A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #8 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-02T11:39:37

#3.098\ \text{unit}^2#

Given that in #\Delta ABC#, #A=\pi/8#, #B=\pi/4#

#C=\pi-A-B#

#=\pi-\pi/8-\pi/4#

#={5\pi}/8#

from sine in #\Delta ABC#, we have

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

#\frac{a}{\sin(\pi/8)}=\frac{b}{\sin (\pi/4)}=\frac{c}{\sin ({5\pi}/8)}=k\ \text{let}#

#a=k\sin(\pi/8)=0.383k#

# b=k\sin(\pi/4)=0.707k#

#c=k\sin({5\pi}/8)=0.924k#

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

#=\frac{0.383k+0.707k+0.924k}{2}=1.007k#

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

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

#8=\sqrt{1.007k(1.007k-0.383k)(1.007k-0.707k)(1.007-0.924k)}#

#8=0.125k^2#

#k^2=64#

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

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

#r=\frac{8}{1.007k}#

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

#=\pi r^2#

#=\pi (8/{1.007k})^2#

#=\frac{64\pi}{1.007^2k^2}#

#=\frac{64\pi}{1.007^2\cdot 64}\quad (\because k^2=64)#

#=3.098\ \text{unit}^2#