Sir Mike Ambrose is the author of the question.
P coordinate is;
P(3, 9)
Q coordinate is;
Q(2, 5), Q(3, 5)
It implies;
Shaded area exactly in its square units simplest form is;
(...
Sir Mike Ambrose is the author of the question.
Calculating the circle's radius, r.
21r + 15r + 12√(2)r = 21*12√(2)sin45
(36+12√(2))r = 252
r = 252/(36+12√(2)) cm
r = 3(3-√(2)) cm
I...
Sir Mike Ambrose is the author of the question.
Let the radius of the ascribed semicircle be 2 units.
Therefore;
Radius of the inscribed circle is 1 unit.
It implies;
Area R is;
Are...
Let a be the side of the square.
cos30 = b/8
b = 4√(3) cm.
sin30 = c/8
c = 4 cm.
a = b+12
a = (12+4√(3)) cm.
Area Square is;
a²
= (12+4√(3))²
= 358.27687752661 cm²
d² = 12²+4²...
Let the single side length of the two congruent inscribed regular pentagon be 1 unit.
Area Green is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Area Shaded.
a =...
Sir Mike Ambrose is the author of the question.
Area Orange exactly in square cm decimal is;
Area triangle with height 1.64419338427 cm and base (4.95325421888sin129.5411167921) cm
= ½*1.644...
a = atan(3/2)°
b = atan(2/3)°
b = Angle BEG.
tan(atan(2/3)) = c/2
⅔ = c/2
c = (4/3) cm.
c = BG.
OG = 2-(4/3)
OG = ⅔ cm.
Notice;
Radius of the circle is 2 cm.
Angle OGH = 180-...
Sir Mike Ambrose is the author of the question
Let the side of the regular Pentagon be 2 units.
Therefore;
Area orange is;
Area triangle with height 1.7013016167 units and base 1.8345763...
a² = 2*12²-2*12²cos120
a = 12√(3) cm.
a = 20.78460969083 cm.
a = AB.
b = 2a
b = 24√(3) cm.
b = 41.56921938165 cm.
b = AC.
c² = 12²+41.56921938165²-24*41.56921938165cos30
c = 31.7490157...
a² =2*12²
a = 12√(2) cm.
a = AC.
12b+12b+12√(2)b = 12*12
b = 144/(24+12√(2))
b = 12-6√(2) cm.
b = 3.51471862576 cm.
b = radius of the inscribed circle.
BF = ½(12√(2))
BF = 6√(2) cm....
