A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #21 #. What is the area of the triangle's incircle?

1 Answer
Binayaka C.
Jun 7, 2016

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

Explanation:

The three angles are #/_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# are equal and their included angle is #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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