Mathematics Question and Solution
Radius of the semi circle r is;
r = ½(10)
r = 5 cm.
a = 1+5
a = 6 cm.
b = 5-1
b = 4 cm.
a² = b²+c²
c = √(6²-4²)
c = √(20)
c = 2√(5) cm.
cosd = 4/6
d = acos(2/3)
d = 48.1896851042°
e = 90-d
e = 90-48.1896851042
e = 41.8103148958°
f = 180-d
f = 180-48.1896851042
f = 131.8103148958°
It implies;
g² = 6²+5²-2*6*5cos131.8103148958
g = 10.0498756211 cm.
(10.0498756211/sin131.8103148958) = (5/sinh)
h = 21.7667200119°
cosi = 1/10.0498756211
i = acos(1/10.0498756211)
i = 84.2894068625°
Therefore;
j = 360-i-e-h-90
j = 270-21.7667200119-84.2894068625-41.8103148958
j = 122.1335582298°
k = ½(j)
k = 61.0667791149°
It implies;
tan61.0667791149 = l/1
l = 1.8090169944 cm.
m = l+c
m = 1.8090169944+2√(5)
m = 6.2811529494 cm.
Therefore, Blue Area is;
½(10m)
= ½*10*6.2811529494
= 31.4057647469 cm²
