Solving for the Dimensions of a Rectangle Given Perimeter and Area

In this article, we will explore how to find the dimensions of a rectangle given its perimeter and area. We will use the example where the perimeter is 12 cm and the area is (frac{5}{2}) cm2.

The Importance of Perimeter and Area in Geometry

Understanding how to find the dimensions of a rectangle using its perimeter and area is crucial in many practical applications, such as in architecture, engineering, and everyday problem-solving. The perimeter is the total length of the boundary of the rectangle, while the area is the space enclosed within that boundary.

Formulas and Given Values

The perimeter of a rectangle is given by:

P 2l 2w

The area of a rectangle is given by:

A l times w

Given:

Perimeter, P 12) cm, Area, A frac{5}{2}) cm2.

Solving for the Dimensions

Let's denote the length by l and the width by w. According to the given perimeter and area, we have two equations:

2l 2w 12 implies l w 6 l times w frac{5}{2}

Step-by-Step Solution

Step 1: Solve for one variable

From the first equation, we can express w in terms of l:

w 6 - l

Step 2: Substitute into the area equation

Substitute w into the area equation:

l(6 - l) frac{5}{2}

Expanding this gives:

6l - l^2 frac{5}{2}

Rearranging leads to:

l^2 - 6l frac{5}{2} 0

Step 3: Multiply through by 2 to eliminate the fraction

2l^2 - 12l 5 0

Step 4: Use the quadratic formula

The quadratic formula is given by:

l frac{-b pm sqrt{b^2 - 4ac}}{2a}

Here, a 2, b -12, and c 5. Substituting these values into the quadratic formula gives:

l frac{12 pm sqrt{(-12)^2 - 4 cdot 2 cdot 5}}{2 cdot 2}

Calculating the discriminant:

144 - 40 104)

Thus,

l frac{12 pm sqrt{104}}{4} 3 pm frac{sqrt{26}}{2}

Step 5: Find w

Using w 6 - l:

If l 3 frac{sqrt{26}}{2}, then: w 6 - left(3 frac{sqrt{26}}{2}right) 3 - frac{sqrt{26}}{2} If l 3 - frac{sqrt{26}}{2}, then: w 6 - left(3 - frac{sqrt{26}}{2}right) 3 frac{sqrt{26}}{2}

Therefore, the dimensions of the rectangle are:

Length: 3 frac{sqrt{26}}{2} cm Width: 3 - frac{sqrt{26}}{2} cm

These dimensions satisfy both the perimeter and area conditions provided.

Alternative Interpretation

Another attempt can be made with:

2ab 12 implies ab 6 ab frac{5}{2}

Solving for b from the first equation, we get:

b frac{6}{a}

Substituting into the second equation:

a left(frac{6}{a}right) frac{5}{2}

This simplifies to:

6 - a frac{5}{2}

Solving this, we get:

6 - a frac{5}{2} implies a 6 - frac{5}{2} frac{7}{2} b 6 - frac{7}{2} frac{5}{2}

This results in a frac{7}{2} and b frac{5}{2}, which don't satisfy the original area condition. Therefore, the first interpretation is correct as it involves solving a quadratic equation.

Conclusion

The dimensions of the rectangle, satisfying both the perimeter and area conditions, are: