Chapter 1.7, Problem 38E

Solution

To find: the value of $x$ .

  $x=-2+2\sqrt{21}$ .

Given:

The base lengths of the trapezoid are $3x-1$ and $x+9$ . Its height is $x$ .

Concept used:

The area of a trapezoid with bases $b_1$ , $b_2$ and height $h$ is $A=\frac{1}{2}\left(b_1+b_2\right)h$ .

Calculation:

The area of the given trapezoid is,

  $\begin{array}{c}\frac{1}{2}x\left[\left(3x-1\right)+\left(x+9\right)\right]=20\\ x\left(4x+8\right)=40\\ 4x\left(x+2\right)=40\\ x^2+2x=10\end{array}$

Now adding ${\left(\frac{2}{2}\right)}^2=1$ on both sides, it follows that:

  $\begin{array}{c}{x}^2+2x+1=10+1\\ {\left(x+1\right)}^2=11\\ x+1=\pm \sqrt{11}\\ x=-1\pm \sqrt{11}\end{array}$

Since $x=-1-\sqrt{11}$ is negative, therefore it is not possible as the length is always positive.

Therefore,

  $x=-1+\sqrt{11}$ .