Given a triangle with vertices A, B and C the area of triangle can be computed using the formula, #A_Delta= 1/2 bar(AC)*bar(AB) sinalpha# where #alpha# is the agnle between the two side #bar(AC)# and #bar(AB)#. We also should note the Area of the Triangle
can be computed using,
#A_Delta= 1/2 r (bar(AC)+bar(AB)+bar(BC))= 1/2rP; P= Perimeter " and " r# the radius of the inscribed circle.
We have 3 equations"
#A_Delta =3 = 1/2 bar(AC)*bar(AB) sin(pi/12)# ===> (1)
#A_Delta =3= 1/2 bar(AB)*bar(BC) sin(pi/4)# ====> (2)
#A_Delta =3= 1/2 bar(AC)*bar(BC) sin((8pi)/12)# ===> (3)
#bar(AC)/bar(BC) = sin(pi/4)/(sin(pi/12))# ====> (4)
#bar(AB)/bar(BC) = sin((8pi)/12)/(sin(pi/12))# ===> (5)
#bar(AB)/bar(AC) = sin((8pi)/12)/(sin(pi/4))# ===> (6)
Using (1) and (6) we write:
#3 = 1/2 bar(AC)*bar(AC) sin((8pi)/12)/(sin(pi/4))sin(pi/12)# ===> (7)
This a binomial with #bar(AC)_1 = 7.12016, bar(AC)_2=−7.12016)#
Since we are calculate a "length" the positive will do, #bar(AC)_1 = 7.12016 # Now we also use (1) to calculate #bar(AB)#
#bar(AB) = 1/sin(pi/12)6/7.12016 = 3.25586# ===> (8)
#bar(BC) = 1/sin(8pi/12)6/7.12016 = 0.97304# ===> (9)
Now use the Triangle Area Formula with r:
#A_Delta=3= 1/2 r (bar(AC)+bar(AB)+bar(BC))= # solve for r
#r = 6/((bar(AC)+bar(AB)+bar(BC))# ===> (10)
#r = 6/((7.12016 )+(3.25586)+(0.97304))= 0.52868#
Now calculate the Area of the circle:
#A_(@) = pir^2 = pi(0.52868)^2 = .2795pi#
