A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{3 π}{8}$ $( 3pi)/8$ , and the triangle's area is $16$ $16$ . 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 $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{3 π}{8}$ $( 3pi)/8$ , and the triangle's area is $16$ $16$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-06-13T09:53:23

The area of the incircle is $= 6.52 u^{2}$ $=6.52u^2$

Explanation:

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The area of the triangle is $A = 16$ $A=16$

The angle $ˆ A = \frac{1}{8} π$ $hatA=1/8pi$

The angle $ˆ B = \frac{3}{8} π$ $hatB=3/8pi$

The angle $ˆ C = π - (\frac{1}{8} π + \frac{3}{8} π) = \frac{4}{8} π = \frac{π}{2}$ $hatC=pi-(1/8pi+3/8pi)=4/8pi=pi/2$

The sine rule is

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k$ $a/sinA=b/sinB=c/sinC=k$

So,

$a = k sin A$ $a=ksinA$

$b = k sin B$ $b=ksinB$

$c = k sin C$ $c=ksinC$

Let the height of the triangle be $= h$ $=h$ from the vertex $A$ $A$ to the opposite side $B C$ $BC$

The area of the triangle is

$A = \frac{1}{2} a \cdot h$ $A=1/2a*h$

But,

$h = c sin B$ $h=csinB$

So,

$A = \frac{1}{2} k sin A \cdot c sin B = \frac{1}{2} k sin A \cdot k sin C \cdot sin B$ $A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB$

$A = \frac{1}{2} k^{2} \cdot sin A \cdot sin B \cdot sin C$ $A=1/2k^2*sinA*sinB*sinC$

$k^{2} = \frac{2 A}{sin A \cdot sin B \cdot sin C}$ $k^2=(2A)/(sinA*sinB*sinC)$

$k = \sqrt{\frac{2 A}{sin A \cdot sin B \cdot sin C}}$ $k=sqrt((2A)/(sinA*sinB*sinC))$

$= \sqrt{\frac{32}{sin (\frac{π}{8}) \cdot sin (\frac{3}{8} π) \cdot sin (\frac{1}{2} π)}}$ $=sqrt(32/(sin(pi/8)*sin(3/8pi)*sin(1/2pi)))$

$= 9.81$ $=9.81$

Therefore,

$a = 9.81 sin (\frac{1}{8} π) = 3.64$ $a=9.81sin(1/8pi)=3.64$

$b = 9.81 sin (\frac{3}{8} π) = 9.06$ $b=9.81sin(3/8pi)=9.06$

$c = 9.81 sin (\frac{1}{2} π) = 9.51$ $c=9.81sin(1/2pi)=9.51$

The radius of the incircle is $= r$ $=r$

$\frac{1}{2} \cdot r \cdot (a + b + c) = A$ $1/2*r*(a+b+c)=A$

$r = \frac{2 A}{a + b + c}$ $r=(2A)/(a+b+c)$

$= \frac{32}{22.2} = 1.44$ $=32/(22.2)=1.44$

The area of the incircle is

$a r e a = π \cdot r^{2} = π \cdot {1.44}^{2} = 6.52 u^{2}$ $area=pi*r^2=pi*1.44^2=6.52u^2$

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{3 π}{8}$ $( 3pi)/8$ , and the triangle's area is $16$ $16$ . What is the area of the triangle's incircle?

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Explanation:

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A triangle has vertices A, B, and C. Vertex A has an angle of π8pi/8 , vertex B has an angle of 3π8( 3pi)/8 , and the triangle's area is 1616 . What is the area of the triangle's incircle?

1 Answer

Explanation:

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{3 π}{8}$ $( 3pi)/8$ , and the triangle's area is $16$ $16$ . What is the area of the triangle's incircle?