A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(7pi)/8 #, and the triangle's area is #8 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-23T14:34:00

#1.0767\ \text{unit}^2#

Given that in #\Delta ABC#, #A=\pi/12#, #B={7\pi}/8#

#C=\pi-A-B#

#=\pi-\pi/12-{7\pi}/8#

#={\pi}/24#

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 ({7\pi}/8)}=\frac{c}{\sin ({\pi}/24)}=k\ \text{let}#

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

# b=k\sin({7\pi}/8)=0.383k#

#c=k\sin({\pi}/24)=0.1305k#

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

#=\frac{0.259k+0.383k+0.1305k}{2}=0.38625k#

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

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

#8=\sqrt{0.38625k(0.38625k-0.259k)(0.38625k-0.383k)(0.38625k-0.1305k)}#

#8=0.006392k^2#

#k^2=1251.634#

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

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

#r=\frac{8}{0.38625k}#

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

#=\pi r^2#

#=\pi (8/{0.38625k})^2#

#=\frac{64\pi}{0.1492k^2}#

#=\frac{1347.699}{1251.634}\quad (\because k^2=1251.634)#

#=1.0767\ \text{unit}^2#