Mathematics Question and Solution
Let the radius of the circle be a.
sinb = a/(a+a)
b = 30°
cos30 = c/2a
c = √(3)a units.
d = 2-c
d = (2-√(3)a) units.
sin30 = e/d
e =½(2-√(3)a) units.
cos30 = f/d
½√(3) = f/(2-√(3)a)
f = ½(2√(3)-3a) units.
g = a+a+a
g = 3a units.
h = f+g
h = ½(2√(3)-3a)+3a
h = ½(2√(3)+3a) units.
h is the diameter of the half circle.
j = ½h
j = ¼(2√(3)+3a) units.
j is the radius of the half circle.
k = a+j
k = a+¼(2√(3)+3a)
k = ¼(2√(3)+7a) units.
l = j-a
l = ¼(2√(3)+3a)-a
l = ¼(2√(3)-a) units.
Therefore;
k² = l²+e²
(¼(2√(3)+7a))² = (¼(2√(3)-a))²+(½(2-√(3)a))²
(12+28√(3)a+49a²)/16 = (12-4√(3)a+a²)/16+¼(4-4√(3)a+3a²)
12+28√(3)a+49a² = 12-4√(3)a+a²+16-16√(3)a+12a²
36a²+48√(3)a-16 = 0
9a²+12√(3)a-4 = 0
a²+⅓(4√(3))a = ⅑(4)
(a+⅓(2√(3)))² = ⅑(4)+(⅓*2√(3))²
(a+⅓(2√(3)))² = ⅑(4)+⅑(12)
(a+⅓(2√(3)))² = ⅑(16)
a = -⅓(2√(3))±√(⅑(16))
a = -⅓(2√(3))±⅓(4)
It implies;
a = ⅓(4-2√(3)) units.
a = 0.17863279495 units.
Again, a is the radius of the circle.
Recall.
j = ¼(2√(3)+3a)
And a = ⅓(4-2√(3))
j = ¼(2√(3)+2(⅓(4-2√(3))))
j = ¼(2√(3)+4-2√(3))
j = ¼(4)
j = 1 unit.
Again, j is the radius of the half circle.
It implies, area half circle is;
½(πj²)
= ½(π*1²)
= ½(π) square units.
