Mathematics Question and Solution
Notice.
The ascribed quadrilateral is regular (square).
Let a be it's side length.
½*ab = 3
ab = 6
b = (6/a) units.
b is the base of the red area.
½*ac = 4
ac = 8
c = (8/a) units.
c is the base of the green area
d = a-(6/a)
d = (a²-6)/a units.
d is the height of the yellow area.
e = a-(8/a)
e = (a²-8)/a units.
e is the base of the yellow area.
Therefore;
Calculating a.
½*d*e = 5
½*(a²-6)/a*(a²-8)/a = 5
((a²-6)(a²-8))/a² = 10
(a⁴-14a²+48) = 10a²
a⁴-24a²+48 = 0
(a²-12)² =(-12)²-48
(a²-12)² = 144-48
a² = 12±√(96)
a² = 12±4√(6)
It implies;
a ≠ √(12-4√(6)) units.
a = √(12+4√(6)) units.
a = 4.66882843668 units.
Again, a is the side length of the ascribed square.
Therefore;
The required area is;
a²-4-3-5
= 4.66882843668²-12
= 9.79795897115 square units.
Or
√(12+4√(6))²-12
= 12+4√(6)-12
= 4√(6) square units.
