Dimension problem

Question

The area of a rectangle is
35 m2
, and the length of the rectangle is
3 m
more than double the width. Find the dimensions of the rectangle.

Answer 1

Hello, Colin

I am going to take a slightly different approach than Christopher.  

Let l represent the length of the rectangle.

Let w represent the width of the rectangle

 

l= 3+ 2w, since the problem states that the length is " 3 more than double the width"

 

We are going to use the formula for the area of a rectangle, then substitute 3+2w for l and 35 for a.

 

a=lw                                     Substituting

35=(3+2w)(w)                      Distributive property

35= 3w+ 2w^^2                   Rearranging 

35= 2w^2+3w                       Subtracting 35 from both sides

-35                 -35

0= 2w^3+3w-35                    Factoring

0= (2w-7)(w+5)                     

 

2w-7 =0  and w+5=0 are the two possible answers.  However, w+5=0 cannot be an answer since if we solve for w, we get w= -5.  And, we know that dimensions cannot be negative.  So, our answer must be w=3.5. (2w=7, w= 3.5).  

 

Now, we need to find l.  So, we are going to substitute 3.5 for w.

l= 3+ 2w

l= 3+ 2(3.5)

l= 3+ 7

l=10 and w= 3.5

 

I hope this helps. Thanks for posting.

D.Y. T.

dont forget to rate my answer :)

Answer 2

The area of a rectangle is (length*width). I abbreviate width with variable 'w'. Length is 3 more than 2 times the width.  I've used substitution for this->length=3+2w.  If we plug in what we have it would be: 35 = (3+2w)w

Now we have to solve for w:

Using an algebraic proptery we have 3w + 2w^2=35

Reording the equation gives us 2w^2 + 3w -35 = 0

This looks like we can use a system of equation to solve the problem. We could also use the quadradic equation : x = -b +- sqrt(b^2 -4ac/2a)

 

This looks like x= -3 +- sqrt(3^2 -4*2*(-35)/2*2)

 

This will give us two answers: our two dimensions! Go ahead and plug this one into a scientific calculator and you'll have your answer.

 

Hope this helps!