Finding the Perimeter of a Rectangle Given Area and Diagonal

The problem asks us to find the perimeter of a rectangle when its area and the length of its diagonal are known. We are given the area is 300 cm² and the diagonal is 25 cm.

Understanding Rectangle Properties

Let the length of the rectangle be \(l\) and the width be \(w\). The key properties of a rectangle relevant to this problem are:

• Area (\(A\)) = length \(\times\) width = \(l \times w\)
• Diagonal (\(d\)): By the Pythagorean theorem, the square of the diagonal is equal to the sum of the squares of the length and width (\(d^2 = l^2 + w^2\)).
• Perimeter (\(P\)) = 2 \(\times\) (length + width) = \(2(l + w)\)

Applying the Given Information

We are given:

• Area \(A = lw = 300\) cm²
• Diagonal \(d = 25\) cm

Using the Pythagorean theorem for the diagonal, we have:

\(d^2 = l^2 + w^2\)

\(25^2 = l^2 + w^2\)

\(625 = l^2 + w^2\)

Calculating the Perimeter

We need to find the perimeter, which is \(P = 2(l + w)\). To do this, we need the sum of the length and width, \((l + w)\).

Consider the algebraic identity: \((l + w)^2 = l^2 + w^2 + 2lw\)

We know the value of \(l^2 + w^2\) from the diagonal (625) and the value of \(lw\) from the area (300).

Substitute these values into the identity:

\((l + w)^2 = 625 + 2(300)\)

\((l + w)^2 = 625 + 600\)

\((l + w)^2 = 1225\)

Now, take the square root of both sides to find \(l + w\):

\(l + w = \sqrt{1225}\)

To find the square root of 1225, we can note that \(30^2 = 900\) and \(40^2 = 1600\). The number ends in 5, so its square root must also end in 5. Let's try 35:

\(35 \times 35 = 1225\)

So, \(l + w = 35\) cm.

Finally, we can calculate the perimeter:

\(P = 2(l + w)\)

\(P = 2(35)\)

\(P = 70\) cm

Verification (Optional)

If \(l + w = 35\) and \(lw = 300\), we can think of \(l\) and \(w\) as the roots of the quadratic equation \(x^2 - (l+w)x + lw = 0\), which is \(x^2 - 35x + 300 = 0\).

We can factor this equation: \((x - 15)(x - 20) = 0\). The roots are \(x = 15\) and \(x = 20\).

So, the length and width are 20 cm and 15 cm (or vice versa). Let's check if these values satisfy the diagonal condition:

\(l^2 + w^2 = 20^2 + 15^2 = 400 + 225 = 625\)

\(d^2 = 25^2 = 625\)

Since \(l^2 + w^2 = d^2\), the values \(l=20\) and \(w=15\) are consistent with the given information.

The perimeter is \(2(l+w) = 2(20+15) = 2(35) = 70\) cm.

Conclusion

Based on the calculations using the area and diagonal of the rectangle, the perimeter is 70 cm.

Revision Table: Rectangle Area, Diagonal, and Perimeter

Additional Information: Geometric Formulas and Identities

This problem demonstrates how basic geometric formulas and algebraic identities can be combined to solve problems. The key identity used was \((a+b)^2 = a^2 + b^2 + 2ab\). In this case, \(a\) and \(b\) were the length and width of the rectangle. Understanding how to relate area, perimeter, and diagonal through the side lengths \(l\) and \(w\) is fundamental for solving such geometry problems.

The Pythagorean theorem is also a cornerstone here, linking the diagonal (hypotenuse) to the sides (legs) of the right triangles formed within the rectangle.

Always double-check calculations, especially square roots, to ensure accuracy in finding the final answer.