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 $\frac{7 π}{12}$ $(7pi)/12$ , and the triangle's area is $22$ $22$ . 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 $\frac{7 π}{12}$ $(7pi)/12$ , and the triangle's area is $22$ $22$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-08-09T15:58:59

Area of Incircle $A_{r} = π \cdot (\frac{22}{13.13}) \approx 541.6$ $A_r = pi * (22/13.13) ~~ 541.6$ sq.units

Explanation:

$ˆ A = \frac{π}{8}, ˆ B = \frac{7 π}{12}, ˆ C = \frac{7 π}{24}, A_{t} = 22$ $hat A = pi / 8, hat B = (7pi)/12, hat C = (7pi)/24, A_t = 22$

Incircle radius $r = \frac{A_{t}}{s}$ $r = A_t / s$ where s is the semi perimeter of the triangle.

Area of triangle $A_{t} = (\frac{1}{2}) b c sin A = (\frac{1}{2}) a c sin B = (\frac{1}{2}) a b sin C$ $A_t = (1/2) bc sin A = (1/2) ac sin B = (1/2) ab sin C$

$b c = \frac{2 A_{t}}{sin A} = \frac{2 \cdot 22}{sin (\frac{π}{8})} = 114.98$ $bc = (2 A_t) / sin A = (2*22)/sin (pi/8) = 114.98$

$c a = \frac{2 A_{t}}{sin A} = \frac{2 \cdot 22}{sin (\frac{7 π}{12})} = 45.55$ $ca = (2 A_t) / sin A = (2*22)/sin ((7pi)/12) = 45.55$

$a b = \frac{2 A_{t}}{sin A} = \frac{2 \cdot 22}{sin (\frac{7 π}{24})} = 55.46$ $ab = (2 A_t) / sin A = (2*22)/sin ((7pi)/24) = 55.46$

$\sqrt{a b \cdot b c \cdot c a} = \sqrt{114.98 \cdot 45.55 \cdot 55.46} = 538.95$ $sqrt(ab * bc * ca) = sqrt(114.98 * 45.55 * 55.46) = 538.95$

$b = \frac{\sqrt{a b \cdot b c \cdot c a}}{c a} = \frac{538.95}{45.45} \approx 11.86$ $b = sqrt(ab * bc * ca) / (ca) = 538.95 / 45.45 ~~ 11.86$

$c = \frac{\sqrt{a b \cdot b c \cdot c a}}{a b} = \frac{538.95}{55.46} \approx 9.72$ $c = sqrt(ab * bc * ca) / (ab) = 538.95 / 55.46 ~~ 9.72$

$a = \frac{\sqrt{a b \cdot b c \cdot c a}}{b c} = \frac{538.95}{114.98} \approx 4.69$ $a = sqrt(ab * bc * ca) / (bc) = 538.95 / 114.98 ~~ 4.69$

$s = \frac{a + b + c}{2} = 13.13$ $s = (a + b + c) / 2 = 13.13$

Incircle radius $r = \frac{A_{t}}{s} = \frac{22}{13.13}$ $r = A_t / s = 22 / 13.13$

Area of Incircle $A_{r} = π \cdot (\frac{22}{13.13}) \approx 541.6$ $A_r = pi * (22/13.13) ~~ 541.6$ sq.units

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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 $\frac{7 π}{12}$ $(7pi)/12$ , and the triangle's area is $22$ $22$ . What is the area of the triangle's incircle?

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Explanation:

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Explanation:

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{7 π}{12}$ $(7pi)/12$ , and the triangle's area is $22$ $22$ . What is the area of the triangle's incircle?