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?
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.
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.
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