Thread: Rectangles

Can someone help me with these, please

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).

Please help

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.

What's with the shouting?

Originally Posted by doomdesire34

Can someone help me with these, please

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

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

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?

sorry, just gimme a few more minutes, thanks for the reply

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

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?

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

Thank you very much for helping, I have a feeling I'm going to be the only person with these done for tommorrow