A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $9$ $9$ . 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 $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $9$ $9$ . What is the area of the triangle's incircle?

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Answer 1 · 2016-03-14T12:53:57

Inscribed circle Area $= 4.37405$ $=4.37405" "$ square units

Explanation:

Solve for the sides of the triangle using the given Area $= 9$ $=9$
and angles $A = \frac{π}{2}$ $A=pi/2$ and $B = \frac{π}{3}$ $B=pi/3$ .

Use the following formulas for Area:

Area $= \frac{1}{2} \cdot a \cdot b \cdot sin C$ $=1/2*a*b*sin C$

Area $= \frac{1}{2} \cdot b \cdot c \cdot sin A$ $=1/2*b*c*sin A$

Area $= \frac{1}{2} \cdot a \cdot c \cdot sin B$ $=1/2*a*c*sin B$

so that we have

$9 = \frac{1}{2} \cdot a \cdot b \cdot sin (\frac{π}{6})$ $9=1/2*a*b*sin (pi/6)$
$9 = \frac{1}{2} \cdot b \cdot c \cdot sin (\frac{π}{2})$ $9=1/2*b*c*sin (pi/2)$
$9 = \frac{1}{2} \cdot a \cdot c \cdot sin (\frac{π}{3})$ $9=1/2*a*c*sin (pi/3)$

Simultaneous solution using these equations result to
$a = 2 \cdot \sqrt[4]{108}$ $a=2*root4 108$
$b = 3 \cdot \sqrt[4]{12}$ $b=3*root4 12$
$c = \sqrt[4]{108}$ $c=root4 108$

solve half of the perimeter $s$ $s$

$s = \frac{a + b + c}{2} = 7.62738$ $s=(a+b+c)/2=7.62738$

Using these sides a,b,c,and s of the triangle, solve for radius of the incribed circle

$r = \sqrt{\frac{(s - a) (s - b) (s - c)}{s}}$ $r=sqrt(((s-a)(s-b)(s-c))/s)$

$r = 1.17996$ $r=1.17996$

Now, compute the Area of the inscribed circle

Area $= π r^{2}$ $=pir^2$
Area $= π {(1.17996)}^{2}$ $=pi(1.17996)^2$
Area $= 4.37405$ $=4.37405" "$ square units

God bless....I hope the explanation is useful.

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $9$ $9$ . 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 $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $9$ $9$ . What is the area of the triangle's incircle?