Mathematics Question and Solution
20² = 12²+a²
a² = 400-144
a = √(256)
a = 16 cm.
b = 16-9
b = 7 cm.
c = atan(3/4)°
Let Alpha be d.
d = ½(c)
d = 0.5atan(¾)°
tan(0.5atan(¾)) = e/16
e = ⅓(16) cm.
tan(0.5atan(¾)) = f/7
f = ⅓(7) cm.
g = 12-⅓(16)
g = ⅓(36-16)
g = ⅓(20) cm.
cos(0.5atan(¾)) = 7/h
h = 7.37864787373 cm.
cos(0.5atan(¾)) = 16/j
j = 16.86548085423 cm.
k = 16.86548085423-7.37864787373
k = 9.4868329805 cm.
l = ½(k)
l = 4.74341649025 units.
¾ = m/7
m = ¼(21) cm.
n = ¼(21)-⅓(7)
n = 2.91666666667 cm.
o² = 2.91666666667²+4.74341649025²-2*2.91666666667*4.74341649025cos(90-0.5atan(¾))
o = 4.71772661824 cm.
(4.71772661824/sin(90-0.5atan(¾)) = (4.74341649025/sinp)
p = 72.52522574369°
q = 180-72.52522574369-(90-0.5atan(3/4))
q = 35.90972307923°
r² = 4.74341649025²+6.66666666667²-2*4.74341649025*6.66666666667cos(90+0.5atan(3/4))
r = 9.32440048713 cm.
(9.32440048713/sin(90+0.5atan(¾))) = (6.66666666667/sins)
s = 42.7093899574°
t = 42.7093899574+35.90972307923
t = 78.61911303663°
u = 180-78.61911303663-72.52522574369
u = 28.85566121968°
(v/sin28.85566121968) = (9.32440048713/sin72.52522574369)
v = 4.71772661823 cm.
Or
(v/sin71.56505117708) = (4.74341649025/sin72.52522574369)
v = 4.71772661824 cm.
tan28.85566121968 = w/12
w = 6.61224489795 cm.
It implies;
Area Turquoise Exactly in Decimal cm² is;
Area triangle with height and base 6.61224489795 cm and 12 cm respectively - Area triangle with height 9.32440048713 cm and base (4.71772661824sin78.61911303663) cm.
= 0.5*12*6.61224489795-0.5*9.32440048713*4.71772661824sin78.61911303663
= 39.6734693877-21.5625
= 18.1109693877 cm²
≈ 18.11 cm² to 2 decimal places.
