Finding the Area of a Rectangle Using Diagonal and Perimeter

In this article, we will explore a problem involving a rectangle where its diagonal and perimeter are given, and we will use this information to find the area. This problem is a real-world application of the Pythagorean theorem and algebraic manipulation.

Problem Statement

Given a rectangle with a diagonal of 13 cm and a perimeter of 34 cm, the task is to find the area of the rectangle.

Step-by-Step Solution

Step 1: Set up the equations

First, let's denote the length and width of the rectangle as l and w, respectively.

Formula for the perimeter of a rectangle: P 2(l w) 34 → l w 17 Formula for the diagonal of a rectangle using the Pythagorean theorem: d √(l2 w2) 13 → l2 w2 169

Step 2: Solve the equations

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

w 17 - l

Substituting w into the second equation:

l2 (17 - l)2 169

Expanding and simplifying the equation:

l2 289 - 34l l2 169

2l2 - 34l 120 0

Dividing the entire equation by 2:

l2 - 17l 60 0

Step 3: Factor the quadratic equation

The quadratic equation can be factored as:

(l - 12)(l - 5) 0

Thus, the solutions for l are:

l 12 l 5

Step 4: Find the dimensions

If l 12, then:

w 17 - 12 5

If l 5, then:

w 17 - 5 12

So the dimensions of the rectangle are l 12 cm and w 5 cm.

Step 5: Calculate the area

The area A of the rectangle is given by:

A l × w 12 × 5 60 cm2

Conclusion

The area of the rectangle is 60 cm2.

Alternative Approach

Another way to approach the problem is to use the fact that half of the perimeter is the width the length. Let's denote the width as W and the length as L. We can use these relations:

P/2 W L → 17 W L d √(W2 L2) 13 → W2 L2 169

From the first equation, we can express L in terms of W:

L 17 - W

Substituting L into the second equation:

W2 (17 - W)2 169

Expanding and simplifying the equation:

W2 289 - 34W W2 169

2W2 - 34W 120 0

Dividing the entire equation by 2:

W2 - 17W 60 0

This quadratic equation can be factored as:

(W - 12)(W - 5) 0

Thus, the solutions for W are:

W 12 W 5

Correspondingly, if W 12, then L 5; if W 5, then L 12.

The area is calculated as:

A W × L 12 × 5 60 cm2

Therefore, the area of the rectangle is 60 cm2.