Question

the perimeter of a rectangle is 100m and the are of the rectangle is 54m^2. find the dimensions of the rectangle?

Answer 1

helpppppppp!

Answer 2

54 = x • y
100 = x + x + y + y

Answer 3

2 Length + 2 Width = Perimeter for rectangle = 100 m

Length * Width = Area (in m\(^2\) here)

You'll get a square, unless you have something to specify the ratio

Answer 4

And yes, a square is a type of rectangle.

Answer 5

100 = 2x + 2y

-2x = 2y - 100
2x = 100-2y
x = 50 - y

Substitute (50 - y) into the first equation for x and solve for y.

54 = x • y
54 = (50 - y)(y)
54 = 50y - y^2

Now solve for y. :)

Answer 6

@agentx5 is right. When you solve for y, so will have either the length or width, unless the ratio between the length and with is specified.

Answer 7

let the length be l and width = w

then for perimeter

2l + 2w =100 or l = 50 - w

for area

54 = lw

or

54 = (50 -w)w

54 = 50w - w^2


rewriting

w^2 - 50w + 54 = 0


you'll need the general quadratic formula

\[x = \frac{50 \pm \sqrt{-50^2 - 4\times1\times 54 }2\times1}\]

Answer 8

its asking me for the length and with! im so confused!

Answer 9

:/

Answer 10

@campbell_st I don't think it was intended to be that challenging... and here, quadratic:
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

From what @campbell_st and @JBrightman3 wrote for you:
a =1
b =-50
c = 54
This is ax^2 + bx + c = 0, the form you HAVE TO re-arrange the equation into in order to use the quadratic formula, which those to gentlemen did for you. (You did try by hand to follow along with what they were showing you yes?)

\(25±\sqrt{571}\)

@mathhelp69 try this now in each of the two formulas (the one for area and the one for perimeter) I gave you back at the start and see what they give you as answers, alright?

Answer 11

He just had an error with his LaTeX causing the formula he was giving you to not show.