Solving for Rectangle Dimensions Using Perimeter and Given Conditions
The problem presented involves finding the dimensions of a rectangle given its perimeter and a relationship between its length and width. Specifically, we are tasked with determining the length and width of a rectangle when the length is one less than four times the width, and the perimeter is 18 units. This problem can be solved using a combination of algebraic reasoning and basic geometry principles.
Setting Up the Equations
Let's denote the width of the rectangle as ( w ) and the length as ( l ). According to the problem, the length is one less than four times the width, which can be expressed mathematically as:
[ l 4w - 1 ]
Additionally, we know that the perimeter of the rectangle is 18 units. The formula for the perimeter of a rectangle is given by:
[ P 2l 2w ]
Substituting the given perimeter, we get:
[ 18 2l 2w ]
Dividing the entire equation by 2 to simplify, we obtain:
[ 9 l w ]
Now, we have two equations:
[ l 4w - 1 ] [ l w 9 ]Solving the Equations
To solve these equations simultaneously, we can substitute the expression for ( l ) from the first equation into the second equation:
[ 4w - 1 w 9 ]
Simplifying this, we get:
[ 5w - 1 9 ]
Adding 1 to both sides of the equation, we have:
[ 5w 10 ]
Dividing both sides by 5, we find:
[ w 2 ]
Now that we have the width, we can substitute ( w 2 ) back into the first equation to determine the length:
[ l 4w - 1 ]
[ l 4(2) - 1 ]
[ l 8 - 1 ]
[ l 7 ]
Thus, the width of the rectangle is 2 units, and the length is 7 units.
Verification
To ensure the solution is correct, we can verify the perimeter calculation:
[ P 2l 2w ]
[ P 2(7) 2(2) ]
[ P 14 4 ]
[ P 18 ]
The perimeter is indeed 18 units, confirming our calculations.
Conclusion
In summary, by using algebraic equations and basic geometry principles, we determined that the width of the rectangle is 2 units, and the length is 7 units. This solution aligns with the given conditions and verifies the perimeter calculation.