Chapter 8, Problem 17CT

To determine

The lengths of the sides, where the area of the triangle shown below is $54\sqrt{6}$ square units.

  

Answer to Problem 17CT

Solution:

The lengths of the sides of the triangle are $15$ units, $18$ units, and $21$ units respectively.

Explanation of Solution

Given information:

The area of the triangle shown below is $54\sqrt{6}$ square units.

  

The area $K$ of a triangle with sides $a,\text{}b,$ and $c$ is $K=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $s=\frac{1}{2}\left(a+b+c\right)$ .

Substitute the values of the sides in the formula $s=\frac{1}{2}\left(a+b+c\right)$ to find $s$

  $s=\frac{1}{2}\left(5x+6x+7x\right)$

  $\Rightarrow s=\frac{1}{2}\left(18x\right)=9x$

  $\Rightarrow s=9x$

Now, substitute the value of $s$ and the sides in the area of the triangle formula.

  $K=\sqrt{9x\left(9x-5x\right)\left(9x-6x\right)\left(9x-7x\right)}$

By simplifying, it gives

  $\Rightarrow K=\sqrt{9x\left(4x\right)\left(3x\right)\left(2x\right)}$

  $\Rightarrow K=\sqrt{36\times 6x^4}$

  $\Rightarrow K=6x^2\sqrt{6}$

By substituting the value of the area $K=54\sqrt{6}$, it gives

  $\Rightarrow 54\sqrt{6}=6x^2\sqrt{6}$

Divide by $6\sqrt{6}$ on both sides.

  $\Rightarrow 9=x^2$

  $\Rightarrow x^2=9$

  $\Rightarrow x=\pm 3$

As the sides cannot be negative, hence $x=3$

Substitute $x=3$, in the expression of sides $5x,6x$, and $7x$ to get the length of the sides of the triangle.

  $5\times 3=15,6\times 3=18,\text{and}7\times 3=21$ .

Therefore, the lengths of the sides of the triangle are $15$ units, $18$ units, and $21$ units respectively.