HELP!! The length of a rectangle is 5cm less than twice the width. The area is 188 sq cm find the dimensions.

Answer 1

Let w = the width and l = length. The length is 2w - 5 cm (5 cm less than twice the width).

Area of a rectangle = l x w, and the area is 188 sq cm, so we have:
l x w = 188. Substitute for l in terms of the width, and you have:
(2w - 5)w = 188, or
2w ^2 - 5w - 188 = 0 if we multiply out and get everything on one side of the equation.

I solved for w by using this site:
http://library.thinkquest.org/C004647/calculator/polynomial/polynomial.html
and plugging in the following for the polynomial:
2x^2-5x-188
(their "x" variable is what we're calling "w").

There are 2 answers from the calculator - We need to use the positive answer, since this is the width of the rectangle and it doesn't make sense to have a negative width. SO, we get width = 11.02561 cm.

Now recall that the length = 2w - 5. Plugging in our width result for w, we get:
length = 2(11.02561) - 5 = 17.05122 cm

Final answer (dimensions are rounded):
width = 11.02561 cm
length = 17.05122 cm

Answer 2

Let the width be xcm. Then the length is 5cm less than twice xcm, i.e.
2x - 5 cm

Then use area = length * width. that is
188 = x(2x - 5)

Can you solve this equation? Remove parentheses, subtract 188 from both sides, and then either factorise the quadratic or use the formula. Try to finish it yourself, but if you want to see my completion scroll down.















































2x^2 - 5x - 188 = 0
x = (5 +/- sqrt(25 + 1504))/4. We want only positive answer, so
Width x = 11.026 approx.
Length =16.052 But this only gives area 177, not 188. See if you can find my error.

Answer 3

Let the width be w (it is always easier to have the smallest as the unknown).

So the length = 2w - 5

Area = length times width

Area = w (2w-5)......now multiply out the bracket

Area = 2w^2 - 5w.....this equals 188

2w^2 - 5w = 188.......subtract188 from both sides to get 0 alone on one side

2w^2 - 5w - 188 = 0........now use the method you've been taught to solve quadratic equations

good luck

Answer 4

let x be length, l and width w.
therefore, l=(2w-5) cm
also, lw=188.
therefore, (2w-5)w=188
2w^2-5w=188
2w^2-5w-188=0
using quadratic formula,

width, w=11 so length, l=188/11=17.1

Answer 5

W=x
L= 2x-5

A=wl

188= x(2x-5)
188= 2x^2-5x
2x^2-5x-188

use quadratic formula to solve for x
x= 11.03

Answer 6

Hmmmm. Several possible approaches to this one.

try difference of two squares:
(x+y)*(x-y) = x^2 - y^2

Let x be (L + W) / 2.
Let Y be L - x = X - w = 2.5
L = x + 2.5
W = x - 2.5

(x + 2.5)(x - 2.5) = x^2 - 6.25 = 188
x^2 = 188 + 6.25 = 194.25

x = sqrt(194.25) = .5*sqrt(777) (about 13.93736)

Length = x + 2.5 = .5*sqrt(777) + 2.5 (about 16.43736)
W = x - 2.5 = .5*sqrt(777) - 2.5 (about 11.43736)

Strange numbers, but there it is. Hope it helps...

Answer 7

W=x
L= 2x-5

A=wl

188= x(2x-5)
188= 2x^2-5x
2x^2-5x-188

use quadratic formular to solve for x
x= 11.03

Answer 8

xy = 188
y = 2x-5
x (2x-5) = 188
2x^2 - 5x - 188 = 0
x = [5 +,- sqrt{25+8(188)}]/4
x = {5 +,- 39.10243} / 4
x = 11.02561
y = 17.05121

Answer 9

L = 2W - 5
A = LW
A = W (2W - 5)
188 = 2W² - 5W
2W² - 5W - 188 = 0
W =10.81882960450232 cm
L = 16.63765920900464 cm

Answer 10

(l*w = 188
(l = 2w -5

(2w - 5)*w = 188
2w² - 5w - 188 = 0

w = 5² - 4.2.-188