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?

2 Answers
Hammer
Jul 13, 2018

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

Explanation:

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#

Harish Chandra Rajpoot
Jul 13, 2018

#0.6346\ \text{unit}^2#

Explanation:

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

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