# AreasThe length of a rectangle is 2 m more than its width. The area of the rectangle is 80 m^2. What is the length and the width of the rectangle?

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The area of a rectangle is given by the product of length and width. We know that the length is 2 more than the width. If length is denoted by L, width is L - 2

L*(L - 2) = 80

=> L^2 - 2L - 80 = 0

=> L^2 - 10L + 8L - 80 = 0

=> L( L - 10) + 8(L - 10) = 0

=> (L + 8)( L - 10) = 0

=> L = 10 , L = -8 is ignored as length is not negative

W = L - 2 = 8

**The length and width of the rectangle are 10 and 8 resp.**

Let L be the length and W be the width. We can say that

L = W + 2

And we can say that

L*W = 80

Now we can substitute the value of L from the first equation into the second.

W*(W+2) = 80

That gives us

w^2 + 2w = 80 or w^2 + 2w - 80 = 0

Now this is a quadratic equation that can be factored down to

(w + 10) (w-8) = 0

Which gives us the values

w = - 10 and w = 8

Of course, the width cannot be negative so **w = 8.**

If the width is 8 and L = 2 + W, then **L = 10**

This shows us that the **Length is 10 and the Width is 8.**

We'll put L the length and W the width.

We'll write mathematically the constraint concerning the length and the width:

L = 2 + W

The area of the rectangle is:

A = Length*Width

From enunciation, we know that the area is A = 80 m^2.

80 = (2 + W)*W

We'll remove the brackets:

80 = 2W + W^2

We'll use symmetric property:

W^2 + 2W - 80 = 0

We'll apply the quadratic formula:

W1 = [-2 + sqrt(4 + 320)]/2

W1 = (-2 + 18)/2

W1 = 8

W2 = (-2-18)/2

W2 = -10

Since a length cannot be negative, we'll reject the negative value. We'll keep the positive value for W = 8 m.

The length is:

L = 2 + W

L = 2 + 8

**L = 10 m**