SOLUTION: Find the dimensions of a rectangle whose area is 180cm^2 and whose perimeter is 54 cm.

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Question 98653: Find the dimensions of a rectangle whose area is 180cm^2 and whose perimeter is 54 cm.
Answer by praseenakos@yahoo.com(507) About Me  (Show Source):
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Question:
Find the dimensions of a rectangle whose area is 180cm^2 and whose perimeter is 54 cm.
Answer:
Given that,
Area+=+180+cm%5E2+ and
+perimeter+=+54+cm+

Perimeter of rectangle is given by the formula, 2( length + width)

==> 2( length + width) = 54

==> ( length + width) = +54%2F2

==> ( length + width) = 27

Now suppose, length = x cm

Then width = 27 - x


Now area is given by the formula, length * width


That is length * width = 180 cm^2

==> x * ( 27-x) = 180

==> +27x+-+x%5E2+=+180+

==> +x%5E2+-+27+x+%2B+180+=+0+ -----------------(1)

This is a quadratic equation and you can solve this equation using quadratic formula for x.


==> x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+


Comparing equation(1) with the standard quadratic equation,
ax%5E2+%2B+bx+%2B+c+=+0, we have,

a = 1, b = -27 and c = 180

==> x+=+%28-%28-27%29+%2B-+sqrt%28%28-27%29%5E2-4%2A1%2A180+%29%29%2F%282%2A1%29+

==> x+=+%28+27+%2B-+sqrt%28729-720%29%29+%2F+2++

==> x+=+%28+27+%2B-+sqrt+9+%29+%2F2++

==> x+=+%28+27+%2B+-+3%29%2F2+


==> either, +x+=+%2827+%2B+3%29%2F2 or +x+=+%2827+-+3%29%2F2


==> either, +x+=+30+%2F2 or +x+=+24%2F2


==> either, +x+=+15 or +x+=+12


that is either length = 15 cm of 12 cm

When length = 15 cm, width = 27 - 15 = 12 cm


And when length = 12cm, width = 27 - 12 = 15 cm.

Hence the solution.


Hope you found the explanation useful.

Regards.

Praseena.