Finding the Dimensions of a Postcard: A Simple Geometry Problem
Suppose you have a postcard with an area of 24 square inches and a perimeter of 20 inches. How do you determine its dimensions? This article will guide you through the process, using basic geometry and algebra, to find the exact length and width of the postcard.
Understanding the Problem
A postcard, like a standard rectangular piece of paper, has an area and a perimeter. The area of the postcard is simply the product of its length and width, while the perimeter is the sum of all its sides.
In this case, we know that the area A is 24 square inches, and the perimeter P is 20 inches.
Setting Up Equations
Let's denote the length of the postcard as L and the width as W. We can now set up two equations based on the given information:
1. The Area Equation
The area of the postcard is given by the equation:
LW 24
2. The Perimeter Equation
The perimeter of the postcard is given by the equation:
2L 2W 20, which simplifies to L W 10
We now have a system of two equations with two unknowns:
LW 24 L W 10Solving the System of Equations
To solve this system, we can express W in terms of L from the second equation and substitute it into the first equation:
W 10 - L
Substituting W in the first equation:
L(10 - L) 24
This simplifies to:
10L - L^2 24
Further rearranging gives us:
L^2 - 10L 24 0
This is a quadratic equation, and we can solve it using the quadratic formula:
L (-b ± sqrt{b^2 - 4ac}) / 2a
Here, a 1, b -10, c 24. Plugging these values in:
L (10 ± sqrt{10^2 - 4 cdot 1 cdot 24}) / (2 cdot 1)
This simplifies to:
L (10 ± sqrt{100 - 96}) / 2
L (10 ± sqrt{4}) / 2
L (10 ± 2) / 2
This gives us two potential solutions for L: L (12 / 2) 6 L (8 / 2) 4
Now, we can find W for each solution:
If L 6, then W 10 - 6 4 If L 4, then W 10 - 4 6Hence, the dimensions of the postcard are 6 inches by 4 inches.
Conclusion
The dimensions of the postcard are 6 inches by 4 inches. This solution satisfies both the area and perimeter conditions given.