+0  
 
+1
72
3
avatar+2712 

Here's a question that I came across:

 

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 \(k\)?

tertre  Mar 30, 2018
 #1
avatar+86890 
+2

C  = (m, 4)

D   = (k, 0 )

Slope of line containing CD

(4 - 0)

_____   =  1

 m  - k

 

4  =  m  - k

m - k  = 4     (1)  

 

Height of trapezoid, H,  = 4

 

Area  =  (1/2)H * (sum of the bases)

34 = (1/2)* 4  (sum of the bases)

34   = 2(sum of the bases)

17  =  sum of the bases

 

But   the length of the bottom base  is    k + 1  and the length of the top base is m

So

 

(k + 1) +  m  =  17

m + k  = 16     (2)

 

And using (1)  and (2)   and we have that

 

m - k  =  4

m + k  =  16    add  these

 

2m  =  20

m  =  10

 

So....k  = 6

 

Note....the equation of the perpendicular line is y  = x - 6 

 

Here's a pic :

 

 

 

cool cool cool

CPhill  Mar 30, 2018
 #2
avatar+2712 
+2

Wow! Thanks so much, CPhill! Great job! Fabulous!

tertre  Mar 30, 2018
 #3
avatar+86890 
0

Thanks, tertre   !!!!

 

 

cool cool cool

CPhill  Mar 30, 2018

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