Question 1199005
<br>
Here is a sketch, showing altitudes DE and CF:<br>
{{{drawing(400,200,0,50,0,20
,line(2,2,42,2),line(2,2,10,17),line(10,17,22,17),line(22,17,42,2)
,line(10,2,10,17),line(22,2,22,17)
,locate(1,2,A),locate(42,2,B),locate(22,19,C),locate(9,19,D),locate(9,2,E),locate(21,2,F)
,locate(4,12,17),locate(16,19,12),locate(32,12,25),locate(11,9,h)
)}}}<br>
The quick path to the answer is to guess that all the numbers in the problem are whole numbers.  So look for Pythagorean triples.  With AD=17, it is likely that AE is 8 and the altitude h is 15; and with the altitude 15, FB would be 20.  And 8+12+20 = 40, as required.<br>
So the bases are 12 and 40, and the height is 15; the area is 15*((12+40)/2) = 15*26 = 390.<br>
ANSWER: 390<br>
If you want a formal algebraic solution....<br>
{{{AE = sqrt(17^2-h^2)}}}
{{{FB = sqrt(25^2-h^2)}}}<br>
Then<br>
{{{sqrt(17^2-h^2)+12+sqrt(25^2-h^2)=40}}}<br>
I won't go through the whole process of solving that equation....<br>
To solve it, isolate one of the square roots on one side of the equation and square both sides.  That will leave you still with one square root; again isolate it and square both sides.  Now you will have no square roots left; and the remaining equation can be solved for the altitude h with only mildly ugly arithmetic.<br>