Question

☻-------SOLVE------☺.The area of a playground is 160 yd^2. The width of the playground is 6 yd longer than its length. Find the length and width of the playground.

A.) length = 22 yd, width = 16 yd

B.) length = 16 yd, width = 10 yd

C.) length = 16 yd, width = 22 yd

D.) length = 10 yd, width = 16 yd

Answer 1

i think the answer is D

Answer 2

Option D.)
`length=10\ yd`,
`width=16\ yd`

Explanation:

we know that

The area of a rectangle (playground) is equal to

`A=LW`

we have

`A=160\ yd^(2)`

so

`160=LW` -----> equation A

`W=L+6` -----> equation B

substitute equation B in equation A and solve for L

`160=L(L+6)`

`L^(2)+6L-160=0`

using a graphing tool

solve the quadratic equation

see the attached figure

The solution is

`L=10\ yd`

Find the value of W

`W=L+6` ----->
`W=10+6=16\ yd`