Chapter 7.3, Problem 79SR

To determine

To Find: The area of the triangle.

Answer to Problem 79SR

  $A=\frac{27\sqrt{15}}{4}$

Explanation of Solution

Given: The area of the triangle with sides of length $a,b,c$ is given by $\sqrt{s(s-a)(s-b)(s-c)}$ Where $s=\frac{1}{2}(a+b+c)$ . If the length of the sides of triangle are $6,9,12ft$

Given: sides $6,9,12ft$

Then

  $\begin{array}{l}s=\frac{1}{2}(a+b+c)\\ s=\frac{1}{2}(6+9+12)\\ s=\frac{27}{2}\end{array}$

i.e

  $s=\frac{27}{2}$

Now

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{\frac{27}{2}\left(\frac{27}{2}-6\right)\left(\frac{27}{2}-9\right)\left(\frac{27}{2}-12\right)}\\ A=\sqrt{\frac{27}{2}\left(\frac{15}{2}\right)\left(\frac{9}{2}\right)\left(\frac{3}{2}\right)}\\ A=\frac{1}{4}\sqrt{27(15)(9)(3)}\\ A=\frac{1}{4}\sqrt{3\times 3\times 3\times 3\times 5\times 3\times 3\times 3}\\ A=\frac{27}{4}\sqrt{15}\\ A=\frac{27\sqrt{15}}{4}\end{array}$

Hence, the area is $A=\frac{27\sqrt{15}}{4}$