sparkles3144 wrote:A rectangle has a perimeter of 28 inches. What are the dimensions of the rectangle?
(1) The area of the rectangle is 40 inches squared.
(2) The length is 6 inches more than the width
I know how to solve this.
However, I am wondering if I can assume that l and w are integers.
It is not stated.
Thanks!
No, we cannot assume that the lengths are integers.
Here's one solution:
Target question:
What are the dimensions of the rectangle?
Given: A rectangle has a perimeter of 28 inches.
Let L = length of rectangle
Let W = width of rectangle
If the perimeter is 28, then 2L + 2W = 28
We can simplify this by dividing both sides by 2 to get:
L + W = 14
Statement 1: The area of the rectangle is 40 inches squared.
In other words, LW = 40
Since we already know that
L + W = 14, we can solve for W to get W = 14 - L
Now take the equation LW = 40, and replace W with (14 - L) to get . . .
L(14 - L) = 40
Expand: 14L - L^2 = 40
Rearrange to get: L^2 - 14L + 40 = 0
Factor: (L - 4)(L - 10) = 0
So, L = 4 or 10
If L = 4, then W = 10, which means
the dimensions are 4 by 10
If L = 10, then W = 4, which means
the dimensions are still 4 by 10
In other words, the
the dimensions must be 4 by 10
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: The length is 6 inches more than the width
In other words, L = W + 6
Since we already know that
L + W = 14, we can solve for W to get W = 14 - L
Now take the equation L = W + 6, and replace W with (14 - L) to get . . .
L = (14 - L) + 6
Rearrange: 2L = 20
Solve: L = 10
If L = 10, then W = 4, which means
the dimensions are 4 by 10
Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer =
D
Cheers,
Brent
. A key point I missed out and was scratching my head .. Thanks a lot for your guidance again
