Mathematics Question and Solution
Calculating R, radius of the inscribed circle.
Let a be the radius of the ascribed semi circle.
b = (2a-4) units.
c = (a-4) units.
a² = c²+d²
a² = (a-4)²+d²
d² = a²-(a²-8a+16)
d² = 8a-16
d = √(8a-16) units.
e = ½(a) units.
f = e+a
f = ½(3a) units.
(½(3a))² = (½(a))²+g²
g² = ¼(9a²)-¼(a²)
g = √(¼(8a²))
g = √(2)a units.
It implies;
Observing similar plane shape (right-angled) side length ratios.
4 - ½(a)
√(8a-16) - √(2)a
Cross Multiply.
½√(8a-16) = 4√(2)
√(8a-16) = 8√(2)
8a-16 = 64*2
a-2 = 16
a = 18 units.
Again, a is the radius of the ascribed semi circle.
Recall.
e = ½(a) units
And a = 18 units.
e = ½(18)
e = 9 units.
e is the radius of the inscribed semi circle.
h = (18-R) units.
(18-R)² = R²+j²
324-36R+R² = R²+j²
j = √(324-36R) units.
k = 9+j
k = (9+√(324-36R)) units.
l = (9+R) units.
Therefore, R, radius of the inscribed circle is;
(9+R)² = R²+(9+√(324-36R))²
81+18R+R² = R²+81+18√(324-36R)+324-36R
54R = 18√(324-36R)+324
3R = √(324-36R)+18
(3R-18)² = 324-36R
9R²-108R+324 = 324-36R
9R²-72R = 0
9R = 72
R = 8 units.
Again, R is the radius of the inscribed circle.
