How to calculate the perimeter of a rectangle if the area is 42 and the legth is one more than the width?

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Let the length of the rectangle be L and let the width be W.

As the length is 1 more than the width L = W + 1

The perimeter of the rectangle is given by 2W + 2L. The area of the rectangle is given by L*W.

L*W = 42

=> (W + 1)W = 42

=> W^2 + W - 42 = 0

=> W + 7W - 6W - 42 = 0

=> W(W + 7) - 6(W + 7) = 0

=> (W - 6)(W + 7) = 0

=> W = 6 or -7

We take only the positive value , W = 6

The length of the rectangle is 7. The perimeter is 6*2 + 7*2 = 12 + 14 = 26

The required perimeter of the rectangle is 26.

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Let us assume that the length of the rectangle is L and the width is W.

Given that the area is 42.

But wer know that the area  L*W

==> L*W = 42 ...............(1)

Now we are given the L is one more than W.

==> L = W+1.............(2)

We will substitute (2) into (1).

==> (W+1) * W = 42

==> w^2 + w = 42

==> w^2 + w - 42 = 0

We will factor.

==> (W+7) ( w-6) = 0

==> w1= -7 ( not valid)

==> w2= 6 ==> L = 6+1 = 7

Then we have L =7 and W= 6, we will calculate the perimeter.

==> We know that P = 2L + 2W = 2*7 + 2*6 = 14+12 = 26

Then the perimeter of the rectangle is 26 units.

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