cos30 = r/(1-r)
Where r is the radius of each of the three congruent inscribed circles.
It implies;
2r = √(3) - √(3)r
r = √(3)/(2+√(3))
r = (2√(3)-3)/(1)
r = (2√(3)-3) unit.
r = 0.46410161514 uni...
Area blue is;
Area square with side 10 cm - Area triangle with height 5 cm and base 10 cm - Area triangle with height 10 cm and base (4.71404520791sin45) cm - Area triangle with height 5 cm and bas...
Calculating r, radius of the Semi Circle.
½(π)r² = 338π
r² = 676
r = 26 cm.
a = ½(48)
a = 24 cm.
Calculating x.
It implies;
Observing Pythagoras Rule.
26² = 24²+x²
x² = 676-576
x² = 100
x = 10...
a = ⅐*180(7-2)
a = ⅐(180*5)
a = ⅐(900)°
a is the single interior angle of the regular heptagon.
Calculating x and y.
b = 360-2*⅐(900)
b = (360/1)-(1800/7)
b = ⅐(2520-1800)
b = ⅐(720)°
c = ½(b)
c...
a = 10+94
a = 104°
a is angle BAC.
b = 24+34
b = 58°
b is angle ABC.
Let AB = 1 unit.
c = 180-58-104
c = 180-162
c = 18°
c is angle ACB.
It implies;
(1/sin18) = (d/sin104)
d =...
a² = 6²+8²
a = √(100)
a = 10 cm.
a is the hypotenuse of the ascribed right-angled triangle.
Let b be the hypotenuse of the blue inscribed right -angled triangle.
It implies, observing similar tria...
a = 4+8
a = 12 cm.
a is the radius of the quarter circle.
tanb = 12/4
b = atan(3)°
c = (60+atan(3))°
(12/sin(60+atan(3))) = (4/sind)
d = 14.4420678686°
e = 180-60-atan(3)-d
e = 12...
b = 8-a
It implies;
27^(a)/3^(8-a) = 81
3^(3a) = 3^(8-a)*3⁴
3a = 8-a+4
3a = 12-a
4a = 12
a = 3.
Calculating b.
b = 8-a
And a = 3
b = 8-3
b = 5
Let the square side be x.
Calculating x.
a = (5-½(x)) units.
b = (5-x) units.
It implies;
5² = (5-½(x))²+(5-x)²
25 = 25-5x+¼(x)²+25-10x+x²
¼(5x²)-15x+25 = 0
¼(x²)-3x+5 = 0
x²-12x+20 = 0
Resolvi...
The side of area green, inscribed square is;
Let it be x.
Therefore;
1.11496246101x + 0.82101781219x = 27.3672604063
1.9359802732x = 27.3672604063
x = 14.1361256544 cm
Therefore;
Exact perimeter...
