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?

Question

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?

Violet Aldrich · Accepted Answer

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

Parker Allen · Answer

undefined To find the area of the triangle's incircle, you can use the formula:

$$ \text{Area} = \text{Semiperimeter} \times \text{Inradius} $$

First, calculate the semiperimeter ($ s $) of the triangle using the formula:

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

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

Since we have the angles of the triangle, we can use trigonometric ratios to find the side lengths. For example, in a right triangle with angle $ \frac{\pi}{2} $ and hypotenuse $ c $, $ c = \frac{AB}{\sin{\frac{\pi}{2}}} = AB $.

Using the same logic, we can find $ AB $ and $ BC $.

Then, calculate $ s $.

Next, use Heron's formula to find the area of the triangle using the side lengths:

$$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $$

Once you have the area of the triangle, you can rearrange the formula to solve for the inradius ($ r $):

$$ r = \frac{\text{Area}}{s} $$

Substitute the values of the area and semiperimeter into this equation to find the inradius, which represents the radius of the incircle.