Question

the perimeter of a rectangle is 30 yd, and the area of a rectangle is 36 yd^2 . find the dimensions of the rectangle ?

length:
width:

Answer 1

2x+2y=30 (perimeter) x=(30-2y)2
x*y=36 (area)
(30-2y)/2*y=36
Solve for y

Answer 2

what the length and the width ?

Answer 3

??

Answer 4

?????

Answer 5

Did you solve for y in the equation above?

Answer 6

i couldn't

Answer 7

its hard

Answer 8

does y equal 5/6 ?

Answer 9

No. You start with the equation where we inserted the expression derived from the perimeter equation.
\[x*(30-2x)/2=36\]
we can simplify (30-2x)/2 to (15-x)

Answer 10

(30/2)-(2x/2)

Answer 11

15-x

Answer 12

ok ?

Answer 13

so now we have \[x*(15-x)=36\]
expand the left side of the equation, then move all values to one side of the equation

Answer 14

x=3,12

Answer 15

what the width and the length ?

Answer 16

correct. and actually if we try to use both these values in our two equations (perimeter and area) you'll find that one of the answers is x and the other is y (length is 3, width is 12)

Answer 17

the length of a rectangle is 5 m more than twice its width ,and the area of the rectangle is 88m^2 . find the dimensions .

length :
width :

Answer 18

??

Answer 19

???

Answer 20

??