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

Question

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

2 Answers

Impact of this question

2286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T21:42:10

The area of the triangle's incircle is $(10 - 4 \sqrt{6}) π$ $(10-4sqrt6)pi$ .

Explanation:

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

$S = π r^{2}$ $S = pir^2$ , where $r$ $r$ 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,

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

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

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

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

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

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

Answer 2 · 2018-07-13T21:46:39

$0.6346\ \text{unit}^2$

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 $\Delta ABC$

$=\pi r^2$

$=\pi (2/{1.1125k})^2$

$=\frac{4\pi}{1.2376k^2}$

$=\frac{10.15336}{16}\quad (\because k^2=16)$

$=0.6346\ \text{unit}^2$

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12pi/12 , vertex B has an angle of (pi)/2 π2(pi)/2 , and the triangle's area is 2 22 . What is the area of the triangle's incircle?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

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