SOLUTION: To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway.If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be t

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway.If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be t      Log On


   

Question 1171867: To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway.If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be the dimensions of the tarpaulin? Please help me answer this question. Arigathanks in advance.
Found 3 solutions by math_tutor2020, Alan3354, MathTherapy:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

35/4 = 8.75
27/2 = 13.5

The area is 8.75 square meters and the perimeter is 13.5 meters.

Let L and W be the length and width
If the area is 8.75 square meters, then
A = L*W
L*W = A
L*W = 8.75

If the perimeter is 13.5 meters, then
P = 2*(L+W)
2*(L+W) = 13.5
L+W = 13.5/2
L+W = 6.75
L = 6.75-W

Plug this into the first equation and get everything to one side
L*W = 8.75
(6.75-W)*W = 8.75
6.75W-W^2 = 8.75
6.75W-W^2-8.75 = 0
-W^2+6.75W-8.75 = 0
100W^2-675W+875 = 0
In the last step I multiplied both sides by -100 to clear out the decimals and to make the leading term positive.

From here we could divide every term by 25 going from
100W^2-675W+875 = 0
to
4W^2-27W+35 = 0

Solving that equation is the same as solving
4x^2-27x+35 = 0

Let's use the quadratic formula to solve
x+=+%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29 or x+=+%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-27%29%2Bsqrt%28%28-27%29%5E2-4%284%29%2835%29%29%29%2F%282%284%29%29 or x+=+%28-%28-27%29-sqrt%28%28-27%29%5E2-4%284%29%2835%29%29%29%2F%282%284%29%29

x+=+%2827%2Bsqrt%28169%29%29%2F%288%29 or x+=+%2827-sqrt%28169%29%29%2F%288%29

x+=+%2827%2B13%29%2F%288%29 or x+=+%2827-13%29%2F%288%29

x+=+%2840%29%2F%288%29 or x+=+%2814%29%2F%288%29

x+=+5 or x+=+1.75

So the width is either W = 5 or W = 1.75

----------------------------------------------------

If the width was W = 5, then,
L = 6.75-W
L = 6.75-5
L = 1.75

Then note how
L*W = 1.75*5 = 8.75
and
2*(L+W) = 2*(1.75+5) = 13.5
which matches with our area and perimeter requirements.

So one possible set of dimensions is
width = 5 meters
length = 1.75 meters

----------------------------------------------------

If the width was W = 1.75, then,
L = 6.75-W
L = 6.75-1.75
L = 5
We basically get the flip of the previous section. So the order of the length and width doesn't matter.

We could say either
length = 5 meters
width = 1.75 meters
OR
length = 1.75 meters
width = 5 meters

Regardless of the order, they will both satisfy the equations we set up earlier, so they fit the requirements your teacher has set.

In reality we only have one unique set of dimensions.
----------------------------------------------------

Answer: The rectangle is 5 meters by 1.75 meters
Note: In fraction form, 1.75 = 7/4

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway. If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be the dimensions of the tarpaulin? Please help me answer this question.
----------------
Multiply the area by 4 and the Perimeter by 2.
---> area = 140 = L*W
L+W = 27
---
Find a pair of factors of 140 with a sum of 27
---> 7*20
============
Divide by 4
--> 1.75 by 5 meters

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway.If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be the dimensions of the tarpaulin? Please help me answer this question. Arigathanks in advance.
Let the width and length be W, and L, respectively

Then we get: 

matrix%281%2C3%2C+LW%2C+%22=%22%2C+35%2F4%29

matrix%281%2C3%2C+%2827%2F4+-+W%29W%2C+%22=%22%2C+35%2F4%29 ------ Substituting 27%2F4+-+W for L

matrix%281%2C3%2C+27W%2F4+-+W%5E2%2C+%22=%22%2C+35%2F4%29

matrix%281%2C3%2C+27W+-+4W%5E2%2C+%22=%22%2C+35%29 ------- Multiplying by LCD, 4



4W(W - 5) - 7(W - 5) = 0

(4W - 7)(W - 5) = 0  

4W - 7 = 0                  OR                W - 5 = 0

4W = 7                      OR                W = 5

Width or highlight_green%28matrix%281%2C6%2C+W%2C+%22=%22%2C+7%2F4%2C+or%2C+1.75%2C+m%29%29

If width = matrix%281%2C2%2C+7%2F4%2C+m%29, length, or 

This can be VICE-VERSA as well, so