Mathematics Question and Solution
Let a be the radius of the inscribed half circle.
b²+8² = (2a)²
b = √(4a²-64) units.
It implies;
Calculating a.
2a ~ 8
a ~ √(4a²-64)
Cross Multiply.
8a = 2a√(4a²-64)
4 = √(4a²-64)
16 = 4a²-64
4a² = 80
a² = 20
a = 2√(5) units.
Again, a is the radius of the inscribed half circle.
tanc = a/2a
c = atan(½)°
cosc = d/8
cos(atan(½)) = d/8
d = 7.155417528 units.
d is the base of triangle APB.
Notice.
a, radius of the inscribed half circle is the height of triangle APB.
Therefore, area triangular APB is,
½(ad)
= 0.5*2√(5)*7.155417528
= 16 square units.
