|
Question 99336This question is from textbook algebra structure and method
: Kirsten built a rectangular corral with a fence on three sides.A side of the barn served as a short side of the corral.She used 130m of fencing.The length of the corral was 20m longer than the width. Find the dimensions of the corral.Choices for the widths:25, 30, 35.
This question is from textbook algebra structure and method
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Since the side of the barn served as one width of the corral, Kirsten's 130 meters of fencing
had to serve as the other width (W) and the two lengths (L). So the 130 meters of fencing
must provide W + L + L. In equation form this is:
.
W + L + L = 130
.
But you are told that L is 20 meters longer than W which means L = W + 20. In the equation
for the total length of the fencing you can substitute W + 20 for L since they are equal. This
makes the equation become:
.
W + (W + 20) + (W + 20) = 130
.
The two sets of parentheses are each preceded by a plus sign, so the parentheses can be
removed without affecting the equation. This makes the equation become:
.
W + W + 20 + W + 20 = 130
.
On the left side combine the three W terms and the two 20's to make the equation:
.
3W + 40 = 130
.
Get rid of the 40 on the left side by subtracting 40 from both sides to reduce the equation
to:
.
3W = 90
.
Finally, solve for W (the width) by dividing both sides by 3 (the multiplier of W) to get:
.
W = 90/3 = 30 meters
.
Since W is 30 meters and L is 20 meters longer than that, L must be 30 + 20 or 50 meters.
.
This says the size of the corral is 30 meters by 50 meters. And 30 meters of the barn wall
is used as one side of this corral. The total amount of fencing used is 50 meters plus
30 meters plus another 50 meters, and this adds up to be 130 meters.
.
Hope this helps you to understand what the problem was asking for and how you can go about
getting the answer.
.
|
|
|
| |
