[SOLVED] Find the area of rectangle ABCD

**fabxx** · September 29th 2008, 04:09 PM

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) $\frac{8}{3}$

**Jhevon** · September 29th 2008, 04:17 PM

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

**Soroban** · September 29th 2008, 04:37 PM

Hello, fabxx!

Did you make a sketch?

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)

**fabxx** · September 29th 2008, 05:08 PM

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?

**Jhevon** · September 29th 2008, 05:21 PM

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

Thread: [SOLVED] Find the area of rectangle ABCD

LinkBack

Thread Tools

Display

[SOLVED] Find the area of rectangle ABCD

Similar Math Help Forum Discussions

Find the area of the rectangle

If ABCD is a rectangle and P is any point its interior, prove" AP²+PC² =PB²+PD²

How Do You Find The Maximum Area Of A Rectangle In An Eclipse?

Find Area of Rectangle

Find rectangle with largest area

Search Tags