rectangle is 4 timesas along as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider

Question

rectangle is 4 times as along 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

Hi,

Original rectangle has length and width of 4x and x.

Length of new rectangle is 4x + 5 and width is x + 2.

(4x + 5)(x + 2) = 270
4x² + 13x + 10 = 270
4x² + 13x - 260 = 0

Solving by the quadratic formula gives:
.............____________
-13 +- √13² - 4(4)(-260)
-------------------------------- =
.............2(4)

............._________
-13 +- √169 + 4160
--------------------------- =
.............8

............._________
-13 +- √169 + 4160
--------------------------- = 6.599
.............8

The width of the original rectangle is 6.599 and the length is 26.396.

I hope that helps!! :-)

Answer 2

First thing is to translate this from English into math notation:

"[A] rectangle is 4 times as long as it is wide."
Let ℓ be the length and let w be the width. Then ℓ = 4w.

"A second rectangle is 5 centimeters longer and 2 centimeters wider than the first." Translation: The dimensions of this second rectangle are ℓ+5 and w+2. If you want these dimensions strictly in terms of w, then substitute 4w for every occurrence of ℓ to get dimensions of 4w + 5 and w + 2.

"The area of the second rectangle is 270 square centimeters greater than the first." The area of the second rectangle will be (4w + 5)(w + 2) and the area of the first rectangle is (4w)(w), so this sentence translates to the equation
(4w+5)(w+2) = (4w)(w) + 270

"What are the dimensions of the original rectangle" means solve for w, then use that value to get ℓ.

We now simplify both sides of the area equation by FOILing the left side to get
(4w)(w) + (4w)(2) + (5)(w) + (5)(2) = (4w)(w) + 270
4w² + 8w + 5w + 10 = 4w² + 270
Although this looks like a quadratic equation, the quadratic term 4w² can be subtracted from both sides to leave a linear equation:
8w + 5w + 10 = 270
13w + 10 = 270

Now we have simplified each side as much as it can be and it is time to move things across the equals sign. Because all the terms in w are on the left hand side, we elect to leave them there and move everything else to the right side. Normally, you undo things in reverse order in which you do them. The add 10 was the last thing done on the left so it is the first thing to be undone. We undo the add 10 with a subtract 10 (or an add -10, if you prefer):

13w + 10 − 10 = 270 - 10
13w = 260

Finally, we undo the multiply by 13 by dividing both sides by 13. This leaves w by itself on the left and its value on the right, namely
w = 20 cm.
We can now find ℓ by multiplying our now-known value of w by 4 to get ℓ = 80 cm. The length units are part of the answer and should be reported along with the numbers for length and width of the original rectangle.

Answer 3

first rectangle is 4w x w
second rectangle is (4w+5) x (w+2) have area 270 cm^2 more than the former hence
(4w+5)(w+2) = 4w*w + 270
Or 13w = 260
Or w = 20
length = 4w = 4*20 = 80
original is 80cm x 20cm

Answer 4

20x80
20x80=1600
22x85=1870
let the shorter side of the first rectangle = x
x * 4x = area of first rectangle = 4x^2
(x+2)(4x+5)= area of first + 270
4x^2 + 13x + 10 = area of first + 270
4x^2 + 13x + 10 - 4x^2 = 270
13x + 10 = 270
x=20
4x=80

Answer 5

l = length = 4 w (widths)
l = 4w
A = 4w^2

l + 5
w +2

(l+5)(w+2) - 4w^2 = 270

but 'l' = 4w
(4w + 5)( w + 2) - 4w^2 = 270
4w^2 + 5w + 2w + 10 - 4w^2 = 270
10w + 10 = 270
10w = 260
w = 26
4w = l = 4 x 26 = 104
So dimensions are 26 x 104.

Answer 6

the dimensions of the orginal rectangle are 80cm by 20cm

Answer 7

20 cm x 80 cm

Answer 8

I think there's information missing.