A, B and C form the vertices of a triangle. BC = 5.7cm and angle CBA = 137 degrees. Given that the area of the triangle is 10.4 cm^2, what is the perimeter of the triangle?

1 Answer
Jul 31, 2018

21.33"cm (2dp)".

Explanation:

Recall that, the Area of DeltaABC=1/2*BC*BA*sin/_CBA.

:. 10.4=(1/2)(5.7)(BA)sin137^@,

i.e., 10.4=2.85*BA*(0.682)=(1.9437)BA.

:. BA=10.4/1.9437=5.35...............(1).

Further, by the cosine-rule,

AC^2=CB^2+BA^2-2*CB*BA*cos/_CBA,

=5.7^2+5.35^2-2(5.7)(5.35)(-0.731),

=32.49+28.62+44.58,

AC^2=105.69.

:. AC=sqrt105.69=10.28...............(2).

Thus, we have, BA=5.35, AC=10.28, CB=5.7.

Hence, the reqd. perimeter=5.35+10.28+5.7=21.33cm.