Question

the length of a rectangle is 3 less than 5 times its width.
Write a simpified algebriac expression for the perimeter of a rectangle.
If the rectangle width is tripled and its length is doubled,the perimeter of new rectangle is 92cm greater than the original perimeter.
Find the area of the original rectangle.

Answer 1

original width ---> x
original width ---> 5x-3

Perimeter = 2x +2(5x-3) = ....

new width --->3x
new length ---> 2(5x-3)

so 2(3x) + 2(5x-3) = 92
solve for x
....

then evaluate x(5x-3)

let me know what you get.

Answer 2

Hi aisha,

Let's start with formulating an algebriac expression for the perimeter.
We know that perimeter of rectangle = 2(length) + 2(width).

Since we don't know length or width, we let x be the width of the rectangle.
Hence length = 5x-3 [because length = (5 X width) - 3].
So perimeter = 2[5x-3]+2(x) = 10x-6+2x= 12x-6. That answers the first part.

If width is tripled, new width = 3x.
If length is doubled, new length = 2(5x-
3) = 10x-6. So new perimeter = 2(3x)+2(10x-6) = 6x+20x-12= 26x-12.

Now, perimeter of new rectangle - perimeter of original rectangle = 92cm.
So [26x-12]-[12x-6] = 92cm
Simplifying the left hand side, 14x-6=92cm
Therefore, 24x=98cm, x=98/14=7cm.

So to find area of original rectangle, it is length X width = (5x-3)(x) = 5x^2 - 3x = 5(7X7) - 3(7) = 224cm^2
Hope I helped! (:

-J

Answer 3

Go with J's solution, I noticed an error near the end of mine,
my equation should have been
2(3x) + 4(5x-3) = 92

Answer 4

Oops sorry! I realised that I also made a mistake near the end.
When I said "...Therefore, 24x=98cm.." I meant 14x (Not 24!) = 98cm. =)

Answer 5

The length of a rectangle is 3cm less than twice the width.The perimeter of the rectangles of the sides of the rectangle is 24cm.Find the lengths of the sides of the rectangle.

Answer 6

In a given rectangle,the length is two less than four times the width,w.Find the expression that represents the area and the perimeter.