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 $\frac{π}{4}$ $( pi)/4$ , 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 $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{4}$ $( pi)/4$ , and the triangle's area is $21$ $21$ . What is the area of the triangle's incircle?

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Answer 1 · 2016-06-07T11:16:33

The area of triangle's incircle is $11.3 (1 d p)$ $11.3(1dp)$ sq.unit

Explanation:

The three angles are $∠ A = \frac{180}{2} = 90^{0}; ∠ B = \frac{180}{4} = 45^{0}; ∠ C = 180 - (90 + 45) = 45^{0}$ $/_A= 180/2=90^0 ; /_B=180/4=45^0 ; /_C=180-(90+45)=45^0$ This is a right angled isocelles triangle. The sides opposite to the angles $∠ B and ∠ C$ $/_B and /_C$ are equal and their included angle is $90^{0}$ $90^0$ . So the area of the triangle $21= 1/2*b^2 :. b =sqrt 42=6.48=c :.a=sqrt(6.48^2+6.48^2)=sqrt 84=9.17$ So semi perimeter of the triangle $s= ((6.48+6.48+9.17)/2) =11.06$ Hence the radius of the incircle is $r=A_t/s=21/11.06=1.9 ; A_t$ is area of triangle. Hence the area of triangles incircle is $pi*r^2 = pi* 1.9^2 =11.3(1dp)$ 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 $\frac{π}{4}$ $( pi)/4$ , 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 $\frac{π}{2}$ $pi/2$ , vertex B has an angle of $\frac{π}{4}$ $( pi)/4$ , and the triangle's area is $21$ $21$ . What is the area of the triangle's incircle?