Question 142911
{{{drawing( 400, 300, -10, 190, -10, 140,
  locate(30, 30, A), locate(150, 30, B), locate(135, 110, C), locate(40, 110, D), locate(130, 30, E),
  line(30, 30, 150, 30),
  line(150, 30, 130, 110),
  line(130, 110, 50, 110),
  line(50, 110, 30, 30),
  line(50, 110, 150, 30),
  line(130, 110, 30, 30),
  line(130, 110, 130, 30)
  )}}}

Let the trapezoid be ABCD. 
Given: AB = 15 units, CD = 9 units, Area of trapezoid ABCD = 27 sq units.
Draw CE, perpendicular to AB from C at E. Let CE = h units.
Observe that CE is the height for both the triangles ABC and ACD as their bases are parallel.


Area of trapezoid ABCD 
= Area of triangle ABC + Area of triangle ACD
= {{{(1/2) x 15 x h + (1/2) x 9 x h}}} sq units [since {{{AREA = 1/2}}}{{{x}}}{{{BASE}}}{{{x}}}{{{ALTITUDE}}}]
= {{{(1/2)(15h + 9h)}}} sq units
= {{{(1/2)24h}}} sq units
= {{{12h}}} sq units


But area of trapezoid ABCD is given to be 27 sq units.
So, {{{12h = 27}}}
{{{h = 9/4 = 2.25}}} units


Thus, required height is 2.25 units.