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 #2 #. 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 #2 #.  What is the area of the triangle's incircle?

Sophia Alfred · Accepted Answer

The area of the triangle's incircle is #(10-4sqrt6)pi#.

Let #S# be the area of the incircle, calculated with the formula

#S = pir^2#, where #r# is the inradius of the triangle.

The simplest formula to find the inradius is

#r= "Area"/"Semiperimeter" = Delta/s#

where #Delta# is the area of the triangle, equal to #2# in this case, and #s# is the semiperimeter:

#s=(a+b+c)/2#

See this answer of mine for a generalised proof of this. As all triangles are tangential, the formula applies here aswell.

The angle of vertex #C# can easily be calculated:

#C = pi - A - B = (5pi)/12#

Our triangle is visualised below.

As it is a right triangle, its area will be equal to

#Delta = (AB xx BC)/2 = (c*a)/2=2 => ac=4#

Using the trigonometric functions, we see that

#sin A = "opposite"_A/"hypotenuse"=a/b#

#sin C = "opposite"_C/"hypotenuse" = c/b#

#=> (ac)/b^2 = sinAsinC => ac = b^2*sinAsinC = 4#

Using the sum and difference formulas, we find out that

#sin C = sin color(black)((5pi)/12) = (sqrt6+sqrt2)/4#

#sin A = sin color(black)(pi/12) = (sqrt6-sqrt2)/4#

Hence

#1/16 b^2 (sqrt6+sqrt2)(sqrt6-sqrt2) = 4#

#1/16 b^2 (sqrt6^2-sqrt2^2) = 4#

#1/16 b^2 *4 =4 => b^2 = 16 => color(blue)(b=4)#

Then,

#color(blue)(a = bsinA = sqrt6-sqrt2)#

#color(blue)(c = bsinC = sqrt6+sqrt2)#

#:. s = (sqrt6-sqrt2+4+sqrt6+sqrt2)/2 =sqrt6+2#

#r = Delta/s = 2/(sqrt6+2) = 2/(sqrt6+2) * color(purple)((sqrt6-2)/(sqrt6-2))=(2sqrt6-4)/2 #

#color(red)( :. r=sqrt6-2)#

#S = pir^2 = pi(sqrt6-2)^2 = pi(6-4sqrt6+4)=(10-4sqrt6)pi#

Sadie Adams · Answer

#0.6346\ ext{unit}^2# Given that in #\Delta ABC#, #A=\pi/12#, #B=\pi/2# #C=\pi-A-B# #=\pi-\pi/12-\pi/2# #={5\pi}/12# from sine in #\Delta ABC#, we have #\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}# #\frac{a}{\sin(\pi/12)}=\frac{b}{\sin (\pi/2)}=\frac{c}{\sin ({5\pi}/12)}=k\ ext{let}# #a=k\sin(\pi/12)=0.259k# # b=k\sin(\pi/2)=k# #c=k\sin({5\pi}/12)=0.966k# #s=\frac{a+b+c}{2}# #=\frac{0.259k+k+0.966k}{2}=1.1125k# Area of #\Delta ABC# from Hero's formula #\Delta=\sqrt{s(s-a)(s-b)(s-c)}# #2=\sqrt{1.1125k(1.1125k-0.259k)(1.1125k-k)(1.1125k-0.966k)}# #2=0.125k^2# #k^2=16# Now, the in-radius (#r#) of #\Delta ABC# #r=\frac{\Delta}{s}# #r=\frac{2}{1.1125k}# Hence, the area of inscribed circle of #\Delta ABC# #=\pi r^2# #=\pi (2/{1.1125k})^2# #=\frac{4\pi}{1.2376k^2}# #=\frac{10.15336}{16}\quad (\because k^2=16)# #=0.6346\ ext{unit}^2#

Ella Armstrong · Answer

undefined To find the area of the triangle's incircle, we can use the formula $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is the semi-perimeter of the triangle.

First, we need to find the lengths of the sides of the triangle using the given angles.

Since the angle at vertex $A$ is $ \frac{\pi}{12} $ and the angle at vertex $B$ is $ \frac{\pi}{2} $, the remaining angle at vertex $C$ can be found by subtracting the sum of the other two angles from $ \pi $:

$$ \pi - \left( \frac{\pi}{12} + \frac{\pi}{2} \right) = \pi - \left( \frac{\pi}{12} + \frac{6\pi}{12} \right) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} $$

Now, we can use the Law of Sines to find the lengths of the sides. Let's denote the side opposite vertex $A$ as $a$, the side opposite vertex $B$ as $b$, and the side opposite vertex $C$ as $c$.

$$ \frac{a}{\sin\left(\frac{\pi}{12}\right)} = \frac{b}{\sin\left(\frac{\pi}{2}\right)} = \frac{c}{\sin\left(\frac{5\pi}{12}\right)} $$

$$ a = \frac{\sin\left(\frac{\pi}{12}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c $$
$$ b = \frac{\sin\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c $$
$$ c = \frac{\sin\left(\frac{5\pi}{12}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c $$

Since $ \sin\left(\frac{\pi}{2}\right) = 1 $, we have:
$$ a = \sin\left(\frac{\pi}{12}\right) \times c $$
$$ b = c $$
$$ c = \sin\left(\frac{5\pi}{12}\right) \times c $$

Now, we can use the given area ($A = 2$) to find the semi-perimeter $s$:

$$ A = rs $$
$$ 2 = r \times \left( \frac{a+b+c}{2} \right) $$
$$ 2 = r \times \left( \frac{\sin\left(\frac{\pi}{12}\right) \times c + c + \sin\left(\frac{5\pi}{12}\right) \times c}{2} \right) $$

Solving for $r$ will give us the radius of the incircle, and then we can use the formula for the area of the incircle ($A_{\text{incircle}} = \pi r^2$) to find the area.

This process may involve some algebraic manipulations and trigonometric calculations, but it will yield the area of the triangle's incircle.