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 $12$ $12$ . What is the area of the triangle's incircle?

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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 $12$ $12$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-05-18T16:46:01

Area of incircle is $3.808$ $3.808$

Explanation:

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

![https://www.quora.com/What-is-the-radius-of-the-incircle-of-the-3-4-5-right-triangle]( useruploads.socratic.org )

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

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

Here we have two angles $\frac{π}{2}$ $pi/2$ and $\frac{π}{12}$ $pi/12$ and third angle is $π - \frac{π}{2} - \frac{π}{12} = \frac{5 π}{12}$ $pi-pi/2-pi/12=(5pi)/12$ and using sine law, we have

$\frac{a}{sin (\frac{π}{12})} = \frac{b}{sin (\frac{5 π}{12})}$ $a/sin(pi/12)=b/sin((5pi)/12)$ or $\frac{a}{b} = \frac{sin (\frac{π}{12})}{sin (\frac{5 π}{12})}$ $a/b=sin(pi/12)/sin((5pi)/12)$

i.e. $\frac{a}{b} = \frac{0.25882}{0.96593} = 0.26795$ $a/b=0.25882/0.96593=0.26795$ and as $a b = 24$ $ab=24$

Hence $b^{2} = (a b) \times \frac{a}{b} = 24 \times 0.26795$ $b^2=(ab)xxa/b=24xx0.26795$

and $b = \sqrt{6.4308} = 2.536$ $b=sqrt6.4308=2.536$ and $a = \frac{24}{2.536} = 9.464$ $a=24/2.536=9.464$

and $c = \sqrt{{2.536}^{2} + {9.464}^{2}} = \sqrt{95.9986} = 9.8$ $c=sqrt(2.536^2+9.464^2)=sqrt95.9986=9.8$

Hence $r = \frac{24}{2.536 + 9.464 + 9.8} = \frac{24}{21.8} = 1.101$ $r=24/(2.536+9.464+9.8)=24/21.8=1.101$

and area of incircle is $π {(1.101)}^{2} = 3.808$ $pi(1.101)^2=3.808$

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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 $12$ $12$ . What is the area of the triangle's incircle?

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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 $12$ $12$ . What is the area of the triangle's incircle?