The length of a rectangle is twice the width. If the length is increased by 4 inches and the width is diminished by 1 inch, a new rectangle is formed whose perimeter is 198 inches. Fine the dimensions of the original rectangle.
l=2w-----1
2{(l+4)+(w-1)}=198-----2
Let's assume the width of the original rectangle is x inches.
According to the problem, the length of the rectangle is twice the width, which means the length is 2x inches.
When the length is increased by 4 inches, the new length becomes 2x + 4 inches.
Similarly, when the width is diminished by 1 inch, the new width becomes x - 1 inches.
We are given that the perimeter of the new rectangle is 198 inches.
The perimeter of a rectangle is given by the formula: P = 2l + 2w, where P is the perimeter, l is the length, and w is the width.
Substituting the values into the formula for the new rectangle, we have:
198 = 2(2x + 4) + 2(x - 1)
Simplifying the equation:
198 = 4x + 8 + 2x - 2
Combining like terms:
198 = 6x + 6
Subtracting 6 from both sides:
192 = 6x
Dividing both sides by 6:
x = 32
Therefore, the width of the original rectangle is 32 inches.
Since the length is twice the width, the length of the original rectangle is 2 * 32 = 64 inches.
Hence, the dimensions of the original rectangle are width = 32 inches and length = 64 inches.
To solve this problem, we can use algebraic equations. Let's denote the width of the original rectangle as "w" inches. Since the length is twice the width, we can represent the length as "2w" inches.
According to the problem, when the length is increased by 4 inches, and the width is diminished by 1 inch, a new rectangle is formed with a perimeter of 198 inches.
The perimeter of a rectangle is given by the formula: P = 2(l + w), where P is the perimeter, l is the length, and w is the width.
So, for the original rectangle, the perimeter is given by:
P1 = 2(2w + w) = 2(3w) = 6w
For the new rectangle, the perimeter is given as 198 inches:
P2 = 198
Since the length of the new rectangle is (2w + 4) inches, and the width is (w - 1) inches, we can rewrite the formula for the new perimeter as:
P2 = 2((2w + 4) + (w - 1)) = 2(3w + 3) = 6w + 6
Now, equating the two perimeters:
P1 = P2
6w = 6w + 6
By subtracting 6w from both sides of the equation, we have:
0 = 6
Since this equation is not true, it means there is no solution that satisfies the given conditions. Therefore, we cannot find the dimensions of the original rectangle that meet the criteria provided in the problem statement.