Question 471797: The length of a rectangle is 2 inches longer than the width. The area of the rectangle is 11 more than the perimeter. Find the width.
Answer by karaoz(32)   (Show Source):
You can put this solution on YOUR website! "The length of a rectangle is 2 inches longer than the width. The area of the rectangle is 11 more than the perimeter. Find the width."
First, translate the problem using a single letter for any unknown value.
It is a good idea to choose the letters so that they will remind us what exactly is represented by them.
So, for this problem, we could use:
L = the Length of a rectangle
W = the Width of a rectangle
A = the Area of a rectangle
P = the Perimeter of a rectangle
Translation of the problem is then:
"L is 2 inches longer than W. A is 11 more than P. Find W."
Next, settle on which units for all the unknown values are most convenient to use. Here, an obvious choice is "inch" for L and W.
While A and P are also unknown values, our knowledge about the geometry of rectangles should help us understand that these values depend on L and W.
Hence, P will also be in "inches" while A will be measured in "inches squared".
By the way, the difference in measurement units for P and A is exactly the reason why the second sentence in the problem statement is rather awkward, making the whole problem to be ill-defined.
Strictly speaking the area of the rectangle cannot be compared to the perimeter of the rectangle since they are never measured in the same units.
Only things that are commensurable can be compared, such as the length and the width of a rectangle.
The author of the problem has probably assumed that we will choose "inches" as the units for L, W and P and "inches squared" for A.
By saying that A is 11 more than P, the author simply wanted to say that A has 11 inches squared more than what P has in inches.
In other words, if we disregard this complication about units and the clumsiness of the statement then we could translate the statement as A = P + 11.
So, if L and W are both in inches then the first sentence can be further reduced into:
L = W + 2.
We have already discussed the second sentence and hence the "translation" of that sentence is:
A = P + 11.
The whole problem statement can be then expressed as:
L = W + 2. A = P + 11. W = ?
where the only thing we failed to convey is that we are talking about inches (or inches squared) and therefore that is the only thing we need to keep in mind until we get to the end of the solution process.
The above translation is now pure algebra but to solve the problem we will also need to use our knowledge about the area and the perimeter of a rectangle.
Namely, we will need the well known formulas: A = WL and P = 2W + 2L, which relate the values of A and P with W and L for any rectangle.
Finally, putting together the translation of the problem together with what we already know about A and P and their relations to L and W, is:
| L = W + 2 | (1) |
| A = P + 11 | (2) |
| P = 2W + 2L | (3) |
| A = WL | (4) |
W = ?
The first four lines now represent the system with 4 equations and 4 unknowns, which should have a solution.
Since we need to find out only the value of W then it is a good idea to use a solution process which will get us to the value of W as soon as possible.
So, in this case, we will eliminate the remaining three variables eventually ending up with one equation where the only unknown is W.
Substituting (3) and (4) into (2) we have:
WL = 2W + 2L + 11 (5)
Now, substituting (1) into (5) we will have only one equation with W as the only unknown.
Simplifying both sides of the equation, we have:
It is still possible to simplify the right hand side of the equation:
We are done with simplifying and it is time to recognize what type of equation we are dealing with.
This is polynomial equation of the second degree, hence we have quadratic equation.
One way to solve it is to transform it into its standard form and then use quadratic formula to get final solution.
Transforming equation
into its standard form can be done by subtracting 4W + 15 from both sides of the equation and then simplifying.
We are ready to use quadratic formula  , where a is the coefficient in front of  , b is the coefficient in front of  and c is a constant term.
Therefore, in our quadratic equation, a = 1, b = (-2) and c = (-15).
So, one solution is:
W = 1 - 4 = -3
and the other solution is:
W = 1 + 4 = 5.
Mathematically, they are both correct but for the given context of the problem we would be having difficulties to imagine rectangle whose widths is -3 inches.
Hence, we will opt to select the second solution as the only one that makes sense to us.
W = 5 inches
|
|
|
