# [SOLVED] Find the area of rectangle ABCD

• Sep 29th 2008, 04:09 PM
fabxx
[SOLVED] Find the area of rectangle ABCD
Hi, this is on the SAT OG Book page 747 #16

In rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\displaystyle \frac{2}{3}$, what is the area of the rectangle ABCD?

a) $\displaystyle \frac{1}{2}$

b)$\displaystyle \frac{3}{4}$

c)$\displaystyle \frac{8}{9}$

d) 1

e)$\displaystyle \frac{8}{3}$
• Sep 29th 2008, 04:17 PM
Jhevon
Quote:

Originally Posted by fabxx
Hi, this is on the SAT OG Book page 747 #16

In rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\displaystyle \frac{2}{3}$, what is the area of the rectangle ABCD?

a) $\displaystyle \frac{1}{2}$

b)$\displaystyle \frac{3}{4}$

c)$\displaystyle \frac{8}{9}$

d) 1

e)$\displaystyle \frac{8}{3}$

note that the area of rectangle ABCD is given by AD*AB

now, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so

=> (BE + AD)*AB = 4/3

but, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that

• Sep 29th 2008, 04:37 PM
Soroban
Hello, fabxx!

Did you make a sketch?

Quote:

In rectangle ABCD, point E is the midpoint of line segment BC.
If the area of $\displaystyle ABED$ is $\displaystyle \frac{2}{3}$, what is the area of $\displaystyle ABCD$?

$\displaystyle (a)\;\frac{1}{2} \qquad(b)\;\frac{3}{4}\qquad (c)\;\frac{8}{9} \qquad (d)\;1 \qquad (e) \;\frac{8}{3}$

Code:

    D *---------------* C       |  *          |       |      *      |       |          *  |     F * - - - - - - - * E       |              |       |              |       |              |     A *---------------* B

Draw median EF.

$\displaystyle \text{Rect }DCEF \:=\:\frac{1}{2}(\text{Rect }ABCD)$

$\displaystyle \Delta DCE \:=\:\frac{1}{2}(\text{Rect }DCEF) \;=\;\frac{1}{4}(\text{Rect }ABCD)$

. . Hence: .$\displaystyle \text{Quad }ABED \:=\:\frac{3}{4}(\text{Rect }ABCD)$

We are told that: .$\displaystyle \text{Quad }ABED \:=\:\frac{2}{3}$

We have: .$\displaystyle \frac{3}{4}(\text{Rect }ABCD) \;=\;\frac{2}{3} \quad\Rightarrow\quad \boxed{\text{Rect }ABCD \:=\:\frac{8}{9}}$ . . . answer (c)

• Sep 29th 2008, 05:08 PM
fabxx
Quote:

Originally Posted by Jhevon
note that the area of rectangle ABCD is given by AD*AB

now, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so

=> (BE + AD)*AB = 4/3

but, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that