# Rectangles

• Apr 24th 2007, 03:46 PM
doomdesire34
Rectangles
Can someone help me with these, please :confused:

1. The length of a rectangle is 6cm less than twice its width. Find the dimensions of a rectangle if its area is 108cm^2 (squared)?

2. Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m^2 (squared).

3. Find the dimensions of a rectangle whose perimeter is 42m and whose area is 104^2 (squared).

EDIT: Please help, I don't understand it at all. Our teacher always starts a new chapter on the day of the previous test and between periods we don't get anytime to comprehend much.
• Apr 24th 2007, 04:00 PM
Jhevon
What's with the shouting?

Quote:

Originally Posted by doomdesire34
Can someone help me with these, please :confused:

1. The length of a rectangle os 6cm less than twice its width. Find the dimensions of a rectangle if its area is 108cm^2 (squared)?

let the width be x
then the length is 2x - 6

now area is length times width, and we are told that is 108
=> length*width = 108
=> (2x - 6)x = 108
=> 2x^2 - 6x = 108
=> 2x^2 - 6x - 108 = 0
=> x^2 - 3x - 54 = 0
=> (x - 9)(x + 6) = 0
so x - 9 = 0 => x = 9
or x + 6 = 0 => x = -6

x cannot be -6, so x = 9

so the dimensions of the rectangle is: width = 9cm, length = 2x - 6 = 12cm

Quote:

2. Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m^2 (squared).
let the sides of the rectangle be x and y
let the perimeter be P
let the area be A

now P = 2x + 2y = 46
and A = xy = 126

so to find x and y, we need to solve the simultaneous equations:

2x + 2y = 46 ...................(1)
xy = 126 ........................(2)

x + y = 23 ..................(3) = (1)/2
xy = 126 .....................(2)

From (2), we see that x = 126/y, substitute 126/y for x in (3), we get:

126/y + y = 23
=> 126 + y^2 = 23y ...........i multiplied through by y
=> y^2 - 23y + 126 = 0
=> (y - 14)(y - 9) = 0
=> y - 14 = 0 => y = 14
or y - 9 = 0 => y = 9

but x + y = 23
=> x = 14 ...........when y = 9
or x = 9 .............when y = 14

x and y represent arbitrary sides, and the numbers are the same, so just pick one

so one side is 14 and the other is 9
you can say x = 14 and y = 9 or x = 9 and y = 14, same thing

Quote:

3. Find the dimensions of a rectangle whose perimeter is 42m and whose area is 104^2 (squared).
this is exactly like the one above. why don't you try it and post your solution?
• Apr 24th 2007, 04:12 PM
doomdesire34
One minute
sorry, just gimme a few more minutes, thanks for the reply
• Apr 24th 2007, 04:16 PM
doomdesire34
ok I have the sides as x and y and the
P = 2x + 2y = 42
A = xy = 104

2x + 2y =42
xy = 104

104y + y = 21
104 + y^2 = 21y
y^2 - 21y + 104 = 0
(y - 13) (y - 8) = 0
y - 13 = 0 y = 13
y - 8 = 0 y = 8

Dimensions: 13m, 8m
• Apr 24th 2007, 04:29 PM
doomdesire34
ok I have the sides as x and y and the
P = 2x + 2y = 42
A = xy = 104

2x + 2y =42
xy = 104

104y + y = 21
104 + y^2 = 21y
y^2 - 21y + 104 = 0
(y - 13) (y - 8) = 0
y - 13 = 0 y = 13
y - 8 = 0 y = 8

Dimensions: 13m, 8mits from above, sorry I just edited the "sorry I'm stuck" one to the correct problem. Is it right?
• Apr 24th 2007, 04:34 PM
Jhevon
Quote:

Originally Posted by doomdesire34
ok I have the sides as x and y and the
P = 2x + 2y = 42
A = xy = 104

2x + 2y =42
xy = 104

104y + y = 21
104 + y^2 = 21y
y^2 - 21y + 104 = 0
(y - 13) (y - 8) = 0
y - 13 = 0 y = 13
y - 8 = 0 y = 8

Dimensions: 13m, 8m

its from above, sorry I just edited the "sorry I'm stuck" one to the correct problem. Is it right?

Yup, that seems ok. good job:D
• Apr 24th 2007, 04:35 PM
doomdesire34
Thank you very much for helping, I have a feeling I'm going to be the only person with these done for tommorrow :D