Finding the Dimensions of a Rectangle Given its Diagonal and Length-Width Relationship

In this article, we will explore the method to determine the dimensions of a rectangle when given its diagonal length and the relationship between its length and width. This is a common application of Pythagoras' theorem.

Problem Statement

The problem is that the diagonal of a rectangle measures 10 cm. The length of the rectangle is 2 cm more than the width. We need to find the dimensions of the rectangle, i.e., the length and the width.

Solving the Problem

Let's denote the width of the rectangle as W and the length as L. We know that the length is 2 cm more than the width, so we can express it as:

[L W 2]

According to the Pythagorean theorem, the diagonal of a rectangle, which forms the hypotenuse of a right-angled triangle formed by the length and the width, can be calculated as follows:

[L^2 W^2 10^2]

Substituting the value of L into the Pythagorean theorem:

[(W 2)^2 W^2 100]

Expanding and simplifying the equation:

[W^2 4W 4 W^2 100]

[2W^2 4W 4 100]

[2W^2 4W - 96 0]

Dividing the entire equation by 2:

[W^2 2W - 48 0]

This quadratic equation can be solved using the quadratic formula:

[W frac{-b pm sqrt{b^2 - 4ac}}{2a}]

Where a 1, b 2, and c -48. Substituting these values:

[W frac{-2 pm sqrt{2^2 - 4 cdot 1 cdot (-48)}}{2 cdot 1}]

[W frac{-2 pm sqrt{4 192}}{2}]

[W frac{-2 pm sqrt{196}}{2}]

[W frac{-2 pm 14}{2}]

So, we get two solutions:

[W_1 frac{12}{2} 6]

[W_2 frac{-16}{2} -8]

Since width cannot be negative, we take W 6 cm. Therefore, the length L W 2 6 2 8 cm.

Verification

To verify, we use the Pythagorean theorem again:

[L^2 W^2 10^2]

[8^2 6^2 100]

[64 36 100]

Thus, the solution is correct with a diagonal of 10 cm.

Conclusion

The dimensions of the rectangle are:

[Width 6 , text{cm}]

[Length 8 , text{cm}]

This method can be used to solve similar problems in geometry involving the diagonal of a rectangle.