a² = 2(1)²
a = √(2) units.
a is the diagonal of the inscribed square and also the side length of the two congruent inside bigger regular triangle.
b = ½(a)
b = ½√(2) units.
c²+b² = a²
c² = √(2)²-...
a = 2(20)
a = 40 units.
b = 60+20
b = 80 units.
c = 2(20)
c = 40 units.
d = atan(80/40)+atan(20/40)
d = 90°
d is angle ACB
Area Triangle ABC is;
½*√(80²+40²)*√(40²+20²)
= ½*√(6400+1600)*√(1600...
Let the inscribed green square side length be a.
2b² = a²
b = ½√(2)a units.
c² = 2(4)²
c = 4√(2) units.
c is the diagonal of the ascribed bigger square.
It implies;
4+a+½(a) = c...
Calculating x, length AD.
c² = x²+3²
c = √(x²+9) units.
c is AC.
d²+2² = c²
d² = √(x²+9)²-4
d² = x²+5
d = √(x²+5) units.
d is AB.
Notice.
a-b = 30
Therefore;
a = (30+b)°
sinb = x/√(x²+9) --- (1...
Let b = 1 unit.
Calculating a.
c² = 2a²
c = √(2)a units.
c is the diagonal of each of the three identical squares.
d = c-a
d = (√(2)a-a) units.
It implies;
1² = a²+(√(2)a-a)²-2a(√(2)a-a)cos45...
Let the single side length of the square be (x+12) units.
Calculating x.
18²=(x+12)²+(x+6)²
324=x²+24x+144+x²+12x+36
324=2x²+36x+180
2x²+36x-144=0
x²+18x-72=0
(x+9)²=72+81
x = -9±√(153)
x ≠...
Calculating Length x.
tana = 5/10
a = atan(½)°
b = 2a
b = 2atan(½)°
c = 180-2b
c = (180-4atan(½))°
It implies;
x² = 2(10)²-2(10)²cos(180-4atan(½))
x² = 144
x = 12 units.
Calculating Length x.
a = 4+6
a = 10 units.
a is the side length of the square.
b = 10-3
b = 7 units.
c = 10-2-3
c = 5 units.
d² = 7²+4²
d = √(49+16)
d = √(65) units.
tane = 4/7...
Calculating Shaded Area.
a = 4(4)
a = 16 units.
a is AB.
b = a+4
b = 20 units.
b is AC.
It implies;
15²+20² = 7²+c²
225+400-49 = c²
c = 24 units.
c is CE.
d = ½(c)
d = 12 units.
d is CD = DE.
d...
Calculating alpha. Let it be a.
Let the three equal lengths be 1 units each.
b = 180-30-7.5
b = 142.5 °
b is angle AKC.
Notice.
Angle CAK = Angle KBC = 7.5°
(1/sin30) = (c/sin7.5)
c = 0.2610523...
