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?

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Answer 1 · 2018-07-31T07:44:02

# 21.33"cm (2dp)"#.

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#.