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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $21$ $21$ . 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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $21$ $21$ . What is the area of the triangle's incircle?

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Answer 1 · 2016-07-11T11:07:48

Area of triangle's incircle is $6.7 (1 d p)$ $6.7(1dp)$ sq.unit

Explanation:

$∠ A = \frac{180}{12} = 15^{0}; ∠ B = \frac{58180}{12} = 75^{0}; ∠ C = 180 - (75 + 15) = 90^{0}$ $/_A = 180/12=15^0 ; /_B= 58180/12=75^0 ; /_C=180-(75+15)=90^0$
Area of triangle $A_{r} = 21$ $A_r=21$ (given). Let a,b,c be the sides of triangle.We know $\frac{a \cdot b \cdot sin C}{2} = A_{r} or a \cdot b = \frac{42}{sin 90} = 42$ $(a*b*sinC)/2=A_r or a*b=42/sin90=42$ similarly
$b \cdot c = \frac{42}{sin 15} = 162.28$ $b*c=42/sin15=162.28$ and $c*a=42/sin75=43.48:. ab*bc*ca=(abc)^2=42*162.28*43.48 or abc=544.38 :.$ $a=544.38/162.28=3.355 ;b =544.38/43.48=12.52 ; c= 544.38/42=12.96$ So semi perimeter $S=(3.355+12.52+12.96)/2 =14.42$ Incircle radius is $A_r/S= 21/14.42=1.46$ Area of triangle's incircle is $pi*1.46^2=6.7(1dp)$ sq.unit [Ans]

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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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $21$ $21$ . 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 pi/12 π12pi/12 , vertex B has an angle of (5pi)/12 5π12(5pi)/12 , and the triangle's area is 21 2121 . 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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $21$ $21$ . What is the area of the triangle's incircle?