Mathematics Question and Solution
πr² = 9π
r² = 9
r = 3 units.
r is the radius of the inscribed circle.
cosa = 6/7
a = acos(6/7)°
a = 31°
b = 90-a
b = (90-acos(6/7))°
b = 59°
tan(0.5*59) = r/c
c = 3/tan(0.5*59)
c = 5.3024820487 units.
d = r+c
d = 3+5.3024820487
d = 8.3024820487 units.
tan59 = e/d
e = 13.8176505263 units.
f = e-7
f = 6.8176505263 units.
sin31 = g/7
g = 3.6052665244 units.
h = 180-59
h = 121°
It implies;
j² = 3.6052665244²+6.8176505263²-2*6.8176505263*3.6052665244cos121
j = 9.2085290772 units.
(9.2085290772/sin121) = (6.8176505263/sink)
k = 39.3914067477°
l = 90-k
l = 50.6085932523°
Therefore, radius of the large circle is;
Let it be m.
sin50.6085932523 = (0.5*8.3024820487)/m
m = (0.5*8.3024820487)/(sin50.6085932523)
m = 5.3714887803 units.
Again, m is the radius of the large circle.
Area large circle is;
πm²
= π(5.3714887803)²
= 90.6440326526 square units
