pairs of parentheses and factorize.
(x³-14x²)+(14x²-196x)+(22x-308) = 0
x²(x-14)+14x(x-14)+22(x-14) = 0
(x²+14x+22)(x-14) = 0
Therefore;
x-14 = 0
x = 14 units.
Again, x is the diameter of the semi circle.
