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The width of the rectangle is 5 units less than the length if the area is 36 square units then find the dimension of the rectangle

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Answer:

The dimensions of the rectangle are approximately 8.54 units by 3.54 units.

Explanation:

Let's start by using algebra to represent the information given in the problem.

Let L be the length of the rectangle.

Then, the width of the rectangle is 5 units less than the length, which means the width can be expressed as L - 5.

The area of a rectangle is given by the formula A = L * W, where A is the area, L is the length, and W is the width. In this case, we are given that the area is 36 square units. So we can write:

A = L * (L - 5)

36 = L * (L - 5)

Expanding the right-hand side of the equation, we get:

36 = L^2 - 5L

Moving all the terms to the left-hand side of the equation, we get:

L^2 - 5L - 36 = 0

Now we can use the quadratic formula to solve for L:

L = (-(-5) ± sqrt((-5)^2 - 4(1)(-36))) / (2(1))

Simplifying this expression, we get:

L = (5 ± sqrt(229)) / 2

We can discard the negative solution since the length of a rectangle cannot be negative.

So, the length of the rectangle is approximately 8.54 units (rounded to two decimal places).

To find the width, we can substitute this value of L into the expression we derived earlier for the width:

W = L - 5

W = 8.54 - 5

W ≈ 3.54 units (rounded to two decimal places)

Therefore, the dimensions of the rectangle are approximately 8.54 units by 3.54 units.

User MonteCristo
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