Question

A rectangle has a perimeter of $46cm$ and an area of $120c{m^2}$. How do you find its dimensions by writing an equation and using the quadratic formula to solve it?

Answer 1

Hint: In order to solve this question, we first need to write what is already given such as the perimeter and the area can be written as $2\left( {l + B} \right) = 46$ and $l \times B = 120$, where l and B represents the length and breadth respectively
After we have established these statements, we use the equation given for area and isolate the variable for breadth. Then we substitute this in the equation for the perimeter. Once we have done this, we can easily form our quadratic equation and then apply our formula to get the required answer.

Formula used: $l = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
In the given question, we are provided with a perimeter and the area of the rectangle. Now as we know the formula for finding the perimeter of a rectangle is given as:
Perimeter= $2\left( {l + B} \right)$ , where $l = $ length and $B = $ breadth
And the formula for area is given as:
Area = $\left( {l \times B} \right)$
Now according to the given:
Perimeter =$46$
Therefore, $2\left( {l + B} \right) = 46....\left( 1 \right)$
Area= $120$
$ \Rightarrow l \times B = 120$
We can also write the above equation as:
$ \Rightarrow B = \dfrac{{120}}{l}$
Let us recall equation (1):
$ \Rightarrow 2l + 2B = 46$
On substituting the value of $B$ that we have derived in the above equation, we get:
$ \Rightarrow 2l + 2\left( {\dfrac{{120}}{l}} \right) = 46$
$ \Rightarrow \dfrac{{240}}{l} + 2l = 46$
Let us multiply both sides of the equation with $l$ :
$ \Rightarrow 240 + 2{l^2} = 46l$
On rearranging the above equation, we get:
$ \Rightarrow 2{l^2} - 46l + 240 = 0$
Now the equation is in the form of a quadratic.
Let us simplify the equation further so that we can solve it easily. Taking $2$ as common factor from the left hand side, we find:
$ \Rightarrow 2\left( {{l^2} - 23l + 120} \right) = 0$
Let us divide both sides of the equation with $2$
Thus, ${l^2} - 23l + 120 = 0$
Now since we have been asked to solve the quadratic using the formula, hence we make use of the formula to find the roots of the equation. In this case, the roots of the equation is represented by $l$ which signifies the length of the rectangle.
As we know that a quadratic equation is expressed in the form of: $a{x^2} + bx + c = 0$
If we compare the standard form to the quadratic in our question, we find the variables to be (do keep in mind that the variable $b$ here is different than the breadth $B$ of the rectangle)
Thus, $a = 1;b = - 23;c = 120$
The quadratic formula is given as$l = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , where $\sqrt {{b^2} - 4ac} $ is known as the discriminant.
Thus, $l = \dfrac{{ - \left( { - 23} \right) \pm \sqrt {{{\left( { - 23} \right)}^2} - 4\left( 1 \right)\left( {120} \right)} }}{2}$
$ \Rightarrow l = \dfrac{{23 \pm \sqrt {529 - 480} }}{2}$
On further simplifying we get:
$ \Rightarrow l = \dfrac{{23 \pm \sqrt {49} }}{2}$
$ \Rightarrow l = \dfrac{{23 \pm 7}}{2}$
Now there can be two roots for $l$, that is
$l = \dfrac{{23 + 7}}{2}$ OR $l = \dfrac{{23 - 7}}{2}$
Thus,$l= \dfrac{{\not{{30}}}}{{\not{2}}} = 15$ or $l = \dfrac{{\not{{16}}}}{{\not{2}}} = 8$

Thus length of the rectangle can either be $15cm$ or $8cm$
When the length is $8cm$
Then breadth = $\dfrac{{120}}{8} = 15cm$
When length is $15cm$,
Then breadth is $\dfrac{{120}}{{15}} = 8cm$

Note: Alternate way of solving this sum is through factorization:
We have our quadratic equation as ${l^2} - 23l + 120 = 0$
Now , in order to do the grouping, we first break the middle term into two such terms which add up to give the middle term and is equal to the product of the first and third term,
$ \Rightarrow {l^2} - 15l - 8l + 120 = 0$
$ \Rightarrow l\left( {l - 15} \right) - 8\left( {l - 15} \right)$
Therefore the two factors are:
$ \Rightarrow \left( {l - 15} \right)\left( {l - 8} \right)$
Thus length could either be:
$ \Rightarrow l = 15,8$
Thus, we get the required answer.