Let the radius of the ascribed semi circle be a.
b = ½(a) units.
c = b+a
c = ½(3a) units.
It implies;
(½(3a))² = (½a)²+6²
¼(9a²) = ¼(a²)+36
¼(8a²) = 36
8a² = 36*4
a² = 18
a = 3√(2...
a = 40+10
a = 50°
a is angle ACB.
b = 180-10-50
b = 120°
b is angle ADC.
(6/sin120) = (c/sin10)
c = 1.2030698654 units.
c is CD.
(6/sin120) = (d/sin50)
d = 5.3073115854 units.
d is...
Let the side length of the inscribed two blue squares be 1 unit.
a² = 2(1)²
a = √(2) units.
a is the radius of the green half circle.
b = a-1
b = (√(2)-1) units.
tanc = 1/(√(2)-1)
c =...
Let the ascribed square side be be 4 units.
Area ascribed square is;
4²
= 16 square units.
Calculating the inscribed red rectangle.
It implies;
a² = 2²+1²
a = √(5) units.
b² = 1²+0.5²
b = √(5/4...
Let 1 be R.
Calculating r.
a = (1-r) units.
tanb = 1/2
b = atan(½)
c = 180-2b
c = (180-2atan(½))°
d = 180-c
d = 2atan(½)°
It implies, calculate r.
sin(2atan(½)) = r/(1-r)
⅘ = r/(1-r)
5r = 4-4...
2a² = √(2)²
2a² = 2
a = 1 unit.
a is the radius of the half circle.
It implies;
x = a+a
x = 2a
x = 2*1
x = 2 units.
x is the diameter of the half circle.
Let a be the radius of the quarter circle.
Notice!
a is R.
a² = a²+2²-2*2*acosb
4acosb = 4
cosb = 1/a --- (1).
4² = 2²+a²+2*2*acosb
12 = a²+4acosb --- (2).
Substituting (1) in (2).
12 = a²+4a(1...
Sir Mike Ambrose is the author of the question.
Let the side length of one of the three congruent squares be 2 units.
Therefore;
Area square is;
2²
= 4 square units.
Area Green is;
Area triangle...
Let the side of the regular heptagon be 2 units.
Calculating PH.
AH is;
AH = 2sin(180/7)÷sin(720/7)
AH = 0.89008373583 unit.
AP is;
AP = √(2-2cos(540/7))
AP = 1.24697960372 unit...
Let a be the side length of the blue inscribed square.
Calculating Area of the blue inscribed square.
tan60 = a/b
b = ⅓(√(3)a) cm.
b = 0.5773502692a cm.
c = a+b
c = 1.5773502692a cm.
Calculating...
