Mathematics Question and Solution
Calculating Area Quadrilateral ABCD.
Notice.
A right-angled triangle is derived considering quadrilateral ABCD with interior angles 40°, 50° and 90°.
AD = 1 unit.
BC = 3 units.
Let AB = CD = x.
sin40 = a/3
a = 3sin40 units.
b = (3sin40-x) units.
sin50 = c/3
c = 3sin50 units.
d = (3sin50-x) units.
Calculating x.
(3sin50-x)²+(3sin40-x)² = 1²
9sin²50-6xsin50+x²+9sin²40-6xsin40+x² = 1
2x²+9-8.45299231683x = 1
2x²-8.45299231683x+8 = 0
x²-4.22649615842x+4 = 0
It implies;
x ≠ 2.79576 units.
x = 1.43074 units.
Again, x is AB = CD.
Recall.
b = (3sin40-x) units.
And x = 1.43074 units.
b = 3sin40-1.43074
b = 0.49762282906 units.
d = (3sin50-x) units.
And x = 1.43074 units.
d = 3sin50-1.43074
d = 0.86739332936 units.
Therefore, Area Quadrilateral ABCD is;
(½*a*c) - (½*b*d)
= (0.5*3sin40*3sin50)-(0.5*0.49762282906*0.86739332936)
= 2.21581744428-0.21581736123
= 2 square units.
