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

Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given

Problem 1

Let  x  be the length of the rectangle, in yards. 
Then its width is (x-2) yards, according to the condition.


The area of the rectangle is 

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


It is equal to 168 square yards, which gives you an equation

x^2 - 2x = 168,   or

x^2 - 2x - 168 = 0.      (1)


Apply the quadratic formula 

 =  =  =  = .


You have two roots:   = 1 + 13 = 14  and   = 1 - 13 = -12.


Only positive root makes sense as the solution to this problem.

So, the length is 14 yards.

Then the width is  14-2 = 12 yards.


Answer.  The dimensions of the rectangle are  14 and  12  yards.

Surely, the equation is the same as (1)


x^2 - 2x - 168 = 0,      (1)


and the method of obtaining this equation remains the same as in the Solution 1.


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 difficult task. Your first wish is to try integer divisors of the number -168 whose sum is 2.

After some trials you will find them as -12 and 14.

Notice that their product is  (-12)*14 = -168  and their sum is  (-12)+14 = 2,

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


Now you get the same answer  x= 14 as in the Solution 1.

Let x be the number (now unknown) which is exactly half way between the width and the length of the playground.


Then, as you understand,  the length = (x+1)  and  the width = (x-1).


The area = 168 = length*width = (x+1)*(x-1).


Thus (x+1)*(x-1) = 168,  which implies   = 168,  i.e.   = 168+1 = 169.


Hence, x =  = 13.


Then the length = 13+1 = 14  and the width = 13-1 = 12.