sin30 = a/10
a = 5 cm.
b = 10+2a
b = 20 cm.
c² = 20²+4²-160cos60
c = 4√(21) cm.
c = 18.3303027798 cm.
(18.3303027798/sin60) = (20/sind)
d = asin(5√(7)/14)°
d = 70.8933946491°
e = 180-d
e = (180-...
Sir Mike Ambrose is the author of the question.
Let the side of square ABCD be x.
Calculating x.
½*x(x+½(x)) = 84+24
½(½(3x²)) = 108
¼(x²) = 36
x² = 36*4
x = 12 cm.
Notice;
The base of blue area...
Blue area is;
8(area quarter circle with radius 4 cm - area triangle with height and base 4 cm each) - 8(area quarter circle with radius 2 cm - area triangle with height and base 2 cm each)
= 8(4π...
Let the red length be a.
Calculating a.
b² = 8²+8²
b = 8√(2) units.
c² = 2²+2²
c = 2√(2) units
d = 8√(2)-2√(2)-2-½(a)
d = ½(12√(2)-4-a) units.
e² = 2a²-2a²cos120
e² = 3a²
e = √(3)a units.
Equ...
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Notice;
The radius, a of the small inscribed circle is;
a = 2 units.
Calculating b, radius of the big circle.
c² = 8²+8²
c = √(128)
c = 8√(2) units.
d² = 2²+2²
d = √(8)
d = 2√(2) units.
It impl...
Let the regular triangle side length be 2 units.
Therefore the square side length is 1 unit.
a² = 2(1²)
a = √(2) units.
b² = 2+1-2√(2)cos105
b = √(3-2√(2)cos105) units.
(√(3-2√(2)cos105)/sin105)...
Sir Mike Ambrose is the author of the question.
Let R be radius of the inscribed circle.
Calculating r.
√(100-25) = r+√(25+r²)
r = ⅓(5√(3)) cm.
Therefore;
Area Shaded as a single fract...
Sir Mike Ambrose is the author of the question.
Let the side length of the regular hexagon be 1 unit.
Area Hexagon is;
(½(6))(1/(2tan(180/6)))
= 3/2tan(30)
= ½(3√(3)) square units.
Notice;
1/sin...
Sir Mike Ambrose is the author of the question.
Let the bigger regular pentagon side length be 1 unit.
a = √(2-2cos180)
a = 1.61803398875 units.
b = sin36/sin108
b = 0.61803398875 units.
c = √(2*...
