A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/3 #, and the triangle's area is #24 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-26T21:04:29

#11.664\ \text{unit}^2#

Given that in #\Delta ABC#, #A=\pi/2#, #B={\pi}/3#

#C=\pi-A-B#

#=\pi-\pi/2-{\pi}/3#

#={\pi}/6#

from sine in #\Delta ABC#, we have

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

#\frac{a}{\sin(\pi/2)}=\frac{b}{\sin ({\pi}/3)}=\frac{c}{\sin ({\pi}/6)}=k\ \text{let}#

#a=k\sin(\pi/2)=k#

# b=k\sin({\pi}/3)=0.866k#

#c=k\sin({\pi}/6)=0.5k#

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

#=\frac{k+0.866k+0.5k}{2}=1.183k#

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

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

#24=\sqrt{1.183k(1.183k-k)(1.183k-0.866k)(1.183k-0.5k)}#

#24=0.2165k^2#

#k^2=110.8545#

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

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

#r=\frac{24}{1.183k}#

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

#=\pi r^2#

#=\pi (24/{1.183k})^2#

#=\frac{576\pi}{1.399489k^2}#

#=\frac{1293.0129}{110.8545}\quad (\because k^2=110.8545)#

#=11.664\ \text{unit}^2#