Mathematics Question and Solution
2πr = 13π
r = ½(13) cm.
r is the radius of the ascribed large circle.
a = 2r
a = 2*½(13)
a = 13 cm.
a is the diameter of the ascribed large circle.
It implies;
c = a-3b
c = (13-3b) cm.
c is the side length of the inscribed square.
b is the radius of the small inscribed circle.
d = c+b
d = (13-3b)+b
d = (13-2b) cm.
Calculating b.
13² = (13-2b)²+(13-3b)²
169 = 169-52b+4b²+169-78b+9b²
0 = 169-130b+13b²
b²-10b+13 = 0
(b-5)² = -13+(-5)²
(b-5)² = 25-13
b = 5±√(12)
b ≠ (5+2√(3)) cm.
b = (5-2√(3)) cm.
b = 1.53589838486 cm.
Recall.
c = (13-3b) cm.
Where c is the side length of the inscribed square.
And b = (5-2√(3)) cm.
Therefore;
c = 13-3(5-2√(3))
c = (6√(3)-2) cm.
c = 8.39230484541 cm.
Therefore, the required length x is;
x = ½(c)
x = ½(6√(3)-2)
x = (3√(3)-1) cm.
x = 4.19615242271 cm.
