Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
2² = 2(3²)-2(3²)cosa
18cosa = 18-4
a = acos(7/9)°
a is angle ABC.
Let b be the radius of the inscribed circle.
Calculating b.
c²+1² = 3²
c = √(8)
c = 2√(2) units.
2b+3b+3b = 2*2√(2)
8b = 4√(2)
b = ½√(2) units.
Again b is the radius of the inscribed circle.
d = 2√(2)-½√(2)
d = ½(3√(2)) units.
e²+(0.5√(2))² = (0.5*3√(2))²
e² = ½(9)-½
e = √(4)
e = 2 units.
tanf = 2/(0.5√(2))
f = atan(2√(2))°
Therefore, area T exactly in fraction is;
2(area triangle with height 2 units and base 0.5√(2) units - area sector with radius 0.5√(2) units and angle atan(2√(2))°
= 2((½*2*½*√(2))-(atan(2√(2))*0.5√(2)*0.5√(2)π/360))
= √(2)-π(atan(2√(2))/360
= (360√(2)-πatan(2√(2)))/360 square units.
= 0.7987338537 square units.
