Mathematics Question and Solution
Let the width of the green inscribed rectangle be a.
Let the length of the green inscribed rectangle be b.
c = (5-a) cm.
d = (5-b) cm.
It implies.
(5-a)(5-b) = 2*5
(5-a)(5-b) = 10 --- (1).
5² = e²+(5-a)²
e² = 25-(25-10a+a²)
e = √(10a-a²) cm.
e+b = 5
b = 5-√(10a-a²) --- (2).
Substituting (2) in (1).
(5-a)(5-(5-√(10a-a²))) = 10
(5-a)√(10a-a²) = 10
(5-a)²(10a-a²) = 100
a = 0.5279 cm.
Again, a is the width of the green inscribed rectangle.
Recall.
(5-a)(5-b) = 10 --- (1).
And a = 0.5279 cm.
Calculating b.
(5-0.5279)(5-b) = 10
5-b = 10/(5-0.5279)
5-b = 2.2360859551
b = 5-2.2360859551
b = 2.7639140449 cm.
Again, b is the length of the green inscribed rectangle.
Area Green Rectangle is;
ab
= 0.5279*2.7639140449
= 1.4590702243 cm²
