Chapter 8.2, Problem 53E

To determine

Find the perimeter and area of triangle.

Answer to Problem 53E

  $1434feet,86,751squqrefeet$

Explanation of Solution

Given information:

Two sides of triangle are $400feetand600feet$ and included angle measures $46°20\text{'}$ .

Calculation:

Consider, the given values,

Side’s length $400feetand600feet$ angle measures $46°20\text{'}$ .

Let perimeter of triangle $P=a+b+c$ .

Apply law of cosines,

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos C\\ c^2={400}^2+{600}^2-2\times 400\times 600\left(\cos 46°20\text{'}\right)\\ c=434.23\end{array}$

Now perimeter of triangle is,

  $\begin{array}{l}P=a+b+c\\ P=400+600+434.23\\ P\approx 1434feet\end{array}$

Find Area of triangle, by Heron’s formula,

  $\begin{array}{l}s=\frac{a+b+c}{2}\\ s=\frac{400+600+434.23}{2}=717\\ A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ A=\sqrt{717\left(717-400\right)\left(717-600\right)\left(717-434\right)}\\ A=\sqrt{717\left(317\right)\left(117\right)\left(283\right)}\\ A\approx 86,751squqrefeet\end{array}$

Hence, the perimeter and area of triangle is $1434feet,86,751squqrefeet$ .