# "George drew a rectangle with an area of 84cm squared. If the lengh is 8cm more than the width, what are the rectangle's dimensions?"

#### davejones

##### New member
Hi guys! Here's the problem,

George drew a rectangle with an area of 84cm squared. If the lengh is 8cm more than the width, what is the length and width of the rectangle?

Now I know the answer is 14 and 6 from just looking at it, 14 x 6 = 84! But I can't work it out. Here's my attempt.

Two equations, L = length , W = width

W + 8 = L and (W+8) x L = 84

So plugging L in to the second equation we have

(W + 8 ) ( W + 8) = 84
W^2 + 16 W + 64 = 84
W^2 +16W = 20

Thanks, Dave

#### tkhunny

##### Moderator
Staff member
Hi guys! Here's the problem,

George drew a rectangle with an area of 84cm squared. If the lengh is 8cm more than the width, what is the length and width of the rectangle?

Now I know the answer is 14 and 6 from just looking at it, 14 x 6 = 84! But I can't work it out. Here's my attempt.

Two equations, L = length , W = width

W + 8 = L and (W+8) x L = 84

So plugging L in to the second equation we have

(W + 8 ) ( W + 8) = 84
W^2 + 16 W + 64 = 84
W^2 +16W = 20

Thanks, Dave
(W+8) x L = 84 <== This is LL, not WL. Please try again.

#### Jomo

##### Elite Member
Hi guys! Here's the problem,

George drew a rectangle with an area of 84cm squared. If the lengh is 8cm more than the width, what is the length and width of the rectangle?

Now I know the answer is 14 and 6 from just looking at it, 14 x 6 = 84! But I can't work it out. Here's my attempt.

Two equations, L = length , W = width

W + 8 = L and (W+8) x L = 84

So plugging L in to the second equation we have

(W + 8 ) ( W + 8) = 84
W^2 + 16 W + 64 = 84
W^2 +16W = 20

Thanks, Dave
To solve (W + 8 ) ( W + 8) = 84 you should NOT foil out (although it is not wrong but rather it is inefficient). Just note that (W+8)^2 =84. Then (W+8) = sqrt(84). So W = sqrt(84) -3.
Also your result of 9.1makes no sense. You have W^2 +16W = 20. Now 9.1^2 >20 and 16*9.1>20 so how can the sum be 20.
As tkhunny has pointed out, your initial equation is wrong

#### HallsofIvy

##### Elite Member
Hi guys! Here's the problem,

George drew a rectangle with an area of 84cm squared. If the lengh is 8cm more than the width, what is the length and width of the rectangle?

Now I know the answer is 14 and 6 from just looking at it, 14 x 6 = 84! But I can't work it out. Here's my attempt.

Two equations, L = length , W = width

W + 8 = L
Yes, this is correct

and (W+8) x L = 84
No, this is NOT correct! Since W+ 8= L, this would be "L x L= 84" which is not a correct formula for area. The area is L x W= 84 so you should have (W+ 8) x W= 84.

So plugging L in to the second equation we have

(W + 8 ) ( W + 8) = 84
W^2 + 16 W + 64 = 84
W^2 +16W = 20