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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 $k$?

Guest Jun 24, 2018

#1**+1 **

Since the line y = -x + 4 has a slope of -1, then the line containing CD has a negative reciprocal slope = 1

Therefore....the slope of this line is

(4 - 0) / ( m - k) = 1

4 = m - k

4 + k = m

Now...the height of the trapezoid is 4

And the length of one base is k - - 1 = k + 1

And the length of the other base is 4 + k

So....the area of the trapezoid can be represented as

(1/2) height [ length of base 1 + length of base 2 ] = 34

(1/2) 4 * [ ( k + 1) + ( 4 + k) ] =34

2 * [ 2k + 5 ] = 34

2k + 5 = 17

2k =12

k = 6

Here is a pic :

CPhill Jun 24, 2018