Let the three equal lengths be 1 unit each.
a² = 1²+1²
a = √(2) units.
b = 150-45
b = 105°
c² = √(2)²+1²+2√(2)cos 105
c = 1.9318516526 units.
(1.9318516526/sin105) = (1/sind)
d = 30...
Let AD be 4 units.
Therefore;
AB will be;
½(AD)+½(½(AD))
AB = ½(4)+½(½(4))
AB = 2+1
AB = 3 units.
Area ABCD is;
4*3
= 12 square units.
Calculating Area Red (Area Trapezoid)....
9² = 16²+17²-2*16*17cosa
81 = 545-544cosa
544cosa = 545-81
cosa = 464/544
a = 31.4669762933°
13² = 16²+11²-2*16*11cosb
169 = 377-352cosb
352cosb = 377-169
cosb = (208/352)
b = 53.7784533...
Let the side length of the red square be a units.
b²+3² = a²
b = √(a²-9) units.
c²+6² = a²
c = √(a²-36) units.
Observing similar plane shape (right-angled) side length ratios.
√(a²-9)...
Notice;
Diameter of the inscribed semi circle is ½ unit.
Therefore radius of the inscribed semi circle is ¼ unit.
Let the radius of the inscribed small circle be r.
Calculating r.
(1...
Radius of the inscribed quarter circle is 36 cm.
Radius of the inscribed semi circle is;
½(½(36))=9 cm.
Radius of the small inscribed circle is;
⅓(⅓(36))=4 cm.
Therefore the side lengths...
Notice!
The square side length is;
4+4+4
= 12 units.
Calculating Area Red.
Area quarter circle with radius 12 units - Area triangle with height and base 12 units respectively - Area quarte...
Let the diameter of the big inscribed half circle be a units.
It's radius is ½(a) units.
Let the diameter of the small inscribed half circle be b units.
It's radius is ½(b) units.
It im...
Equation of the curve is;
y = ½(x^(3/2))
It implies;
PR = √(4²+6²)
PR = 2√(13) units
QR = (2√(13))/(cos(90-atan(⅔)))
= 13 units.
Therefore;
Shaded Area is;
(Area under the cu...
sin60 = 5/a
a = 5/(√(3)/2)
a = ⅓(10√(3)) units.
a is the side length of the inscribed regular triangle.
b = ½(a)
b = ⅓(5√(3)) units.
sin60 = c/⅓(5√(3))
√(3)/2 = 3c/(5√(3))
6c = 15
c =...
