A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(pi)/2 #, and the triangle's area is #12 #. 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/12 #, vertex B has an angle of #(pi)/2  #, and the triangle's area is #12 #.  What is the area of the triangle's incircle?

Evelyn Arboleda · Accepted Answer

Area of incircle is #3.808#

Let us consider a right angled triangle in general and an incircle within it as shown below.

Observe that centre of incircle #O# makes three triangles with sides #a#, #b# and #c# and as area of triangle is #1/2xx"base"xx"height"# #-# and hence area of these triangles is #(ar)/2#, #(br)/2# and #(cr)/2#. As area of complete triangle is #1/2axxb#, we have

#r/2(a+b+c)=(ab)/2# and as area of triangle is #12#, we have #a+b+c=24/r# or #r=24/(a+b+c)#.

Here we have two angles #pi/2# and #pi/12# and third angle is #pi-pi/2-pi/12=(5pi)/12# and using sine law, we have

#a/sin(pi/12)=b/sin((5pi)/12)# or #a/b=sin(pi/12)/sin((5pi)/12)#

i.e. #a/b=0.25882/0.96593=0.26795# and as #ab=24#

Hence #b^2=(ab)xxa/b=24xx0.26795#

and #b=sqrt6.4308=2.536# and #a=24/2.536=9.464#

and #c=sqrt(2.536^2+9.464^2)=sqrt95.9986=9.8#

Hence #r=24/(2.536+9.464+9.8)=24/21.8=1.101#

and area of incircle is #pi(1.101)^2=3.808#

Parker Allen · Answer

undefined To find the area of the triangle's incircle, we can first calculate the lengths of the sides of the triangle using trigonometry. Given that vertex A has an angle of $ \frac{\pi}{12} $ and vertex B has an angle of $ \frac{\pi}{2} $, we can use the law of sines to relate the sides to these angles.

Let's denote the sides as $ a $, $ b $, and $ c $, where side $ a $ is opposite vertex A, side $ b $ is opposite vertex B, and side $ c $ is opposite vertex C.

Using the law of sines:

$$ \frac{a}{\sin(\frac{\pi}{12})} = \frac{b}{\sin(\frac{\pi}{2})} = \frac{c}{\sin(\pi - \frac{\pi}{12} - \frac{\pi}{2})} $$

Simplifying, we have:

$$ \frac{a}{\sin(\frac{\pi}{12})} = \frac{b}{1} = \frac{c}{\sin(\frac{11\pi}{12})} $$

Since $ \sin(\frac{\pi}{2}) = 1 $ and $ \pi - \frac{\pi}{12} - \frac{\pi}{2} = \frac{11\pi}{12} $.

Now, we can solve for $ c $ in terms of $ a $ using the relation between the sides and angles:

$$ c = \frac{a \cdot \sin(\frac{11\pi}{12})}{\sin(\frac{\pi}{12})} $$

Next, we can find the semiperimeter $ s $ of the triangle using the formula:

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

Since we know the area of the triangle is 12, we can use Heron's formula to write:

$$ 12 = \sqrt{s(s-a)(s-b)(s-c)} $$

Substitute the expressions for $ s $, $ a $, $ b $, and $ c $ into the equation above and solve for $ s $. Once you have $ s $, you can find the lengths of the sides $ a $, $ b $, and $ c $.

Finally, use the formula for the area of the incircle:

$$ \text{Area of incircle} = \frac{\text{Area of triangle}}{s} $$

Substitute the known values to find the area of the triangle's incircle.