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?

Question

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?

Caroline Aucoin · Accepted Answer

#3.098\ ext{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\ ext{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\ ext{unit}^2#

Noah Atkinson · Answer

undefined To find the area of the incircle of the triangle, we can use the formula:

$$ A = rs $$

Where:
- $ A $ is the area of the triangle
- $ r $ is the radius of the incircle
- $ s $ is the semi-perimeter of the triangle

The semi-perimeter $ s $ can be calculated as:

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

Where $ a $, $ b $, and $ c $ are the side lengths of the triangle.

Using the given angles, we can calculate the side lengths of the triangle using trigonometric ratios. Then, we can find the semi-perimeter and use it to calculate the radius of the incircle. Finally, we use the formula $ A = \pi r^2 $ to find the area of the incircle.