Question

For the following figure, given that $ABCD$is a trapezoid with median $\overline{MN}$.

If $DC=3x,AB=2x^2,EF=7$ then find the value of x.

Short answer

The value is $x=3.5$.

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

Since $MN$is a median therefore

$\begin{array}{l}ME=\frac{1}{2}DC\left(1\right)\\ MF=\frac{1}{2}AB\left(2\right)\end{array}$

Step 3. Step description.

From the equation (1) and (2) as follows:

$\begin{array}{l}EF=MF-ME\\ EF=\frac{1}{2}AB-\frac{1}{2}DC\\ EF=\frac{1}{2}\left(AB-DC\right)\end{array}$

Step 4. Step description.

Substitute all the values in the formula $EF=\frac{1}{2}\left(AB-DC\right)$and simplify as follows:

$\begin{array}{c}EF=\frac{1}{2}\left(AB-DC\right)\\ 7=\frac{1}{2}\left(2x^2-3x\right)\\ 14=2x^2-3x\\ 2x^2-3x-14=0\\ x\left(2x-7\right)+2\left(2x-7\right)=0\\ \left(x+2\right)\left(2x-7\right)=0\\ x=-2,\frac{7}{2}\\ x=-2,3.5\end{array}$

Since a side cannot be negative therefore the value of x is $3.5$.