Mathematics Question and Solution
Let a be the side length of the regular hexagon.
3² = 1²+a²-2acosb
2acosb = a²-8
cosb = (a²-8)/(2a) --- (1).
c²+(a²-8)² = (2a)²
c² = 4a²-(a⁴-16a²+64)
c = √(-a⁴+20a²-64) units.
sinb = c/(2a)
sinb = √(-a⁴+20a²-64)/(2a) --- (2).
(2√(3))² = 1²+a²-2acos(240-b)
11 = a²-2a(cos240cosb+sin240sinb)
11 = a²-2a(-½(cosb)-½√(3)sinb)
11 = a²+acosb+a√(3)sinb --- (3).
Substituting (1) and (2) in (3).
11 = a²+a((a²-8)/(2a))+a√(3)(√(-a⁴+20a²-64)/(2a))
11 = a²+½(a²-8)+½√(3)√(-a⁴+20a²-64)
22 = 2a²+a²-8+√(3)√(-a⁴+20a²-64)
30-3a² = √(3)√(-a⁴+20a²-64)
(30-3a²)² = √(3)²(-a⁴+20a²-64)
900-180a²+9a⁴ = -3a⁴+60a²-192
12a⁴-240a²+1092 = 0
2a⁴-40a²+182 = 0
a⁴-20a²+91 = 0
Let a² = p
It implies;
p²-20p+91 = 0
Calculating p.
(p-10)² = -91+(-10)²
p = 10±√(9)
p = 10±3
Therefore;
p = 7 units.
Or
p = 13 units.
Recall.
a² = p
Therefore;
For p = 13
a ≠ √(13)
a ≠ 3.60555127546 units.
For p = 7
a = √(7)
a = 2.64575131106 units.
Again, a is the side length of the regular hexagon.
Calculating area regular hexagon.
sin60 = d/√(7)
d = ½√(21) units.
d is half the height of the regular hexagon.
cos60 = e/√(7)
e = ½√(7) units.
f = 2e+a
f = 2*½√(7)+√(7)
f = 2√(7) units.
Area regular hexagon is;
2(area isosceles trapezoid with parallel lengths √(7) units and 2√(7) units respectively, and height ½√(21) units).
= 2*½(√(7)+2√(7))*½√(21)
= 3√(7)*½√(21)
= ½(21√(3)) square units.
