 Here we want to find the area of this irregular quadrilateral, ABCD, and we know that the length AC is 35 units, BF is 18, and DE is 10. So the thing to recognize with this is that ABCD can be broken up into two triangles. There's triangle ABC and triangle ADC. So if we consider the red triangle, that is triangle ABC, for that triangle we see that its base could be AC, in other words its base is 35, and the height, remember the height is perpendicular to the base, and so the height can be BF, so times 18. And one half times 35 times 18 is 315 square units. Now alternatively for the triangle down below for triangle ADC, even though AC is not on the bottom of that triangle, we can use it as the base because DE is a known length and DE is perpendicular to AC. So if we use AC as the base, then we have one half times 35 once again times DE, which is 10, and so that gives us 175, and so the total area of the two triangles is the sum of the two, sorry, the total area of the quadrilateral ABCD is the sum of 315 and 175. And so our total area is 490 square units, and so all we had to do is split it up into two triangles. We had the red triangle, which is this entire triangle, and the blue triangle, which is sort of an upside down triangle with the base of AC.