Dimensions Of A Rectangular Parallelepiped: A Step-by-Step Guide

by ADMIN 65 views

Hey guys! Today, we're diving into a cool geometry problem: figuring out the dimensions of a rectangular parallelepiped. If you're scratching your head looking at ABCDA'B'C'D' and a bunch of side lengths, don't worry! We'll break it down together. We are given that AC = 40 cm, AB' = 2√337 cm, and AD' = 30 cm. Let's get started and make those dimensions crystal clear!

Understanding the Problem

Before we jump into calculations, let's make sure we're all on the same page. So, what exactly is a rectangular parallelepiped? Think of it as a fancy name for a box – a 3D shape where all angles are right angles. Imagine a brick or a cereal box; that's the kind of shape we're dealing with. The key here is that we're given some lengths, but not the direct dimensions (length, width, height). Instead, we have the diagonal of the base (AC) and diagonals of two of the faces (AB' and AD'). Our mission, should we choose to accept it (and we do!), is to use this information to find the actual length, width, and height of our 3D box.

Why is this important? Well, geometry problems like these aren't just about abstract shapes. They teach us how to visualize in three dimensions and how to use the Pythagorean theorem in a practical way. Plus, it's just plain satisfying to crack a tricky puzzle! The beauty of this problem lies in its blend of spatial reasoning and algebraic manipulation. We’ll need to carefully apply the Pythagorean theorem in three dimensions and set up a system of equations. Remember, the main goal is to relate the given diagonal lengths to the dimensions of the parallelepiped.

Let's think about our strategy. We know that the diagonals are related to the sides through the Pythagorean theorem. This means we'll be setting up equations that involve squares of lengths. Don't be intimidated by the square roots and larger numbers; we'll take it one step at a time. Grab your pencils, paper, and maybe a calculator – it's time to unravel this parallelepiped mystery!

Setting Up the Equations

Alright, let's translate our geometric information into algebraic equations. This is where the Pythagorean theorem becomes our best friend. Remember, it states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (a² + b² = c²). We're going to use this in 2D and 3D to connect the given lengths to the dimensions of our parallelepiped.

Let's label the dimensions: let AB = length = l, AD = width = w, and AA' = height = h. These are the three unknowns we're trying to find. Now, let's look at the given information, starting with the base diagonal AC. Triangle ADC is a right-angled triangle, so by the Pythagorean theorem:

AC² = AD² + DC²

Since ABCD is a rectangle, DC = AB = l, and we know AC = 40 cm. So, our first equation is:

40² = w² + l²

1600 = w² + l²

Next, let's consider the face diagonals AB' and AD'. Triangle ABB' is a right-angled triangle, so:

AB'² = AB² + BB'²

Since BB' = AA' = h, and we know AB' = 2√337 cm, we get:

(2√337)² = l² + h²

4 * 337 = l² + h²

1348 = l² + h²

Similarly, triangle ADD' is a right-angled triangle, so:

AD'² = AD² + DD'²

Since DD' = AA' = h, and we know AD' = 30 cm, we get:

30² = w² + h²

900 = w² + h²

Now we have a system of three equations:

  1. l² + w² = 1600
  2. l² + h² = 1348
  3. w² + h² = 900

This might look intimidating, but don't worry! We've broken down the geometry into manageable algebra. The next step is to solve this system of equations to find the values of l, w, and h. We'll use techniques like substitution and elimination to get there.

Solving the System of Equations

Okay, we've got our three equations, and now it's time to put on our algebra hats and solve for l, w, and h. This part might seem a little like a puzzle, but that's what makes it fun! We'll use a combination of substitution and elimination to isolate our variables.

Let's start by labeling our equations for easy reference:

  1. l² + w² = 1600
  2. l² + h² = 1348
  3. w² + h² = 900

One strategy we can use is elimination. Let's subtract equation (2) from equation (1). This will eliminate l² and give us a relationship between w² and h²:

( l² + w² ) - ( l² + h² ) = 1600 - 1348

w² - h² = 252

Now, let's call this new equation (4):

  1. w² - h² = 252

Notice that we now have two equations involving w² and h²: equation (3) and equation (4). Let's write them together:

  1. w² + h² = 900
  2. w² - h² = 252

We can eliminate h² by adding equations (3) and (4):

( w² + h² ) + ( w² - h² ) = 900 + 252

2w² = 1152

w² = 576

Taking the square root of both sides, we get:

w = 24 cm

Great! We've found the width. Now, we can substitute this value back into equation (3) to find h²:

24² + h² = 900

576 + h² = 900

h² = 324

Taking the square root of both sides, we get:

h = 18 cm

Awesome! We've found the height. Finally, we can substitute the value of w² back into equation (1) to find l²:

l² + 24² = 1600

l² + 576 = 1600

l² = 1024

Taking the square root of both sides, we get:

l = 32 cm

We did it! We've solved the system of equations and found all three dimensions: length, width, and height. It’s like we've decoded the secret dimensions of this parallelepiped! The process involved careful algebraic manipulation, strategic elimination, and a good understanding of the Pythagorean theorem.

The Dimensions Revealed

Alright, after all that algebraic sleuthing, we've arrived at the grand finale: the dimensions of our rectangular parallelepiped! Remember, we set out to find the length (l), width (w), and height (h) given some diagonal lengths. And guess what? We nailed it!

Here's what we found:

  • Length (l) = 32 cm
  • Width (w) = 24 cm
  • Height (h) = 18 cm

So, our rectangular parallelepiped has a length of 32 cm, a width of 24 cm, and a height of 18 cm. Pretty cool, right? We started with a seemingly complex problem involving diagonals and square roots, and we systematically broke it down into smaller, manageable steps. We used the Pythagorean theorem, set up a system of equations, and then employed algebraic techniques to solve for our unknowns.

This problem is a great example of how math can be used to describe and understand the world around us. Three-dimensional geometry might seem abstract, but it has real-world applications in fields like architecture, engineering, and even computer graphics. By solving problems like this, we're not just flexing our math muscles; we're also developing our spatial reasoning and problem-solving skills.

Conclusion

So, there you have it! We've successfully navigated the world of rectangular parallelepipeds and discovered how to find their dimensions given some tricky diagonal information. Remember, the key to tackling these kinds of problems is to break them down into smaller steps, use the right tools (like the Pythagorean theorem), and don't be afraid to get your hands dirty with some algebra. This entire process highlights the power and elegance of mathematical problem-solving.

I hope this step-by-step guide has been helpful and maybe even a little bit fun. Geometry problems can be challenging, but they're also incredibly rewarding when you finally crack the code. Keep practicing, keep exploring, and who knows? Maybe you'll be the one designing the next architectural masterpiece or solving the unsolvable geometric puzzle. Until next time, keep those math muscles strong!