Mathematics Question and Solution
Let BF be a.
b² = 18²+a²
b = √(324+a²) units.
b is AB.
c = (a+14) units.
It implies;
calculating a.
(a+14) ~ √(324+a²)
√(324+a²) ~ 18
Cross Multiply.
18(a+14) = 324+a²
18a+252 = 324+a²
a²-18a+72 = 0
a²-12a-6a+72 = 0
a(a-12)-6(a-12) = 0
(a-12)(a-6) = 0
It implies;
a ≠ 12 units.
a = 6 units.
Again, a is BF.
Recall.
b = √(324+a²)
And a = 6 units.
b = √(324+36)
b = √(360)
b = 6√(10) units
Again, b is AB.
Recall Again.
c = (a+14)
And a = 6 units.
c = 6+14
c = 20 units
Again, c is BH.
It implies;
(CH)² = 20²-(6√(10))²
CH = √(400-360)
CH = √(40)
CH = 2√(10) units.
Let d be the radius of the inscribed circle.
Calculating d.
20d+6√(10)d+2√(10)d = 2√(10)*6√(10)
(20+8√(10))d = 120
d = 120/(20+8√(10))
d = 2.64911064067 units.
Therefore area inscribed circle is;
πd²
= π(120/(20+8√(10)))²
= π(260-80√(10))
= 20π(13-4√(10)) square units.
= 22.0470286697 square units.
