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 =...
Calculating Area Blue.
a = 4+8
a = 12 cm.
a is the radius of the ascribed quarter circle.
b² = 12²+4²
b = √(144+16)
b = √(160)
b = 4√(10) units.
tanc = 12/4
c = atan(3)°
(12/sin(60+atan(3))) =...
Let FC=x.
Let CE=y
Therefore;
5+x=y+4
x=y-1 ------- (1).
Notice;
Triangle AEF is similar to triangleEFC.
It implies;
5+x=y
4=x, cross multiply.
x²+5x=4y ----- (2).
Subs...
a = (r+9) units.
b = (r-9) units.
a² = b²+c²
(r+9)² = (r-9)²+c²
c² = r²+18r+81-r²+18r-81
c² = 36r
c = 6√(r) units.
d = (r+4) units.
e = (r-4) units.
f²+e² = d²
f² =...
Sir Mike Ambrose is the author of the question.
Let the side of the regular hexagon be 1 unit.
Therefore;
Length Blue is;
√(2-2cos108)
= 1.61803398875 unit.
Length Red is;
√(1²+...
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
b = 180-a-42
b = 180-150
b = 30°
(12/sin108) = (c/sin30)
c = 6.30877334543 cm.
c is the side length of th...
Let r be the radius of the half circle.
a²+2² = r²
a = √(r²-4) units.
b = 2a
b = 2√(r²-4) units.
c = (r-2) units.
Calculating r.
√(r²-4) ~ 2√(r²-4)
(r-2) ~ 9
Cross Multiply....
2πr = 13π
r = ½(13) cm.
r is the radius of the ascribed large circle.
a = 2r
a = 2*½(13)
a = 13 cm.
a is the diameter of the ascribed large circle.
It implies;
c = a-3b
c = (13-3b) c...
Calculating R, radius of the circle.
R² = (½(9))²+a²
a² = R²-(81/4)
a = √(R²-(81/4)) units.
b = ½(9)+3
b = ½(15) units.
c = a-4
c = (√(R²-(81/4))-4) units.
There, R is;
R² = b²+c...
a² = 2²+2²
Where;
a is radius of the ascribed half circle.
2 is the side length of the inscribed blue square.
a² = 8
a = 2√(2) cm.
Again, a is the radius of the circle.
Therefore, leng...
