SOLUTION: Hi outstanding tutors, I have a question that is in the quadratic equation solving chapter of my book. It is geometry related, however. Here it is: Is it possible for a rectang

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Question 1014558: Hi outstanding tutors,
I have a question that is in the quadratic equation solving chapter of my book. It is geometry related, however. Here it is:
Is it possible for a rectangle with a perimeter of 52 centimeters to have an area of 148.75 square centimeters? Explain. A diagram of the rectangle is given with x representing the with and 26 - x representing the length.
I know that P=2(l+w) and A=wl. I think somehow a quadratic equation is implemented to solve this. Please help! Thanks!!
Found 3 solutions by ikleyn, addingup, MathTherapy:
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.
Hi outstanding tutors,
I have a question that is in the quadratic equation solving chapter of my book. It is geometry related, however. Here it is:
Is it possible for a rectangle with a perimeter of 52 centimeters to have an area of 148.75 square centimeters? Explain. A diagram of the rectangle is given with x representing the with and 26 - x representing the length.
I know that P=2(l+w) and A=wl. I think somehow a quadratic equation is implemented to solve this. Please help! Thanks!!
----------------------------------------------------------------

Answer. Yes, it is possible.

Solution 

Let x be one side measure of a rectangle, then (26 -x) is the measure of an adjacent / consecutive side, and S = x*(26-x) is the rectangle area.

Complete the square:

S = .

So, the maximum of S is 169 .

The area can not be more than 169 .

Since you are asking about the smaller value 142.75 , the solution does exist.


Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
No, it's not possible. Of the square/rectangular shapes, the square is the one with the largest area (try it, a rectangle 3 by 5 would have a perimeter of 16 and an area of 3*5=15. Now a square 4*4, same perimeter of 16 has an area also 16).
On that basis, sqrt148.75= 12.20 is each side of our square and the smallest side we can have in order to get the area. And 12.2*4= 48.8 would be the perimeter.
Your formula would be:
2x+2(26-x)= 52 Go ahead, knock yourself out. It can't be solved.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Hi outstanding tutors,
I have a question that is in the quadratic equation solving chapter of my book. It is geometry related, however. Here it is:
Is it possible for a rectangle with a perimeter of 52 centimeters to have an area of 148.75 square centimeters? Explain. A diagram of the rectangle is given with x representing the with and 26 - x representing the length.
I know that P=2(l+w) and A=wl. I think somehow a quadratic equation is implemented to solve this. Please help! Thanks!!
From your description length = x

Let width be W

Since Perimeter = 52 cm, we get: 2(x + W) = 52______2(x + W) = 2(26)____x + W = 26______W = 26 - x

testing whether or not 148.75 can be the area, we get: x(26 - x) = 148.75





Using the quadratic equation formula, completing the square, or factoring (the trinomial would need to contain all INTEGERS), we find that:

x = 17.5 or x = 8.5, and so, the dimensions of this rectangle would be: , so it's  

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