Mathematics Question and Solution
Let a be the radius of the semi circle.
b²+24² = (2a)²
b = √(4a²-576) units.
b is BC.
c = b-BP
c = (√(4a²-576)-7) units.
c is CP.
It implies;
Calculating a.
2a - (√(4a²-576)-7)
√(4a²-576) - a
Cross Multiply.
2a² = √(4a²-576)*(√(4a²-576)-7)
2a² = 4a²-576-7√(4a²-576)
7√(4a²-576) = 2a²-576
49(4a²-576) = (2a²-576)²
196a²-28224 = 4a⁴-2304a²+331776
4a⁴-2500a²+360000 = 0
Therefore;
a = 20 units.
Again, a is the radius of the half circle.
Recall.
b = √(4a²-576) units.
And a = 20 units.
b = √(4*20²-576)
b = 32 units.
Again, b is BC.
d = BC-BP
d = 32-7
d = 25 units.
Again, d = c = CP.
Therefore, x = OP, the required length is;
25² = x²+20²
x² = 625-400
x = √(225)
x = 15 units.
Again, x = OP, is the required length.
