Sir Mike Ambrose is the author of the question.
Area R ÷ Area Square exactly in its simplest decimal form is;
Area R is;
½(5x7.5) - ½(√(81.25)x√(65.25))sin(tan–¹(5/7.5) - tan–¹(3/7.5)) - ½((4....
Sir Mike Ambrose is the author of the question.
Let the square side be 1 unit.
Therefore;
Area B is;
Area triangle with height 1 unit and base sin(53.13010235416) unit - Area triangle with...
a*9 = 6*3
a = 2 units.
b = ½(9+2)
b = 5.5 units.
c = ½(6+3)
c = 4.5 units.
d = c-3
d = 1.5 units.
Calculating e, radius of the circle.
e² = b²+d²
e² = 5.5²+1.5²
e = 5.70087712...
a²+8² = 10²
a = 6 units.
b = ½(10)
b = 5 units.
b is the radius of the ascribed semi circle.
6² = 5²+5²-2*5*5cosc
50cosc = 50-36
cosc = 14/50
c = acos(7/25)°
c = 73.7397952917°
d =...
a² = 24²+7²
a = 25 units.
a is AC.
b = 7+a
b = 32 units.
c² = b²+24²
c = √(32²+24²)
c = 40 units.
c is the diameter of twice the quarter circle (semi circle (2AB)).
d = ½(c)
d = 20...
(½(r))²+(3/2)² = r²
(9/4) = r²-¼(r²)
(9/4) = ¾(r²)
r² = 3
r = √(3) units.
Where r is the radius of the circle.
Let the square side be a.
b = 45-20
b = 25°
c² = 2a²
c = √(2)a
Calculating a.
a² = 1²+(√(2)a)²-2*1*√(2)acos25
a² = 1+2a²-2.5634255282a
a²-2.5634255282a+1 = 0
It implies;
a = 2.0834...
a² = 2²+2²
a = √(8) cm.
a = 2√(2) cm.
b = 4+2
b = 6 cm.
c = 2+6-4
c = 4 cm.
d² = 6²+4²
d = √(52) cm.
d = 2√(13) cm.
Therefore, shaded portion perimeter is;
4+4+2√(2)+6+2√(13)...
tana = 16/8
a = atan(2)°
b² = 16²+8²
b = 8√(5) cm.
Let BC = c.
Therefore;
c+3c = b
4c = 8√(5)
c = 2√(5) units.
Again, c is BC.
AB = 3c
AB = 3*2√(5)
AB = 6√(5) cm.
d = 180-30-at...
Let the diameter of the inscribed half circle be 2 units.
It radius is 1 unit.
Calculating x, angle alpha.
tana = 1/0.5
a = atan(2)°
b = 2a
b = 2atan(2)°
c = 180-b
c = (180-2atan(...
