Mathematics Question and Solution
Calculating Area triangle ABC.
Let a be the radius of the inscribed circle.
b² = 2a²
b = √(2)a units.
b is EG.
c² = a²+(2+a)²
c² = a²+4+4a+a²
c = √(2a²+4a+4) units.
c is CE.
Therefore;
c² = b²+2²-2*2*bcos135
√(2a²+4a+4)² = (√(2)a)²+4-4*√(2)acos135
2a²+4a+4 = 2a²+4+4
4a = 4
a = 1 unit.
a is the radius of the inscribed circle.
AC = CG+AG
AC = 2+1
AC = 3 units.
Notice!
CG = CH = 2 units.
And a (circle's radius) = DH = DG = AG = DE = AE = 1 unit.
tanb = (DH/CH)
tanb = (1/2)
b = atan(½)°
b is angle DCH = angle DCG.
c = 2b = angle DCH + angle DCG
c = 2atan(½)°
c is angle ACB.
It implies;
tan(2atan(½)) = d/3
(4/3) = d/3
d = 4 units.
d is AB.
It implies, Area Triangle ABC is;
½*(AB)(AC)
= ½*4*3
= 6 square units.
