Mathematics Question and Solution
Calculating the chord length.
Let x be the radius of the half circle.
a = (x-4) units.
b = (x-9) units.
c = (x-7) units.
d = a+b
d = (x-4)+(x-9)
d = (2x-13) units.
d is the length of the chord.
e = ½(d) units.
e = ½(2x-13) units.
f = x-4-½(2x-13)
f = ½(2x-8-2x+13)
f = ½(5) units.
It implies;
7² = g²+(½(2x-13))²
g² = 7²-¼(4x²-52x+169)
g² = ¼(196-4x²+52x-169)
g² = ¼(52x+27-4x²)
g = ½√(52x+27-4x²) units.
Calculating x.
(½(5))²+(½√(52x+27-4x²))² = (x-7)²
¼(25)+¼(52x+27-4x²) = x²-14x+49
25+52x+27-4x² = 4x²-56x+196
8x²-108x+144 = 0
2x²-27x+36 = 0
2x²-24x-3x+36 = 0
2x(x-12)-3(x-12) = 0
(2x-3)(x-12) = 0
It implies;
x ≠ 3/2 units.
x = 12 units.
Again, x is the radius of the half circle.
Calculating the chord length.
Recall.
d = (2x-13) units.
d is the length of the chord.
And x = 12 units.
Therefore;
d = 2(12)-13
d = 24-13
d = 11 units.
Again, d is the length of the chord.
