Mathematics Question and Solution
Let r be the radius of the inscribed circle.
Calculating r.
tan30 = r/(6-r)
√(3)r = 6-r
r = 6/(√(3)+1) cm.
r = 3(√(3)-1) cm.
sin30 = r/a
½ = 3(√(3)-1)/a
a = 6(√(3)-1) cm.
Where a is BO, O is the center of the inscribed circle.
BP = 6(√(3)-1)-r
BP = 6√(3)-6-3√(3)+3
BP = 3√(3)-3
BP = 3(√(3)-1) cm.
cos30 = 6/(BH)
BH = 12/√(3)
BH = 4√(3) cm.
PH = BH-BP
PH = 4√(3)-3√(3)+3
PH = (√(3)+3) cm.
tan30 = (DH)/6
DH = 6/√(3)
DH = 2√(3) cm
HR = 2√(3)-3√(3)+3
HR = (3-√(3)) cm.
(PR)² = (√(3)+3)²+(3-√(3))²-2(3-√(3))(√(3)+3)cos60
PR = 4.24264068712 cm.
(4.24264068712/sin60) = ((3-√(3))/sinb)
b = 15°
Where b is angle HPR.
c = 180-(180-15-15)
c = 180-150
c = 30°
Where c is angle QOR, O is the centre of the inscribed circle.
Therefore shaded region, triangle PQR is;
0.5*4.24264068712*2(3√(3)-3)sin15
= 2.41154273188 cm²
