SOLUTION: The perimeter of a rectangle is 64cm. The area is 240cm square. What are its dimension?

Question 27770: The perimeter of a rectangle is 64cm. The area is 240cm square. What are its dimension?
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is 64cm. The area is 240cm square.
What are its dimensions?
Let the dimensions of the rectangle be length= a cms and width = b cms
Formula for area of a rectangle is
Area = lenght X width = (a)X(b) = (ab) sqcms
Formula for Perimeter of a rectangle is
Perimeter = 2(length + width)= 2(a+b)cms
By data 2(a+b)= 64 which means (a+b) = 32 ----(1) (on division by 2)
and (ab) = 240 ----(2)
We have by formula (a-b)^2 = (a+b)^2-4(ab) ----(*)
Therefore using (1) and (2) in (*)
(a-b)^2 = (a+b)^2-4(ab) = (32)^2 - 4X(240) = 1024 - 960 = 64
(a-b)^2 = 64
which implies (a-b) = 8 ----(3) (taking the positive root)
Now consider (a+b) = 32 ----(1)
Adding (3) and (1)
[(a-b) + (a+b) ] = (8 + 32)
a-b+a+b = 40
a+a-b+b = 40 (by additive associativity)
2a + 0 =40
2a = 40
a = 40/2 = 20
Putting a= 20 in (1)
a+b = 32
20 + b = 32 which means b = (32-20) = 12
Therefore length= 20 cms and width = 12 cms
Verification: Area = length X width = 20X12 = 240 which is correct
And Perimeter = 2(lenght + width) = 2X(20 +12) = 2X32 = 64 which is correct.
Note: In the above, from (a-b)^2 = 64
we have taken the positive root 8 for (a-b).
What if we consider the negative root (-8) because (-8)^2 also = 64
Okay, let us consider (a-b) = -8 say ----(4)
Now adding (4) and (1), 2a = 32-8 = 24 which means a = 12 which implies b = 20
So we get the dimensions the same but only the tags are different.
So in such problems it it enough if we consider the positive root.

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