the area of a rectangle is 63ft^2, and the length is 11 ft more than double the width,find the dimensions of the rectangle

Question

I've tried this but i can't seem to get the right answer,please help and thanks

Answer 1

We know the total area (63 square feet) and we know the formula for area of a rectangle is length times width. 

 The L is 11 feet more than double the width (this is written in English, we want to write it in math!). The L  is (=) 11 more than (+) double (2 times) the width (w) , or in "math" L = 11 + 2w 

 

Going back to the formula for area of a rectangle = L times w , plug in the L we know, and use w for the width. 

 

so Area = L * w = (11 +2w) * w multiply this out to get Area = 11w + 2w^2

 

We know the area is 63 (from above). So 63=11w+2w^2 this is a quadratic equation! We can solve by factoring or using the quadratic formula. Solve for w to get your two answers (ignore a negative answer if you get one, because the width of a rectangle can't be negative). The positive answer you get will be the width! Plug in the width into the Length expression we found (11+2w) to solve for the length.

 

2w^2+11w-63=0  using quadtraic formula x = +/- (square root of b^2 - 4ac ) / 2a, plug in a=2, b=11, and c=-63. You'll get 3.5 and -9 for answers. Ignore -9 for reason listed above, so 3.5 is the width! Do you think you can figure out the length from there? Try it now!

Answer 2

A = l * w

l = 2w + 11 _{(double the length, plus another 11 ft}

 

63 = (2w + 11) w

0 = (2w + 11) w - 63

0 = 2w² + 11w - 63

factors of 63: 3, 3, 7

 

0 = (2w - 7)(w + 9)

due to:

2w * 9 = 18w

-7 * w = 7w

18w - 7w = 11w

-7 * 9 _{(which is 3*3)} = -63

 

0 = 2w - 7

7 = 2w

⁷/₂ = w = 3.5

 

We don't have to worry about the other part because that would become -9 = w, and we can't have a negative length.

 

Now to find l:

2 * 3.5 + 11 = l

7 +11 = l

18 = l

 

we have l, we have w, so let's plug them in and see if it works:

18 * 3.5 = 63