Mathematics Question and Solution
Notice.
r, radius of the ascribed quarter circle is (25/√(2)) units.
r = ½(25√(2)) units.
(AD)² = 2(½(25√(2)))²
(AD)² = 2*¼*625*2
AD = √(625)
AD = 25 units.
It implies;
AB+BC+CD = AD
10+BC+4 = 25
BC = 25-14
BC = 11 units.
a = 25-2(4)
a = 17 units.
b = ½(a)
b = ½(17) units.
sinc = ½(17)/(½(25√(2)))
c = asin(17√(2)/50)°
c = 28.7397952917°
d = 2c
d = 2asin(17√(2)/50)°
e = ½(90-d)
e = (45-asin(17√(2)/50))°
f = ½(180-e)
f = ½(180-45+asin(17√(2)/50))
f = (0.5(135+asin(17√(2)/50)))°
g = f-45
g = ½(45+asin(17√(2)/50))°
It implies;
tan(½(45+asin(17√(2)/50))) = h/4
h = 4tan(½(45+asin(17√(2)/50)))
h = 3 units.
h is CE.
j = 25-2(10)
j = 5 units.
sink = (0.5(5)/(0.5*25√(2)))
k = asin(0.5(5)/(0.5*25√(2)))
k = asin(√(2)/10)°
l = 2k
l = 2asin(√(2)/10)°
m = ½(90-2asin(√(2)/10))
m = (45-asin(√(2)/10))°
n = ½(180-(45-asin(√(2)/10)))
n = ½(135+asin(√(2)/10))
o = ½(135+asin(√(2)/10))-45
o = ½(45+asin(√(2)/10))°
It implies;
tan(½(45+asin(√(2)/10))) = p/10
p = 10tan(½(45+asin(√(2)/10)))
p = 5 units.
p is FB.
Therefore;
FB+BC+CE is;
5+11+3
= 19 units.
