Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
sin30 = a/12
a = 6 cm.
sin60 = b/12
b = 6√(3) cm.
Let the inscribed square side be c.
Calculating c.
(c/sin60)+(csin30) = 6
(⅓(2√(3))+½)c = 6
⅙(4√(3)+3)c = 6
c = 36/(4√(3)+3) cm.
(a) Area Blue exactly is;
c²
= (36/(4√(3)+3))²
= 1296/(4√(3)+3)²
= 1296/(48+24√(3)+9)
= 1296/(57+24√(3))
= 432/(19+8√(3))
= 432(19-8√(3))/169 cm².
(b) Calculating Shaded area to 2 decimal places.
Notice;
Blue square side is;
3.62603375102 cm.
tan60 = c/d
d = 12(4-√(3))/13 cm.
d = 2.09349156224 cm.
e = 6-d
e = 6(5+2√(3))/13 cm.
e = 3.90650843776 cm.
(6/sin45) = (3.90650843776/sinf)
f = 27.41204622623°
g = 180-45-f
g = 107.58795377377°
sin30 = h/3.62603375102
h = 1.81301687551 cm.
cos30 = h/3.62603375102
h = 3.14023734336 cm.
Therefore;
Area Shaded to 2 decimal places is;
(107.58795377377)π/10 - 3*3.90650843776sin107.58795377377 - 0.5(3.62603375102)² - 0.5*3.62603375102*2.09349156224 - 0.5*1.81301687551*3.14023734336 - ⅙(36π) + ½(36)sin60
= 33.79975251904 - 11.17168694814 - 6.57406038177 - 3.79553553108 - 2.84665164831 - 18.84955592154 + 15.58845726812
= 6.15071935632 cm²
≈ 6.15 cm²
