<p><a href="http://img137.imageshack.us/my.php?image=abe3qv.png%5B/url%5D">http://img137.imageshack.us/my.php?image=abe3qv.png</a>
not drawn to scale!</p>
<p>BD = 6, <strong><em>AD = 8</em></strong>, BC = 4</p>
<p>what is the area of ABE?</p>
<p>edit: AD = 8. ooops! LOL</p>
<p>yea........</p>
<hr>
<p>it's 8. ...</p>
<p>cmon...</p>
<p>the quality is horrible</p>
<p>Ok. The answer is 8.</p>
<p>It helps to redraw the figure to scale. By the Pythagorean Theorem, AB = 10 (6^2 + 8^2 = 10^2 or 2<em>3, 2</em>4, 2*5 trangle). Angle BEC is equal to angle DEA (vertical angles). Both angle EBC and angle EDA are right angles, so they are also congruent. Because they are right angles, BC is parallel to AD. So angle EAD is equal to angle ECB. Now we know that all of the corresponding angles are equal, so triangle ADE is similar to triangle CBE.</p>
<p>We know BD = 6. So we label BE x and DE (6-x). Now we set up the proportion AD/DE = BC/BE. 8/(6-x) = 4/x.
By crossmultiplication, 8x = 4(6-x), so <em>x = 24 -4x, 12x = 24, and x = 2. x is the length of BE, which is the base of AEB (our target triangle). Base</em>Height/2 = Area, so 2*8/2 = 8, which is the area of AEB.</p>
<p>good job flipsta_G. this is definitely a hard one.</p>
<p>Another approach (just as an illustration of multiple ways to tackle SAT questions).</p>
<p>1.
Area of ABD:
A(ABD) = 1/2 (6) (8) = 24.</p>
<p>2.
Right triangles AED and BEC are similar because they have one pair of corresponding congruent acute angles (AED and BEC).
BE:ED = BC:AD = 4:8</p>
<p>3.
Triangles ABE and AED have the same height AD, therefore
A(ABE) : A(AED) = BE:ED = 4:8
[Helpful fact to know. Here's the proof:
(1/2 BE AD) : (1/2 ED AD) = BE:ED]</p>
<p>4.
Since A(ABE) + A(AED) = A(ABD) = 24 and
A(ABE) : A(AED) = 4:8 = 8:16,
A(ABE) = 8.</p>
<p>By the way, 3:4:5 = 6:8:10 triangle BDE is a good exapmple of distracting SAT data.</p>
<p>OOPS</p>
<p>read triangle BDA instead in my previous post.</p>
<p>Speaking of distractions!</p>
