Mathematics Question and Solution
Let a be the radius of the inscribed pink circle.
b = (a+7) units.
c = (7-a) units.
d = (a-2) units.
Calculating a.
It implies;
a ~ (a+7)
(a-2) ~ (7-a)
Cross Multiply.
a(7-a) = (a-2)(a+7)
7a-a² = a²+7a-2a-14
2a²-2a-14 = 0
a²-a-7 = 0
(a-½)² = (7)+(-½)²
(a-½)² = ¼(29)
a = ½±√(¼(29))
a = ½±½√(29)
It implies;
a ≠ ½(1-√(29))
a = ½(1+√(29))
a = 3.19258240357 units.
Again, a is the radius of the inscribed pink circle.
Recall.
b = (a+7) units.
And a = 3.19258240357 units.
b = 3.19258240357+7
b = 10.19258240357 units.
c = (7-a) units.
And a = 3.19258240357 units.
c = 7-3.19258240357
c = 3.80741759643 units.
d = (a-2) units.
And a = 3.19258240357 units.
d = 3.19258240357-2
d = 1.19258240357 units.
Therefore;
e² = a²-d²
e = √(3.19258240357²-1.19258240357²)
e = 2.96147422989 units.
f² = b²-c²
f = √(10.19258240357-3.80741759643)
f = 9.45475051495 units.
Therefore, the required length, length of the rectangle is, let it be g.
g = 2+e+f+7
g = 2+2.96147422989+9.45475051495+7
g = 21.4162247448 units.
