Question

Find the dimensions of a rectangle whose width is 6 miles less than its length and whose area is 112 square miles.

Answer 1

Answer:

Step-by-step explanation:

Be the sides x$x, x+6$ you know $x(x+6) = 112$ or $x^2 +6x -112 = 0$$x= -3 \pm \sqrt{9+112} = -3\pm 11$ the solution is $x=8, x=-14$ The negative answer is to be discarded, your rectangle has sides 8 and $8+6=14$ miles

Answer 2

The length of the rectangle is 14 miles and the width of the rectangle is 8 miles.

How to find the area and the perimeter of a rectangle?

For a rectangle with length and width L and W units, we get:

Area of the rectangle = (L × W) unit^2

Perimeter of the rectangle = 2(L + W) units

Let the width of the rectangle be represented by W, while the length is represented by L.

Given the width is 6 miles less than its length, therefore, the width can be written as,

W = (L - 6) miles

Now given the area of the rectangle is 112 sq. miles, therefore, the area of the rectangle can be written as,

Area of the rectangle = Length × Width

112 = L × W

112 = L × (L-6)

112 = L²-6L

L = 14, -8

Since the length, L can not be negative, the value of L is 14 miles, therefore, the width will be 8 miles.

Hence, the length of the rectangle is 14 miles and the width of the rectangle is 8 miles.

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