"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

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Feb 20, 2019
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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

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Apr 12, 2005
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9,789
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
Joined
Dec 30, 2014
Messages
3,018
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
Joined
Jan 27, 2012
Messages
4,795
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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