Finding The Length Of BE In An Equilateral Triangle

Hey there, math enthusiasts! Let's dive into a geometry problem that's a bit of a classic. We're given an equilateral triangle and asked to find the length of a line segment within it. Sound fun? Let's break it down, step by step, so you can totally nail this type of problem next time! We're talking about Q.116 $\triangle ABC$! Remember how we talked about these equilateral triangles? Let's refresh our memories and solidify that base knowledge with this problem! This article will thoroughly explore this mathematical problem! We'll not only solve it but also understand the 'why' behind each step. Let's start with the basics.

Understanding the Problem and Setting Up the Scene

Okay, guys, the problem presents us with a scenario: we have an equilateral triangle, $\triangle ABC$, and we know its sides are each 12 cm long. Now, because it's an equilateral triangle, all three angles are equal too, right? Each angle in an equilateral triangle is 60 degrees. Cool, that's our starting point. The problem introduces $CD$, which is the angle bisector of $\angle C$. This angle bisector cuts the side $AB$ at point $D$. A key thing to remember here is that in an equilateral triangle, the angle bisector also acts as a median and an altitude. This means $CD$ not only splits $\angle C$ in half but also cuts $AB$ into two equal parts, making $AD = DB = 6$ cm. Then, we have point $E$, which is the midpoint of $CD$. And the question asks us to find the length of $BE$. That's the challenge! So, we have all the things in a triangle! That's it! Let's now establish all the knowledge we need to solve the problem and apply it!

To tackle this problem, we're going to use a combination of geometric properties and the Pythagorean theorem. First off, because $CD$ is an altitude in an equilateral triangle, we can calculate its length. The height of an equilateral triangle with side $s$ is given by $h = (\sqrt{3}/2) * s$. In our case, $s = 12$ cm, so the length of $CD$ is $({\sqrt{3}/2}) * 12 = 6\sqrt{3}$ cm. Also, since $E$ is the midpoint of $CD$, we know $CE = ED = (1/2) * CD = 3\sqrt{3}$ cm. This part is crucial! Now that we have all the information, let's get into the main stage. That's when we start constructing our equations to find the solutions! Let's do it!

Applying the Pythagorean Theorem and Solving for BE

Alright, folks, the Pythagorean theorem is our best friend here. We're going to focus on the right-angled triangle $\triangle BDE$. Why this one? Because we want to find $BE$, and $BE$ is the hypotenuse of this triangle. We already know $BD = 6$ cm. We calculated $ED = 3\sqrt{3}$ cm. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, $BE^2 = BD^2 + ED^2$. This is it! Let's solve it and get the final answer! Don't let yourself get confused here. Just follow the equations and you'll get it right! Believe in yourself and keep going, you are almost there!

Let's plug in the values! $BE^2 = 6^2 + (3\sqrt{3})^2$. Calculating this, we get $BE^2 = 36 + 27 = 63$. Taking the square root of both sides, we find that $BE = \sqrt{63}$ cm. We can simplify this further because $63 = 9 * 7$, so $BE = \sqrt{9 * 7} = 3\sqrt{7}$ cm. And there you have it, guys! We've successfully found the length of $BE$. It is an interesting problem, isn't it? Geometry is full of puzzles like this, and each solution brings a sense of achievement and a deeper understanding of mathematical principles. This journey of discovery is the best thing about math! I'm pretty sure you enjoyed it!

Conclusion: The Answer and Key Takeaways

So, the correct answer is $3\sqrt{7}$ cm. That corresponds to option 2 in the multiple-choice question. Congratulations if you got it right! The key takeaways from this problem are: understanding the properties of an equilateral triangle, knowing that the angle bisector, median, and altitude are the same line, and being comfortable with the Pythagorean theorem. These concepts are fundamental in geometry and will come in handy in many other problems. Remember to always draw a diagram to visualize the problem, break down the problem into smaller parts, and apply the relevant formulas and theorems. Practice makes perfect, so keep solving more problems, and you'll become a geometry pro in no time! Keep practicing the concepts that you've learned here and don't be afraid to try some more complicated ones!

Additional Tips for Solving Geometry Problems

• Draw a Clear Diagram: A well-labeled diagram is your best friend. It helps you visualize the problem and identify relationships between different elements.
• Identify Knowns and Unknowns: Clearly list what you know and what you need to find. This helps you plan your solution strategy.
• Break Down Complex Problems: Divide the problem into smaller, more manageable parts. Solve each part separately and then combine the results.
• Use the Right Tools: Make sure you're familiar with relevant formulas, theorems, and properties. Knowing the right tools is half the battle.
• Check Your Work: Always double-check your calculations and make sure your answer makes sense in the context of the problem.

This is just one example of how to approach a geometry problem. There are many other types of problems, each requiring a slightly different approach. But the general principles remain the same: understand the problem, plan your solution, apply the appropriate formulas, and check your work. So, keep practicing, keep learning, and keep enjoying the world of mathematics. Good luck with your future math endeavors, and keep exploring! And if you want to explore more, feel free to dive deeper into the world of geometry! There is always something new to learn and something to be amazed at.

I hope this step-by-step guide was helpful. Feel free to ask any further questions. Keep up the excellent work, and always keep exploring the world of mathematics! And don't forget to practice so you can nail these problems in the future. See you next time, guys! Keep learning and stay curious!