SOLUTION: 1)The perimeter of a rectangle is 28 cm. find the range of possible values of the width of the rectangle if the diagonal is less than 10 cm. 2)The area of the rectangle is 12 cm

Hi, there--

YOUR PROBLEM:
1)The perimeter of a rectangle is 28 cm. find the range of possible values of the width of 
the rectangle if the diagonal is less than 10 cm. 

2)The area of the rectangle is 12 square cm. Find the range of possible values of the width 
of the rectangle if the diagonal is more than 5 cm.

SOLUTION #1:
1)The perimeter of a rectangle is 28 cm. find the range of possible values of the width of 
the rectangle if the diagonal is less than 10 cm.

Let D be the diagonal of the rectangle.
Let W be the width of the rectangle.
Let L be the length of the rectangle.

The formula for the perimeter of a rectangle is 2(W+L). We know that the perimeter is 28 cm, 
so we can write an expression for the length in terms of W.

2(W + L) = 28
W + L = 14
L = 14 - W

Now consider the diagonal of the rectangle. If we trace the diagonal from the top left corner 
to the bottom right corner, then up the width and across the length, we see that we have
traced a right triangle. We can use the Pythagorean Equation to find the range of possible 
values for the width.

The Pythagorean Equation is . In our problem,

a = the width, or W
b = the length, or 14 - W
c = the diagonal which is less than 10 cm.







Solve the inequality by factoring.


Since the the product of these two factors is less than zero, it must be negative. Thus one 
factor must be negative and the other positive ("A negative times a positive is a negative.") 
We have two cases:

Case I: W - 6 < 0 and W - 8 > 0  OR  Case 2: W- 6 > 0 and W - 8 < 0

Case 1:
W - 6 < 0 and W - 8 > 0
W < 6 and W > 8
Is is not possible for a number to be less than 6 and greater than 8. Case 1 has no solutions.

Case II:
W - 6 > 0 and W - 8 < 0
W > 6 and W < 8
The numbers that are greater than 6 and less than 8 are all the numbers between 6 and 8. 
Symbolically, we write this as 6

a is the width, or W
b is the length, or  12/W
c is the diagonal which is greater than 5.




Multiply every term by W^2 to clear the denominator.




Solve by factoring.



We again have several cases to consider.
Case 1:  AND  OR
Case 2:  AND 

In Case 1:
 AND 

Then,
[W<4 and W>-4 or W>4 and W<-4] AND  [W>3 and W>-3  or W<3 and W<-3]

So,
W<4 and W>-4 means W is between -4 and 4, or -43 and W>-3 implies that W>3
W<3 and W<-3 implies that W<-3

Taken together, we have
-43 or W<-3]

Both sides of the AND must be true, so W is between -3 and -4 or between 3 and 4.
We cannot have a rectangle with a negative width, so W is between 3 and 4.

In Case #2:
Case 2:  AND 

Then,
[W>4 and W>-4 or W<4 and W<-4] AND  [W>3 and W<-3  or W<3 and W>-3]

So,
W>4 and W>-4 implies W>4

On the right side, 
W cannot be greater than 3 and less than -3
W<3 and W>-3 implies that W is between -3 and 3.

Both sides of the AND statement must be true.
W cannot be greater than 4 and between -3 and 3.
Thus Case II is not a solution.

In the context of our problem, the width is between 3 and 4 cm.

WOW! I hope this helps.
Ms. Figgy