what is the area of a rectangle whose are is 90m^2 and with perimeter of 38m?

Question

help

i'm sorry i mistyped the question
my question is what is the width and length
thanks to those who answered!

Answer 1

I guess you mean which are the sides! (You gave area)

x*z=90
2x+2z=38 <==> x+z=19 <==> x=19-z

x*z=90 <==> z* (19-z)=90 <==> 19*z - z^2 =90 <==>

z^2- 19*z +90 =0 This is a 2nd grade equation,

a*(z^2) + b *z +c=0

Where,in our case, a= 1, b = -19, c=90

It's solution requires calculating D,which is

D= b^2 -4*a*c =

=19^2-4*1*90=361-360=1

Now there are 2 solutions:

z= (-b +\- SQR D)/2*c => (19 +/- SQR 1 /) 2*1=

z1= 20/2 = 10 m
z2 = 18/2 =9 m

x=19-z => x1=10 and x2=9
So one side is 10 m and other is 9 m

Answer 2

what is the AREA??? The area of the rectangle is already given which is 90m^2. Perhaps you you meant what are the DIMENSIONS of the rectangle.

let w be the width and L be the length
Area = Width * length
Perimeter = 2(width + length)

Area = 90
perimeter = 38

90 = wL
38 = 2(w + L)


solve
38 = 2(w + L)
19 = w + L
L = 19 - w

90 = wL
90 = w(19 - w)
90 = 19w - w^2
w^2 - 19w + 90 = 0
(w - 10) (w - 9) = 0
w = 9 or 10

L = 19 - w
L = 19 - 9
L = 10

L = 19 - 10
L = 9

we are a pair of (9,10) and (10,9) which are exactly the same if we talk about dimensions.

so the width is 9m and the length is 10m

hope this helps!

Answer 3

Surely the 90m^2 is the area.

Did you mistype the question ?