Last question.... Algebra 1?

Question

A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle.

Answer 1

Dimensions of original rectangle:
W, 4W

Dimensions of second rectangle:
(W+2), (4W + 5)

Area of second rectangle is 270 cm^2 greater than the first:

W*4W + 270 = (W+2)(4W+5)
4W^2 + 270 = 4W^2 + 8W + 5W + 10
270 - 10 = 13W
260 = 13W
20 = W

The original rectangle is W by 4W, which is 20cm by 80cm.

Just to check, let's check out the areas:

W*4W = 20*80 = 1,600 square cm

The other rectangle is 2cm wider (20+2 = 22), and 5 cm longer (80 + 5 = 85), so it is 22cm by 85cm:

22 * 85 = 1,870 square cm

The difference between 1,870 and 1,600 is indeed 270 square cm.

Answer 2

So let the first rectangle be R1 and the second be R2 (the lengths, widths, and areas will be denoted as L, W, and A respectively).

So L1=4W1

L2=L1+5

W2=W1+2

A2=A1+270=(L1)(W1)+270=(L2)(W2)

(L1)(W1)+270=(L1+5)(W1+2)
L1=4W1, so

(4W1)(W1)+270=(4W1+5)(W1+2)

4(W1)^2+270=4(W1)^2+13W1+10

13W1=260

W1=20

So the width of the original rectangle is 20.
L1=4W1=(4)(20)=80

So the length of the original rectangle is 80.

The dimensions are 20 x 80.

Answer 3

4x^2+270=(4x+5)(x+2)
4x^2+270=4x^2+13x+10
13x=260
x=20
4x=80
The original rectangle was a 20 X 80 rectangle.

Answer 4

let l be the long of first rectangle
let w be the wide of first rectangle
then l=4w
the second rectangle has dimensions l+5, w+2
the area of second rectangle is
(l+5)(w+2)=(4w+5)(w+2)=4w^2+13w+10
the area of the first rectangle is l*w=4w^2
The diff. between area of rectangles is
13w+10=270
w=20
l=80