Question 1150783
<br>
Let Q be the midpoint of BC, so that PQ is the midsegment of the trapezoid.<br>
Let AB=2x and CD=2y; and let the height of the trapezoid be 2h.<br>
Then the altitudes of triangles ABP and CDP are both h.<br>
The area of triangle ABP is one-half base times height: {{{(1/2)(2x)(h) = xh = 5}}}<br>
The area of triangle CDP is one-half base times height: {{{(1/2)(2y)(h) = yh = 17}}}<br>
The area of the trapezoid is the average of the bases times the height: {{{((2x+2y)/2)(2h) = (2x+2y)(h) = 2(xh+yh) = 2(5+17) = 2(22) = 44}}}<br>
ANSWER: The area of the trapezoid is 44<br>