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 $\frac{π}{3}$ $( pi)/3$ , 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 $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $12$ $12$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-10T08:21:10

$Area of inscribed circle A_{r} = 0.0086 sq units$ $color(brown)("Area of inscribed circle " A_r = 0.0086 " sq units"$

Explanation:

The distances from the incenter to each side are equal to the inscribed circle's radius. The area of the triangle is equal to $12\timesr\times(the triangle's perimeter), 1 2 \times r \times ( the triangle's perimeter )$ $color(crimson)("12×r×(the triangle's perimeter), 1 2 × r × ( the triangle's perimeter ) "$ , where r is the inscribed circle's radius.

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

$ˆ A = \frac{π}{8}, ˆ b = \frac{π}{3}, ˆ C = \frac{13 π}{24}, A_{t} = 12$ $hat A = pi/8, hat b = pi/3, hat C = (13pi)/24, A_t = 12$

$a b = \frac{2 \cdot A_{t}}{sin C} = \frac{2 \cdot 12}{sin (\frac{13 π}{24})} = 24.21$ $ab = (2 * A_t) / sin C = (2 * 12) / sin ((13pi)/24) = 24.21$

$b c = \frac{2 \cdot A_{t}}{sin A} = \frac{2 \cdot 12}{sin (\frac{π}{8})} = 62.72$ $bc = (2 * A_t) / sin A = (2 * 12) / sin (pi/8) = 62.72$

$c a = \frac{2 \cdot A_{t}}{sin B} = \frac{2 \cdot 12}{sin (\frac{π}{3})} = 27.71$ $ca = (2 * A_t) / sin B = (2 * 12) / sin (pi/3) = 27.71$

$a = \frac{\sqrt{a b \cdot b c \cdot c a}}{b c} = \frac{\sqrt{24.21 \cdot 62.72 \cdot 27.71}}{62.72} = 3.27$ $a = sqrt(ab * bc * ca) / (bc) = sqrt(24.21 * 62.72 * 27.71) / 62.72 = 3.27$

$b = \frac{\sqrt{a b \cdot b c \cdot c a}}{b c} = \frac{\sqrt{24.21 \cdot 62.72 \cdot 27.71}}{27.71} = 7.4$ $b = sqrt(ab * bc * ca) / (bc) = sqrt(24.21 * 62.72 * 27.71) / 27.71 = 7.4$

$c = \frac{\sqrt{a b \cdot b c \cdot c a}}{a b} = \frac{\sqrt{24.21 \cdot 62.72 \cdot 27.71}}{24.21} = 8.47$ $c = sqrt(ab * bc * ca) / (ab) = sqrt(24.21 * 62.72 * 27.71) / 24.21 = 8.47$

$Perimeter p = a + b + c = 3.27 + 7.4 + 8.47 = 19.14$ $"Perimeter " p = a + b + c = 3.27 + 7.4 + 8.47 = 19.14$

$I n \circ \leq r a d i u s r = \frac{A_{t}}{12 \cdot p} = \frac{12}{12 \cdot 19.14} = 0.0522$ $Incircle radius " r = A_t / (12 * p) = 12 / (12 * 19.14) = 0.0522$

$Area of inscribed circle A_{r} = π r^{2} = π \cdot {0.0522}^{2} = 0.0086 sq units$ $color(brown)("Area of inscribed circle " A_r = pi r^2 = pi * 0.0522^2 = 0.0086 " sq units"$

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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 $\frac{π}{3}$ $( pi)/3$ , 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 $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{3}$ $( pi)/3$ , and the triangle's area is $12$ $12$ . What is the area of the triangle's incircle?