Question

The perimeter of a rectangle is $30$ inches and its area is $52$ square inches. How do you find the length of the longest side of the rectangle?

Answer 1

Hint: In this question we are asked to find the longest side of the rectangle. To proceed with this question we need to be clear of the terms used in the question. Rectangle is a quadrilateral with four sides, with opposite sides equal in length. Also, all the angles of a rectangle are \[{90^ \circ }\]. Rectangle is also called a parallelogram with \[{90^ \circ }\]. From four sides of a rectangle, two are shorter. Perimeter is the length of the path around the figure and area is the region covered by the figure.

Complete step by step solution:
We are given,
Perimeter of rectangle $ = 30\;inches$
Area of rectangle$ = 52\;inches$
We know,
Perimeter of rectangle $ = 2(l + b)$
Area of rectangle $= l \times b$
Where,
$l = $Length (longer side of rectangle)
$b = $Breadth (shorter side of rectangle)
After combining the equations,
\[ \Rightarrow 2(l + b) = 30\]
\[ \Rightarrow l + b = 15\] (equation1)
$ \Rightarrow l \times b = 52$(equation2)
Now we have to eliminate $b$
So rewriting equation $1$ as,
$ \Rightarrow b = 15 - l$
Substituting this in equation $2$ ,
$ \Rightarrow l \times (15 - l) = 52$
$ \Rightarrow 15l - {l^2} = 52$
$ \Rightarrow {l^2} - 15l + 52 = 0$
We’ll solve the question by using quadratic formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow l = \dfrac{{ - ( - 15) \pm \sqrt {{{( - 15)}^2} - 4 \times 1 \times 52} }}{{2 \times 1}}$
$ \Rightarrow l = \dfrac{{15 \pm \sqrt {225 - 208} }}{2}$
$ \Rightarrow l = \dfrac{{15 \pm \sqrt {17} }}{2}$
Now we’ll have two terms, one with plus and one with minus
$ \Rightarrow l = \dfrac{{15 + \sqrt {17} }}{2}$ and $l = \dfrac{{15 - \sqrt {17} }}{2}$
$ \Rightarrow l = \dfrac{{15 + 4.12}}{2}$ and $l = \dfrac{{15 - 4.12}}{2}$
$ \Rightarrow l = \dfrac{{19.12}}{2}$ and $ \Rightarrow l = \dfrac{{10.88}}{2}$
$ \Rightarrow l = 9.56inches\;or\;5.44inches$

Note: The area of rectangle is expressed in square units whereas perimeter is expressed in linear units. Area and perimeter of the rectangle can be used in real life. Like to find the paint required to paint the wall or tiles required to cover the floor, area of the rectangle has to be calculated. Perimeters can be used to find the length of fencing required to cover the boundary of a garden or finding the length of the ribbon required to cover a table.