+4622 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 \(k\)?
+129899 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 :
