Notice!
Radius of the ascribed semi circle is 6 cm.
It implies;
Radius, a of the inscribed bigger circle is;
a = ½(6)
a = 3 cm.
Calculating b, radius of the smaller inscribed circle.
c = (6-b) c...
Calculating the area of the inscribed rectangle.
Let the length and width of the inscribed rectangle be x and y respectively.
Therefore, observing similar plane shapes (right-angled) side length ra...
Calculating area of the inscribed blue square.
Let the inscribed blue square side be x.
a² = 2x²
a = √(2)x units.
a is the diagonal of the inscribed blue square.
x² = 2b²
b² = x²/2
b = ½√(2)x uni...
Notice!
AQ = 6 cm.
a² = 6²+3²-2*6*3cos108
108° is the single interior angle of the regular pentagon.
a = 7.4916361229 cm.
a is AP.
cosb = 6/7.4916361229
b = acos(6/7.4916361229)°
b is angle PAQ....
(4+5+3)² = 2a²
a² = 72
a = 6√(2) cm.
a is the congruent adjacent sides of the right-angled triangle.
b² = √(72)²+3²-2*√(72)*3cos45
b² = 81-36
b = √(45)
b = 3√(5) cm.
(3√(5)/sin45) = (3/sinc)
c = 1...
Let the base of the ascribed triangle be 1 unit.
a = 180-20-20-10-10
a = 120°
(1/sin120) = (b/sin20)
b = 0.3949308436 units.
c = 180-120-20
c = 40°
(0.3949308436/sin40) = (d/sin120)
d = 0.532088...
a = 180-20-50-40
a = 70°
It implies the triangle is isosceles.
Let the base of the isosceles triangle be 1 unit.
1² = 2b²-2b²cos40
1 = 0.4679111138b²
b = √(1/0.4679111138)
b = 1.4619022...
Notice!
R is the radius of the ascribed semi circle.
Calculating R.
a² = 2*√(2)²
a² = 4
a = 2 units.
a is the diagonal of the inscribed red and white square.
√(2)² = 2b²
b² = 2/2
b = 1 unit.
It...
a² = 2*5²
a = √(50)
a = 5√(2) cm.
tanb = 10/5
b = atan(2)°
c = b-45
c = (atan(2)-45)°
c = 18.4349488229°
d = 90-b
d = (90-atan(2))°
d = 26.5650511771°
e = 180-2d
e = 180-2*26.56...
Calculating x.
Notice!
AD = 100 cm.
Therefore;
Area triangle ABD + Area triangle ACD = Area triangle ABC
0.5*2x*100sin60+0.5*x*100sin60 = 0.5*2x*xsin120
100x+50x = x²
x = 100+50
x = 150 cm.
It i...
