Mathematics Question and Solution
6² = a²+(2+3)²
a² = 36-25
a = √(11) units.
b² = 3²+c²
b = √(9+c²) units.
b is the diameter of the circle.
d = c-a
d = (c-√(11)) units.
e² = 2²+(c-√(11))²
e = √(4+c²-2c√(11)+11)
e = √(c²-2c√(11)+15) units.
Therefore;
√(9+c²)² = 6²+√(c²-2c√(11)+15)²
9+c² = 36+c²-2c√(11)+15
2√(11)c = 51-9
2√(11)c = 42
√(11)c = 21
c = (21√(11))/11 units.
c = 6.3317382361 units.
Notice!
PQ = e = √(c²-2c√(11)+15)
PQ = √((6.3317382361)²-2(6.3317382361)√(11)+15)
PQ = 3.6181361349
Or
d = (c-√(11))
And c = 6.3317382361 units.
d = 6.3317382361-√(11)
d = 3.0151134457 units.
It implies, length PQ is;
(PQ)² = 2²+3.0151134457²
PQ = √(13.0909090904)
PQ = 3.6181361349 units.
