Question

The perimeter of a rectangle is $80$ feet. How do you find the dimensions if the length is $5$ feet longer than four times the width? What is the area of the rectangle?

Answer 1

Hint:Start by mentioning the definition of perimeter of a rectangle. Then we will mention the formula for the perimeter of the rectangle and substitute values in the formula. Then finally evaluate the conditions and solve for the dimensions of the rectangle and area of the rectangle.

Complete step by step answer:
First we will start off by mentioning the definition of perimeter. So, perimeter is a path that surrounds a two-dimensional shape. This term may be used either for the path, or its length in one dimension.
Now, the perimeter of a rectangle is given by, $2(l + b)$.
Now let us consider the length of the rectangle as $x$ and the breadth of the rectangle as $2x$.
And according to the question, the perimeter of the rectangle is $80$ feet.
Since the perimeter of a rectangle is $2$ times the length plus the width:
\[
2(l + w) = 80 \\
l + w = 40 \\
l = 40 - w \\
\]
We also know that the length is $5$ feet longer than four times the width.
$l = 4w + 5$
Now if we combine these two we have,
$
4w + 5 = 40 - w \\
5w = 35 \\
w = 7 \\
$
Now if we reuse the earlier equation, which is $l = 40 - w$ we will get,
$
l = 40 - 7 \\
\,\, = 33 \\
$
Now we will evaluate the area of the rectangle.
Area of the rectangle is given by $l \times w$.
Now, we will substitute values in the formula.
$
A = l \times w \\
A = 33 \times 7 \\
A = 231 \\
$
Hence, the area of the rectangle is $231$ sq. feet.

Note: While mentioning the definition, make sure to mention the terms properly. Reduce the terms by factorisation. While choosing any variable, for any unknown term, choose according to the conditions.