A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(5pi)/12 #, and the triangle's area is #7 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-27T08:42:13

#2.221\ \text{unit}^2#

Given that in #\Delta ABC#, #A=\pi/12#, #B={5\pi}/12#

#C=\pi-A-B#

#=\pi-\pi/12-{5\pi}/12#

#={\pi}/2#

from sine in #\Delta ABC#, we have

#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#

#\frac{a}{\sin(\pi/12)}=\frac{b}{\sin ({5\pi}/12)}=\frac{c}{\sin ({\pi}/2)}=k\ \text{let}#

#a=k\sin(\pi/12)=0.259k#

# b=k\sin({5\pi}/12)=0.966k#

#c=k\sin({\pi}/2)=k#

#s=\frac{a+b+c}{2}#

#=\frac{0.259k+0.966k+k}{2}=1.1125k#

Area of #\Delta ABC# from Hero's formula

#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#

#7=\sqrt{1.1125k(1.1125k-0.259k)(1.1125k-0.966k)(1.1125k-k)}#

#7=0.125k^2#

#k^2=56#

Now, the in-radius (#r#) of #\Delta ABC#

#r=\frac{\Delta}{s}#

#r=\frac{7}{1.1125k}#

Hence, the area of inscribed circle of #\Delta ABC#

#=\pi r^2#

#=\pi (7/{1.1125k})^2#

#=\frac{49\pi}{1.23766k^2}#

#=\frac{124.3787}{56}\quad (\because k^2=56)#

#=2.221\ \text{unit}^2#