Mathematics Question and Solution
a = ½(4√(6)-2√(6))
a = √(6) cm.
tan60 = b/√(6)
b = 3√(2).
b is the height of the trapezoid.
c² = (√(6))²+(3√(2))²
c² = 6+18
c = √(24)
c = 2√(6) cm.
Or
cos60 = √(6)/c
c = √(6)/(½)
c = 2√(6) cm.
c is the forward and backward equal slant lengths of the trapezoid.
d = 180-45-60
d = 75°
e = 75+60
e = 135°
tanf = 3√(2)/(3√(6))
f = 30°
g = 180-135-30
g = 15°
(4√(6)/sin15) = (h/sin30)
h = 18.9282032303 cm.
(4√(6)/sin15) = (j/sin135)
j = 26.7685217196 cm.
k = 60-30
k = 30°
It implies;
Area Blue is;
Area triangle with height 4√(6) cm and base 26.768521719sin30 cm + Area triangle with height 2√(6) cm and base 26.768521719sin30cm - Area trapezoid with parallel sides 4√(6) cm and 2√(6) cm respectively and height 3√(2) cm.
= 0.5*4√(6)*26.768521719sin30+0.5*2√(6)*26.768521719sin30-0.5*3√(2)(4√(6)+2√(6))
= 65.5692193816+32.7846096908-31.1769145362
= 67.1769145362 cm²
