# [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 $\frac{2}{3}$, what is the area of the rectangle ABCD?

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

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

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

d) 1

e) $\frac{8}{3}$
• 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 $\frac{2}{3}$, what is the area of the rectangle ABCD?

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

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

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

d) 1

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

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

(1/2)(BE + AD)*AB = 2/3

=> (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

=> (3/2)AD*AB = 4/3

=> AD*AB = 8/9
• 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 $ABED$ is $\frac{2}{3}$, what is the area of $ABCD$?

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

Code:

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

Draw median EF.

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

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

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

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

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

(1/2)(BE + AD)*AB = 2/3

=> (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

=> (3/2)AD*AB = 4/3

=> AD*AB = 8/9

No figure is given for this. If i were to draw one myself, would it look like the attachment below?
• Sep 29th 2008, 05:21 PM
Jhevon
Quote:

Originally Posted by fabxx
No figure is given for this. If i were to draw one myself, would it look like the attachment below?

yes, it should look like that. note that it is the same as the diagram Soroban drew, except it is turned on its side. i imagined it as Soroban did. it will all work out the same. if your diagram was like the given one or Soroban's you will get the right answer