+248 Trapezoid $ABCD$ has vertices $A(-1,0)$, $B(0,4)$, $C(m,4)$ and $D(k,0)$, with $m>0$ and $k>0$. The line $y = -x + 4$ is perpendicular to the line containing side $CD$, and the area of trapezoid $ABCD$ is 34 square units. What is the value of $k$?
Since we are given that the line y=−x+4 is perpendicular to the line containing side CD, we know that the slope of the line CD is −1. Therefore, we can use the point-slope form of linear equations to find the equation of the line containing side CD.
We know that one point on the line is (m,4), and we know that the slope of the line is −1. So, we get the following equation: \begin{align*} y - 4 &= -1(x - m)\ y &= -x + m + 4 \end{align*} We are also given that the area of the trapezoid is 34 square units, so we can use the formula for the area of a trapezoid to get the following equation: \begin{align*} \frac{1}{2}(1 + m)(4 + 0) &= 34\ 2 + m &= 34\ m &= 32 \end{align*} Now that we know m=32, we can plug it back into the equation y=−x+m+4 to get the equation of the line containing CD: \begin{align*} y &= -x + 32 + 4\ y &= -x + 36 \end{align*} Finally, we can plug the coordinates of point D, (k,0), into this equation to solve for k: \begin{align*} 0 &= -k + 36\\ k &= \boxed{36} \end{align*}
