<p>Can someone help me with number 16 on page 747. It is a math problem and is not located in the consolidated math thread or any other posts. I am very good with the math SAT questions, but this one I found challenging. </p>

<p>In rectangle ABCD, point E is the midpoint of BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD?</p>

<p>A- 1/2
B- 3/4
C-8/9
D-1
E-8/3</p>

<p>I was able to knock out A automatically since the area of the rectangle has to be more than the area of the quad. I also knocked out E because it was way off. So... that left me with B,C,or D. I am not sure mathematically how to solve it and I cant find any tricks. Help.</p>

<p>I'm not sure if this is right, but i just split the rectangle into 4 equal triangles equal to EDC by drawing a line to the midpoint of AD and A, both from E. Since ABED is made up of 3 of these triangles u get get 3X=2/3 so the area of 1 of the triangles is 2/9. Multiply that by how many triangles there are u get 8/9, C.</p>

<p>If you draw quadrilateral ABED, you will find that it consists of one smaller rectangle and half of a smaller rectangle. (The two smaller rectangles, both equal make up rectangle ABCD.) So therefore I went to find what the area the triangle was of quadrilateral ABED. It was (2/3)/3, which was 2/9. Now, if you can visualize that the rectangle can be cut to 4 equal trangles, and the area of one triangle is 2/9, 4(2/9) equals 8/9. My work mainly focused on that in quadrilateral ABED, you could form one triangle and a rectangle, and so I focused on the triangle.</p>

<p>Drop a perpendicular EF from E onto AD.
ABED consists of three equal triangles
ABE, EFA, and EFD, since BE=EC and AF=FD,
and ABCD has 4 of them
(ABE, EFA, EFD, and ECD).
Proportion:
3 triangles --------- area 2/3
4 triangles --------- area x.</p>

<p>x = 2/3 * 4 / 3 = 8/9.</p>

<p>Drawing an additional element in difficult geometry questions often helps.</p>