# "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
As a quadratic this is like 9.1 or something. I must be missing something simple! Please help!

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
As a quadratic this is like 9.1 or something. I must be missing something simple! Please help!

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
As a quadratic this is like 9.1 or something. I must be missing something simple! Please help!

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
As a quadratic this is like 9.1 or something. I must be missing something simple! Please help!

Thanks, Dave
You need to start thinking more and not rely on "formulas" so much. If the equation were really (W+ 8)(W+ 8)= 84, you should certainly NOT multiply it out and then apply the quadratic formula. You should immediately recognize "(W+ 8)(W+ 8)" as (W+ 8)^2 and take the square root of both sides.

Since the correct formula is W(W+ 8)= W^2+ 8W= 84, I would "complete the square". Adding (8/2)^2= 16 to both sides gives W^2+ 8W+ 16= (W+ 4)^2= 84+ 16= 100. Taking the square root of both sides, W+ 4= +/- 10. With the "+", W= 10- 4= 6 cm so that L= W+ 8= 14 cm. The "length is 8cm more than the width" and the area of the rectangle is LW= 6(14)= 84 square cm as needed.

(The other solution to the quadratic equation would take "-10": W+ 4= -10 so W= -14. Since the width of a rectangle is never negative, we discard that solution.)

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