Solving For W In A Geometric Figure Equilateral Triangle And Square
In this comprehensive guide, we will walk you through the process of determining the value of 'w' in a geometric figure composed of an equilateral triangle and a square. This problem, often encountered in mathematics, requires a solid understanding of geometric principles and algebraic manipulation. We'll break down the problem step-by-step, ensuring you grasp every concept along the way.
Understanding the Problem
Before we dive into the solution, let's clearly define the problem. Our geometric figure, as depicted in Figure 5, comprises an equilateral triangle ABC and a square BCDE. A crucial piece of information is that DE, a side of the square, is equal to 2w cm. We are also given that the perimeter of the entire figure is 140 cm. Our objective is to find the numerical value of 'w'.
This problem cleverly combines the properties of two fundamental geometric shapes: the equilateral triangle and the square. To solve it effectively, we need to recall and apply key characteristics of each shape, especially their sides and perimeters. Let's refresh our understanding of these concepts before proceeding.
Equilateral Triangles and Their Properties
An equilateral triangle is a special type of triangle where all three sides are of equal length. This property is not just a visual characteristic; it has significant implications for other aspects of the triangle, such as its angles. In an equilateral triangle, all three angles are also equal, each measuring 60 degrees. This uniformity makes equilateral triangles particularly easy to work with in geometric problems.
In our problem, triangle ABC is identified as an equilateral triangle. This means that sides AB, BC, and CA are all of the same length. This is a crucial piece of information that we will use to relate the triangle's dimensions to the square's dimensions and, ultimately, to the overall perimeter of the figure.
Squares and Their Properties
A square is another fundamental geometric shape, characterized by four equal sides and four right angles (90 degrees). The equality of sides is a defining feature of a square, and it simplifies many calculations involving perimeter and area. The right angles ensure that the square's sides meet at perfect corners, contributing to its symmetrical appearance.
In our figure, BCDE is a square. This tells us that sides BC, CD, DE, and EB are all equal in length. Moreover, since DE is given as 2w cm, we immediately know that all sides of the square are 2w cm. This connection between the variable 'w' and the square's dimensions is a key step in bridging the gap between the given information and the solution we seek.
Connecting the Triangle and the Square
The beauty of this problem lies in the connection between the equilateral triangle and the square. They share a common side, BC. This shared side serves as a bridge, linking the dimensions of the triangle to the dimensions of the square. Because triangle ABC is equilateral, BC is equal in length to AB and CA. And since BCDE is a square, BC is also equal to CD, DE, and EB. This shared side provides a crucial equation that we will use to solve for 'w'.
Setting Up the Equation
Now that we understand the individual shapes and their connection, we can formulate an equation to represent the perimeter of the entire figure. The perimeter is the total length of the outer boundary of the shape. In our case, the outer boundary consists of sides AB, AC, CD, DE, and EB. We know the perimeter is 140 cm, and we have expressions for the lengths of some of these sides in terms of 'w'.
Let's break down the perimeter calculation:
- AB is a side of the equilateral triangle. Since BC is a side of the square and equal to 2w cm, and all sides of the equilateral triangle are equal, AB is also 2w cm.
- AC is another side of the equilateral triangle, so it is also 2w cm.
- CD is a side of the square, and we know all sides of the square are 2w cm.
- DE is given as 2w cm.
- EB is the final side of the square, and it is also 2w cm.
Therefore, the perimeter can be expressed as the sum of these sides: 2w + 2w + 2w + 2w + 2w. This simplifies to 10w. We know the perimeter is 140 cm, so we can set up the equation:
10w = 140
This equation is the cornerstone of our solution. It directly relates 'w' to the given perimeter and allows us to solve for its value.
Solving for 'w'
With our equation established, solving for 'w' is a straightforward algebraic process. We have the equation 10w = 140. To isolate 'w', we need to divide both sides of the equation by 10.
Dividing both sides by 10, we get:
w = 140 / 10
w = 14
Therefore, the value of 'w' is 14. This means that the side DE of the square, which is 2w cm, is 2 * 14 = 28 cm. All sides of the square are 28 cm, and all sides of the equilateral triangle are also 28 cm.
Verifying the Solution
It's always a good practice to verify our solution to ensure accuracy. We can do this by plugging the value of 'w' back into the perimeter calculation and checking if it matches the given perimeter of 140 cm.
We found that w = 14. Let's substitute this value into our perimeter expression: 10w.
10 * 14 = 140
Our calculation confirms that the perimeter is indeed 140 cm when w = 14. This reinforces the correctness of our solution.
Conclusion
In this problem, we successfully determined the value of 'w' by combining our knowledge of equilateral triangles, squares, and perimeter calculations. We set up an equation based on the given perimeter and the side lengths of the shapes, then solved the equation using basic algebraic principles. The solution, w = 14, provides a concrete value for a variable that defines the dimensions of the geometric figure.
This type of problem highlights the interconnectedness of geometric concepts and the power of algebraic tools in solving geometric challenges. By carefully analyzing the properties of shapes and formulating equations, we can unlock solutions to seemingly complex problems. Remember, practice is key to mastering these skills. Work through similar problems to solidify your understanding and build your problem-solving confidence.
Practice Problems
To further enhance your understanding, try these practice problems:
- Suppose the perimeter of the figure is 160 cm. What is the value of 'w'?
- If DE = 3w cm and the perimeter is 150 cm, find the value of 'w'.
- What happens to the value of 'w' if the equilateral triangle is replaced by an isosceles triangle with two sides equal to BC?
By tackling these problems, you'll gain valuable experience and deepen your grasp of geometric problem-solving techniques.
Key Takeaways
- Understanding the properties of geometric shapes is crucial.
- Setting up equations based on given information is a key step in solving geometric problems.
- Algebraic manipulation is a powerful tool for finding unknown values.
- Always verify your solution to ensure accuracy.
- Practice is essential for mastering problem-solving skills.
By keeping these takeaways in mind, you'll be well-equipped to tackle a wide range of geometric challenges. Remember, mathematics is a journey of learning and discovery. Embrace the challenges, and enjoy the process of finding solutions!
