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 $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $72$ $72$ . 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 $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $72$ $72$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-02-12T11:33:31

Area of triangle's incircle is $35.04 (2 d p)$ $35.04(2dp)$ sq.unit.

Explanation:

$∠ A = \frac{π}{2} = 90^{0}; ∠ B = \frac{π}{6} = 30^{0}; ∠ C = 180 - (90 + 30) = 60^{0}$ $/_A= pi/2=90^0 ; /_B= pi/6=30^0 ; /_C =180-(90+30)=60^0$

This is a $30 - 60 - 90$ $30-60-90$ triangle. So if $A C = x$ $AC=x$ then $A B = x \cdot \sqrt{3};; B C = 2 x$ $AB=x*sqrt3; ;BC=2x$ , Area of triangle is $1/2*AB*AC =72 or x*sqrt3*x=72*2 or x^2=144/sqrt3 :.x = 9.12(2dp) ; 2x=18.24(2dp) ; sqrt3x=15.79$
Sides of triangle are $9.12,15.79 &18.24$ unit
Perimeter of triangle is $P= 9.12+15.79 +18.24 =43.15$ unit
Area of triangle is $A_t=72$ sq.unit
Radius of triangle's incircle is $r=(2*A_t)/P=(2*72)/43.15= 3.34(2dp)$ unit
Area of triangle's incircle is $A_r= pi* *r^2=pi*3.34^2 = 35.04(2dp)$ sq.unit. [Ans]

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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 $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $72$ $72$ . 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 $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $72$ $72$ . What is the area of the triangle's incircle?