If the rectangles width is trippled and its length is doubled the perimeter of the new rectangle is 92 centimeter greater than the original perimeter what is the area of the original rectangle?
original p = 2 w + 2 L
final p = 2 (3w) + 2 (2L)
2 (3w) + 2 (2L) = 92 + 2 w + 2 L
3 w + 2 L = 46 + w + L
2 w + L = 46
That is all I can say unless I know something more, for example that the original or final rectangle is a square.
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To solve this problem, let's start by defining some variables.
Let:
- The original rectangle's width be "w"
- The original rectangle's length be "l"
- The original rectangle's perimeter be "P1"
According to the given information:
1. If the width is tripled, the new width would be 3w.
2. If the length is doubled, the new length would be 2l.
3. The new perimeter is 92 centimeters greater than the original perimeter, so the new perimeter would be P1 + 92.
We can calculate the new perimeter using the formulas for the perimeter:
The perimeter of the original rectangle, P1, can be calculated as:
P1 = 2w + 2l
The perimeter of the new rectangle (with tripled width and doubled length), P2, can be calculated as:
P2 = 2(3w) + 2(2l) = 6w + 4l
Since the new perimeter is 92 centimeters greater than the original perimeter, we can write the equation:
P2 = P1 + 92
Substituting the values of P1 and P2, we have:
6w + 4l = 2w + 2l + 92
Simplifying the equation, we get:
4w + 2l = 92
To continue solving for the area of the original rectangle, we need additional information. Are there any constraints or relationships between the width (w) and length (l) of the rectangle given? Let me know if you have any additional details that can help solve the problem.