Lesson Three methods to find the dimensions of a rectangle when its perimeter and the area are given

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<H2>Three methods to find the dimensions of a rectangle when its perimeter and the area are given</H2>

<H3>Problem 1</H3>A rectangle has an area of &nbsp;247 &nbsp;cm2 and a perimeter of &nbsp;64&nbsp; cm. &nbsp;What are its dimensions?

<U>1-st method</U>

First method is to create a corresponding quadratic equation and then solve it using the quadratic formula.

<pre>
Let  x  be the length of the rectangle, in centimeters, and 
let  y  be its width.

Then 2x + 2y = 64.

Hence,  2y = 64 - 2x,  and  y = {{{(64-2x)/2}}} = 32 - x.

The area of the rectangle is

x*y = x*(32-x) = - x^2 + 32x.

It is equal to 247 cm^2, which gives you an equation

-x^2 + 32x = 247,   or

x^2 - 32x + 247 = 0.      (1)

Apply the quadratic formula

{{{x[1,2]}}} = {{{(32 +- sqrt(32^2-4*247))/2}}} = {{{(32 +- sqrt(36))/2}}} = {{{(32 +- 6)/2}}} = {{{16 +- 3}}}.

You have two roots:  {{{x[1]}}} = 16 + 3 = 19  and  {{{x[2]}}} = 16 - 3 = 13.

Usually, the greater dimension is called the length.

So, in this problem the length is 19 centimeters.

Then the width is  32-19 = 13 centimeters.

<U>Answer</U>.  The dimensions of the rectangle are  19 centimeters and  13 centimeters.
</pre>

Regarding this method, &nbsp;I'd like specially to highlight that the quadratic formula works for any quadratic equation.

It works like an army tank and provides a solution to any quadratic equation.

<U>2-nd method</U>

Second method is to create a corresponding quadratic equation and then solve it using factoring.

<pre>
Surely, the equation is the same as (1)

x^2 - 32x + 247 = 0,      (1)

and the method of obtaining this equation remains the same as in the <U>Solution 1</U>.

Now we want to factor the left side in the form

(x - ?)*(x - ?) = 0.       (2)

and now your goal is to find the numbers to put them instead of the question marks in order for to get the same equation (1).

It not very easy task. Your first wish is to try integer divisors of the number 247.

After some trials you will find them as 13 and 19.

Notice that their product is  13*19 = 247  and their sum is  13+19 = 32,

so (2) becomes equivalent to equation (1).

Now you get the same answer  x= 19 as in the <U>Solution 1</U>.
</pre>

Regarding this method, &nbsp;I'd like specially to highlight that it works good and fast when the coefficients of your original equation 
are small and good enough to that extent that you can mentally find the required combination quickly.

Another situation when it works good is when you know the solution in advance.

<U>3-rd method</U>

It is most beautiful method. &nbsp;It uses a &nbsp;<U>reduced</U> &nbsp;quadratic equation.

<pre>
You are given that the perimeter of the rectangle is 64 cm;  hence, the sum of the length and the width is one half of that:

x + y = 32.

Then the average of the length and the width is one half of 32, i.e. 16.

It is clear that the values of x and y are remoted in the same value/distance "u" from 16, so we can write

x = 16 + u,
y = 16 - u.

Then the area is  xy = (16+u)*(16-u) = {{{256 - u^2}}}, and it is equal to 247, according to the condition.

Hence,  {{{256 - u^2}}} = 247,  which gives  {{{u^2}}} = 256 - 247 = 9,  and then  u = {{{sqrt(9)}}} = 3.

Thus the length is  x = 16 + u = 16 + 3 = 19,

and  the width  is  x = 16 - u = 16 - 3 = 13.

We get the same answer quickly and mentally.
</pre>

You may find the third method in some recreational books.

Sometimes in math circles the teachers explain it to students.

In the school Math, &nbsp;the standard method of solution is to reduce the problem to a quadratic equation

and then solve it using the quadratic formula or factoring.

I presented the third method here to make your horizon wider.

If you demonstrate this method in the class, &nbsp;your teacher and your classmates will be impressed !

So,  &nbsp;it was as if you visited a Math circle today . . .

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