Solving for the Dimensions of a Rectangle with a Given Area and Length

When dealing with geometric shapes, one of the common problems is to determine their dimensions (length and width) based on given properties like area. This article will walk you through the process of solving for the dimensions of a rectangle given that its length is three times its width and its area is 192 square centimeters. We will solve this problem step by step, illustrating the mathematical steps involved.

Formulating the Problem

Let's denote the width of the rectangle as w and the length as l.

The length is three times the width: l 3w The area of the rectangle is given by: Area l × w

Substituting the expression for length into the area formula, we get:

192 3w × w

This simplifies to:

192 3w2

Solving the Equation

To find the width (w), we first divide both sides of the equation by 3:

w2 192 / 3 64

Then, we take the square root of both sides:

w √64 8 cm

Now that we have the width, we can find the length by substituting this value back into the equation for length:

l 3w 3 × 8 24 cm

Conclusion

The dimensions of the rectangle are:

Width: 8 cm Length: 24 cm

This method can be applied to other similar problems, such as:

Let the width of the rectangle w cm. Then, 7w2 175 or w 5 cm. Therefore, the rectangle is 35 cm x 5 cm. A LW 192 cm2, L 3W, A W3W 3W2 192, W2 192/3 64, W √64 8 cm, L 3W 38 24 cm. Check: A 8×24 192 cm2 check. L 3W, A 60 in2, 60 W3W 3W2, W2 60/3, W √20 4.472, L 3×4.472 13.416, Check: 13.416×4.472 60 check. The area is L × W 3W × W 3W 192 cm2. Then, divide 192 by 3 and then the square root of the interim result to get the value of W: 192 / 3 64, √64 8. So W 8 cm and L 3W 24 cm. The perimeter P 2 × 8 2 × 24 64 cm. The diagonal D √(82 242) √640 ≈ 25.298 cm. The width of the rectangle x and the length 3x. Its area 3x2 192 or x2 192/3 64 or x 8. The rectangle is 24 cm x 8 cm. Suppose the width is x cm and the length is 4x cm. Then, 4x × x 196, x2 49, x 7 cm. The dimensions of the rectangle 28×7. If we let w be the width and 3w the length, 3w2 192, 3w2 192, w2 64, w √64 8, 3w 24. The rectangle is 8 units wide and 24 units long.

Conclusion

By understanding and applying the basic principles of algebra and geometry, we can easily solve for the dimensions of a rectangle given certain conditions. Whether it's a school assignment or a real-world problem, knowing how to work with these formulas can help you find solutions more efficiently. Practice with different examples and scenarios to enhance your problem-solving skills.