Mathematics Question and Solution
Let the side length of the inscribed regular triangle be a.
2b² = a²
b = √(a²/2)
b = ½√(2)a units.
c = 2b
c = √(2)a units.
c is the side length of the ascribed square.
d² = b²+c²
d² = (√(2)a)²+(½√(2)a)²
d² = 2a²+¼(2a²)
d² = ¼(10a²)
d = ½√(10)a units.
tane = √(2a)/(½√(2)a)
e = atan(2)°
f = 180-45-60-atan(2)
f = 11.5650511771°
Calculating a.
0.5*a*0.5√(10)asin11.5650511771 = 24
0.15849364905a² = 24
a² = 151.425625846
a = 12.3055120107 units.
Again, a is the is the side length of the inscribed regular triangle.
Recall.
b = ½√(2)a
And a = 12.3055120107 units.
b = 0.5√(2)*12.3055120107
b = 8.70131098874 units.
g = 180-45-60-60
g = 15°
cos15 = 8.70131098874/h
h = 9.00826000498 units.
Therefore area triangle x is;
0.5*9.00826000498*12.3055120107sin60
= 48 square units.
