Help with finding height of trapezoid with triangles.
I need to find the value for h. I know I need to use cos or sin or tan but Im just not really sure where to start. I am pretty sure that the answer will be in between 30 and 40. Given: that AB is 40 and CD is 60. M and N are the midpoints of AB and CD respectively. Lines that appear horizontal and vertical are. Please show work! Thank you! :)
here is a link to the picture of it is to small: http://oi45.tinypic.com/2vnkqv4.jpg
Re: Help with finding height of trapezoid with triangles.
You have shown no attempt of your own to do this so I have no idea what formulas or methods you might have to do such a problem and all I can do is say how I would do such a problem- with "analytic geometry". If we take the distance from B to D to be "L", we can set up a coordinate system so that B is at (0, 0) and D is at (L, 0). Then M is at (0, 20) and C is at (L, 60) so the line through M and C is given by y= (40/L)x+ 20. Also A is at (0, 40) and N is at (L, 30) so the line through A and N is given by y= (-10/L)x+ 40. Those lines intersect when (40/L)x+ 20= (-10/L)x+ 40 or (50/L)x= 20, x= (2/5)L. So h= (-10/L)(2/5)L+ 40= 36. As a check h= (40/L)(2/5)L+ 20= 36.
(Initially, I thought we could not do this without knowing the distance from B to D, L. But notice that L cancels out.)