Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
tan30 = a/3
a = √(3) units.
b = a+3
b = (3+√(3)) units.
tan30 = c/b
⅓√(3) = c/(3+√(3))
c = ⅓(3√(3)+3) units.
c = (1+√(3)) units.
It implies;
(1+√(3)) ~ 3
d ~ √(3)
Cross Multiply.
3d =(√(3)+3)
d = ⅓(√(3)+3) units.
Let e be the side length of the smallest inscribed square.
f = d-e
f = (⅓(√(3)+3)-e) units.
Calculating e.
e ~ 3
(⅓(√(3)+3)-e) ~ √(3)
Cross Multiply.
√(3)e = √(3)+3-3e
(3+√(3))e = (3+√(3))
e = 1 unit.
f² = e²+c²
f² = 1²+(√(3)+1)²
f² = 1+3+2√(3)+1
f² = 5+2√(3)
f = √(5+2√(3)) units.
It implies;
√(5+2√(3)) ~ (√(3)+1)
1 ~ g
Cross Multiply.
g = (√(3)+1)/√(5+2√(3)) units.
Again.
1 ~ (√(3)+1)/√(5+2√(3))
(√(3)+1) ~ h
Cross Multiply.
h = (√(3)+1)²/√(5+2√(3))
h = (4+2√(3))/√(5+2√(3)) units.
Therefore, area orange as a single fraction is;
½(gh)
= ½((√(3)+1)/√(5+2√(3)))*(4+2√(3))/√(5+2√(3))
= ½(4√(3)+6+4+2√(3))/(5+2√(3))
= ½(10+6√(3))/(5+2√(3))
= ½((10√(3)+14)/13)
= (5√(3)+7)/13 square units.
= 1.20463492599 square units.
