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

Answer:

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

Step-by-step 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

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