<p><a href="http://www.majortests.com/sat/testpics/p001-10.gif%5B/url%5D">http://www.majortests.com/sat/testpics/p001-10.gif</a>
In the figure above AD = 4, AB = 3 and CD = 9. What is the area of triangle AEC?</p>
<p>Triangle ABE is similar to Triangle CED, so since AB =3 and CD = 9 (3x larger), AE = 1 and ED = 3 (and AD=4 as given). </p>
<p>Right triangle ADC has area of square root [(4<em>4) + (9</em>9)] and right triangle CED has area of square root [(3<em>3) + (9</em>9)] </p>
<p>Area of Triangle ACD - Area of triangle ECD = Area of triangle AEC:
sqrt(97) - sqrt(90) = 9.85-9.49 = .36 sq. units</p>
<p>^^ um im getting something much different.</p>
<p>okie we know the givens plus we have to know that AB=ED which is 3, and if ED is 3, then AE must be 1, since AD = 4. Correct?, now</p>
<p>Now, AB is not needed so i guess its just there to throw us off??..but yeah anyways..::</p>
<p>Area of AEC= ACD - ECD (because AEC is 1 part of ACD and ECD is the other part..by taking the whole ACD, subtracting from one part, ECD, we can find the other part), we can easily find the areas of ACD and ECD by the forumula (.5 B * h), (u dunt need to use any pythag therom)</p>
<p>so we get...</p>
<p>.5(4)(9) - .5(3)(9) = 18-13.5 = 4.5</p>
<p>now is that one of the choices? i think thats right thats how i would do it, especially if its a SAT I math problem, it shouldnt be that complicated.</p>
<p>^^how did you get the three for .5(3)(8)..I'm getting a differen't answer.</p>
<p>oh. I just found the answer..went to the site. Yeah it's 4.5</p>
<p>eg1, you were subtracting the hypotenuses, not the areas.</p>
<p>
</p>
<p>In triangle AEC:
AE = 1 ...... base,
CD = 9 ...... height
Area(AEC) = (AE) (CD) / 2 = 4.5</p>
<p>Dunkin Donuts - Yes - you are right - I wasn't paying enough attention. Oops.</p>