Finding the Dimensions of a Rectangle Given Its Area and a Length-Width Relationship
When faced with a problem involving the dimensions of a rectangle given its area and a relationship between its length and width, one can use algebraic techniques, particularly the concept of quadratics, to find the solution. This article will guide you through the process of determining the dimensions of a rectangle where the length is 6 cm longer than the width, and the area is 187 square cm.
The Problem
A rectangle is 6 cm longer than it is wide. If its area is 187 square cm, what are the dimensions of the rectangle?
Solving the Problem
Let's denote the width of the rectangle as w cm. Given that the length is 6 cm longer than the width, the length can be expressed as (w 6) cm.
The area of a rectangle is given by the formula:
[ text{Area} text{length} times text{width} ]
Substituting the expressions for length and width into the area formula, we get:
[ 187 w(w 6) ]
Expanding and rearranging the equation to form a standard quadratic equation:
[ 187 w^2 6w ]
[ w^2 6w - 187 0 ]
To solve for w, we can use the quadratic formula,
[ w frac{-b pm sqrt{b^2 - 4ac}}{2a} ]
where a 1, b 6, and c -187.
Calculating the discriminant:
[ b^2 - 4ac 6^2 - 4 cdot 1 cdot (-187) 36 748 784 ]
Substituting back into the quadratic formula:
[ w frac{-6 pm sqrt{784}}{2 cdot 1} frac{-6 pm 28}{2} ]
Calculating the two potential solutions:
[ w frac{-6 28}{2} frac{22}{2} 11 ] [ w frac{-6 - 28}{2} frac{-34}{2} -17 ] (not a valid solution, as dimensions cannot be negative)Therefore, the width of the rectangle is 11 cm.
Calculating the Length
Given that the length is 6 cm longer than the width:
[ text{Length} w 6 11 6 17 , text{cm} ]
The Dimensions of the Rectangle
The dimensions of the rectangle are:
[ text{Width} 11 , text{cm} ] [ text{Length} 17 , text{cm} ]This completes the solution. The problem demonstrates how to use quadratic equations to solve real-world geometry problems involving the area of a rectangle.
Additional Examples
As an additional example, let's solve a similar problem. Suppose the area of a rectangle is 112 cm2 and the length is 6 cm longer than the width:
[ text{Area} 112 , text{cm}^2 ]
[ text{Length} text{Width} 6 , text{cm} ]
[ (w 6) times w 112 ]
[ w^2 6w - 112 0 ]
Using the quadratic formula, we find:
[ w frac{-6 pm sqrt{6^2 - 4 cdot 1 cdot (-112)}}{2 cdot 1} ]
[ w frac{-6 pm sqrt{36 448}}{2} ]
[ w frac{-6 pm sqrt{484}}{2} ]
[ w frac{-6 pm 22}{2} ]
[ w 8 ] (negative solution is not valid)
[ text{Length} w 6 8 6 14 , text{cm} ]
Hence, the dimensions are:
[ text{Width} 8 , text{cm} ] [ text{Length} 14 , text{cm} ]The process remains similar, but the numbers are different, demonstrating the flexibility of the method.