# [MOVED] Geom. figure: What are dimensions of original rect?

#### vonsmiley

##### New member
Does someone have time to help me with this?

The width of a rectangle is three-fourths of its length. If we increase the length by eigth feet and the width by six feet, the area is increased by ninety-six square feet. What are the dimensions of the orignal rectangle?

I understand:

the length is (L) so the w must = (L times 3/4)

but is this is true how can the with and length be increased by different amounts and have this remain unchanged.
How do I set this up?

#### stapel

##### Super Moderator
Staff member
The original width is 3/4 the original length. Obviously this will be different once the measurements are changed.

Draw the rectangle. Label the length and the length with your variable and expression.

What is the expression for the area?

Draw the longer length and wider width along the sides of the original rectangle, so you have an added "L" shape wrapped around two sides. Write expressions, in terms of the original variable and expression, for the new width and length.

What is the expression for the total area?

The increase is the difference between the original area and the new area. So subtract the old from the new, and set the resulting expression equal to "96". Solve.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you.

Eliz.

#### vonsmiley

##### New member
Code:
|   8L
|
|
|
------------------
6(8L)
|
| L
|
-------------
3/4L
3/4L times L = AREAorig
8L times 6(8L) = AREAnew

do you mean 3/4L times L minus 8L times 6(8L) = 96

I just can't get it to equal length of rect =4ft
the width of the rect equal = 3ft

#### stapel

##### Super Moderator
Staff member
I'm sorry, but I can't follow your drawings...?

The original area A<sub>orig</sub> is the product of the original length L and the original width (3/4)L. That is:

. . . . .A<sub>orig</sub> = (3/4)L<sup>2</sup>

The new area A<sub>new</sub> is the product of the new length and the new width.

Since the length is increased by eight feet (so it is eight feet longer), I'm not sure why you're multiplying by 8...? Or is there a typo in your post, and the new length is actually "increased to eight times" the original length...? Same question regarding the new width.

Eliz.

#### vonsmiley

##### New member
Showing my ignorance:

It's (+) not (x) and I changed it but I still cann't slove it. I just don't understand why you say set 96 = to (L + 8) x 3/4(L + 8) + 6 #### stapel

##### Super Moderator
Staff member
vonsmiley said:
96 = to (L + 8) x 3/4(L + 8) + 6
The new width is six more than the old width, not six more than 3/4 of the new length. there should not be any "plus eight" in the new width. Then find the difference of the two areas:

. . . . .(L + 8)((3/4)L + 6) - (L)((3/4)L) = 96

To find the dimensions, you just need to solve the equation.

Eliz.

#### Denis

##### Senior Member
I think Stapel misread...area increases by 96...

L(3L/4) + 96 = (L + 8)(3L/4 + 6)
Start by mulyiplying through by 4 to get rid of fractions:
L(3L) + 384 = (L + 8)(3L + 24) ; do the multiplications:
3L^2 + 384 = 3L^2 + 48L + 192

Finish it.

#### vonsmiley

##### New member
The way you wrote this [L(3L/4) + 96 = (L + 8)(3L/4 + 6) ] is confusing.

and why did you leave out (+8) from the second half with the (+6) and why did you only us 96 on one half of the equation instead of the whole thing?

It doens't help me to finish the problem if I don't unterstand how you got there.

#### vonsmiley

##### New member
Even though I can't get it to work, shouldn't the formula look like this
(L + 8)(3/4(L+8)+6)+96=Area

#### stapel

##### Super Moderator
Staff member
vonsmiley said:
The way you wrote this [L(3L/4) + 96 = (L + 8)(3L/4 + 6) ] is confusing.
Why? It says that the old area, plus 96 units, equals the new area. Isn't that what the exercise states?

vonsmiley said:
and why did you leave out (+8) from the second half with the (+6)
Because (as was pointed out earlier) the new width is six more than the old width, not six more than the new length.

It might help if you labelled your variables and expressions, so you could keep track of what they mean.

vonsmiley said:
...shouldn't the formula look like this
(L + 8)(3/4(L+8)+6)+96=Area
This translates as:

. . . . .(the new length) times (six more than 3/4 the new length) plus (96) is (another variable)

Is that what the exercise says?

Eliz.

#### Denis

##### Senior Member
Smiley, let's change your problem a little, to this:

The length of a rectangle is twice its width. If we increase the length by 8 feet
and the width by 6 feet, the area is increased by 88 square feet.
What are the dimensions of the orignal rectangle?

Let width of original rectangle = w
Then length of original rectangle = 2w
So area of original rectangle = w * 2w = 2w^2 ; you ok with that :?:

length of new rectangle = 2w + 8 : right?
width of new rectangle = w + 6 : right?
So area of new rectangle = (2w + 8) * (w + 6) = 2w^2 + 20w + 48 : OK??

Since area of original rectangle is increased by 88, then:
2w^2 + 88 = 2w^2 + 20w + 48 : still with me?
so: 20w = 40 : w = 2
So width original rectangle = 2 and length = 4, right?

And width new rectangle = 2+6 = 8, length = 4+8 = 12, right?

Area new rectangle = 8 * 12 = 96
Area original rect. = 2 * 4 = 8
96 - 8 = 88 : OK?

Now go do yours...[you're lucky Ottawa Senators beat Buffalo earlier!]