+0  
 
+1
40
1
avatar

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
avatar+87293 
+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 :

 

 

 

 

cool cool cool

CPhill  Jun 24, 2018

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