Calculating shaded blue area.
The radius, r if the small circle will be;
(5+r)²=2(5-r)²
25+10r+r²=50-20r+2r²
Therefore
r²-30r+25=0
(r-15)²=-25+225
r = 15±√(200)
r = 15±10√(2)
I...
Calculating shaded inscribed area.
a = 6+6
a = 12 units.
a is the side length of the ascribed square.
b = a-4
b = 8 units.
tanc = 8/6
c = atan(4/3)°
d = 180-2c
d = 2atan(3/4)°
e = ½(d)
e = atan...
Calculating area blue.
a²+1² = 2²
a = √(3) units.
a is the diameter of the inscribed white circle.
b = ½(a)
b = ½√(3) units.
b is the radius of the inscribed white circle.
cos30 = 1/c...
a² = 9²+12²
a = √(81+144)
a = √(225)
a = 15 cm.
It implies;
Calculating r.
(9*3r)+(12*r)+(15*r) = (9*12)
27r+12r+15r = 9*12
54r = 9*12
6r = 12
r = 2 cm.
Let XM be x.
Let AZ be y.
b = (8-x) units.
b is MY.
c = (4-y) units.
c is AY.
Therefore;
½(4x) = ½(8y)
2x = 4y
x = 2y --- (1).
Recall.
b = 8-x
And x = 2y
b = (8-2y) units...
Let the radius of the quarter circle be 1 unit.
2a² = 1
a² = ½
a = ½√(2) units.
a is the side length of the inscribed square.
b = 1-a
b = 1-½√(2)
b = ½(2-√(2)) units.
b is the side length of the g...
a = ½(10)
a = 5 units.
a is the radius of the ascribed half circle.
cosb = 4/5
b = 36.8698976458°
b is angle alpha.
c²+8² = 10²
c = √(100-64)
c = 6 units.
sin(2acos(4/5)) = d/10
d = 9.6 units.
c...
Sir Mike Ambrose is the author of the question.
Let the single side length of the square be 2 units.
Calculating area S.
It will be;
Area trapezoid of two parallel side lengths 2 units an...
Using similar triangle approach.
It implies;
n ~ 1
n+1 ~ n
Cross multiply.
Therefore;
n²=n+1
n²-n-1=0
(n-½)²=1+¼
n =½±½(√(5))
Therefore;
n ≠ ½(1-√(5))
n = ½(1+√(5))...
Let length AB be x.
A careful analysis carried out on the given composite plane shape lead to the derivation of right-angled tiangle EQX with hypotenuse QX ⅓(4x) and the two adjacent side of the tr...
