Finding The Distance: Point C To Line FH In A Cube

Hey guys! Let's dive into a classic geometry problem. We're going to figure out the distance from point C to line FH in a cube. This is super common in math class, so understanding it will really help you out. We will use the original question: Pada kubus ABCD.EFGH, jika panjang rusuknya 5 cm, maka jarak titik C ke garis FH adalah ... cm. Let's break it down step by step and make sure you get it!

Understanding the Problem and Key Concepts

Alright, first things first, let's make sure we're all on the same page. We're dealing with a cube named ABCD.EFGH. Imagine a perfect cube, like a dice or a box. Each side of this cube has a length of 5 cm. What we need to find is the shortest distance from point C to the line that goes through points F and H. This shortest distance is always a perpendicular line. Think of it like this: if you want to get from your house (point C) to a road (line FH) in the quickest way, you'd walk in a straight line, at a right angle, to the road. That's the distance we're looking for.

Now, before we get into the math, let's talk about some key concepts. We'll be using the Pythagorean theorem, which is super important for solving problems with right-angled triangles. Also, we will use the properties of a square, and how diagonals behave. Remember that the diagonal of a square creates two 45-45-90 triangles. And the diagonal of a face of the cube will be, since the sides are 5cm long, $5 imes ext{sqrt}(2)$ cm long. These are some tips that are useful for understanding geometry problems. Finally, understanding of how to visualize the problem in 3D space is critical. So, imagine a cube, and try to picture the line FH and the point C. This way you'll be able to see the relationship between these things.

To solve this, we're essentially looking for the length of a line segment. This is going to connect point C to the line FH, and it will form a right angle with FH. So, the distance from C to FH is really the length of this perpendicular line.

Visualizing the Cube and Identifying Relevant Triangles

Okay, let's get visual! Imagine our cube ABCD.EFGH. Now, let's focus on the face CDHG. If you look at this face, you'll see a square. The line FH is actually a diagonal of the cube. We are interested in the triangle CFH. It is within the cube, and we need to determine its properties. The line we're trying to find (the distance from C to FH) is going to be part of this triangle. Specifically, it is the altitude of this triangle if FH is the base. Note that since the cube's sides are equal, the face diagonals also have the same length. So, the triangle CFH is an equilateral triangle. In an equilateral triangle, all sides are equal. This helps simplify the calculations! If we can determine the length of CF and FH, we can determine the length of CH. And remember, all sides of the cube have a length of 5cm!

Let's get the lengths of each line. We know that the side of the cube is 5 cm. Since the sides of the cube are equal in length, that means that all the sides of the face are equal. Then, the diagonal of the face of the cube will be $5 imes ext{sqrt}(2)$ cm. Since FH is the diagonal of the face, it has the length $5 imes ext{sqrt}(2)$ cm. The length of FC is also $5 imes ext{sqrt}(2)$ cm. The same goes for the length of CH. Finally, since the three sides of the triangle are all equal, we can conclude that triangle CFH is an equilateral triangle, with all sides equal to $5 imes ext{sqrt}(2)$ cm. An important property of an equilateral triangle is that the altitude bisects the base. That means that the point where the altitude from C meets FH is the middle point of FH. We will call that point M. The altitude will be CM.

Calculating the Distance (CM)

Alright, time for some calculations! We know that the line segment CM is perpendicular to FH, and CM is the height of the equilateral triangle CFH. Now, we're going to use the area of the triangle. The area of a triangle is given by the formula: (1/2) * base * height. In our case, the base is FH, and the height is CM (the distance we're trying to find). The formula to calculate the area of an equilateral triangle with side 's' is: $( ext{sqrt}(3)/4) * s^2$. So, if we calculate the area using this formula, and also using the formula (1/2) * base * height, we can find the distance CM.

Let's denote the side of the triangle as 's' and the distance CM as 'h'. We know that s = $5 imes ext{sqrt}(2)$ cm. Let's calculate the area using the formula: $( ext{sqrt}(3)/4) * s^2$. We have: Area = $( ext{sqrt}(3)/4) * (5 imes ext{sqrt}(2))^2$ = $( ext{sqrt}(3)/4) * 50$ = $25 imes ext{sqrt}(3)/2$ cm$^2$

Now, let's express the area using the base and height. The base is FH = $5 imes ext{sqrt}(2)$ cm, and the height is CM = h. So we have: Area = (1/2) * FH * h $25 imes ext{sqrt}(3)/2$ = (1/2) * $5 imes ext{sqrt}(2)$ * h

Now, solve for h (CM): h = ($25 imes ext{sqrt}(3)/2$) / (2.5 * $ ext{sqrt}(2)$) h = $(5 imes ext{sqrt}(3)) / ext{sqrt}(2)$

To rationalize the denominator, we multiply both the numerator and denominator by $ extsqrt}(2)$, which gives h = $(5 imes ext{sqrt(6)) / 2$

So, the distance from point C to line FH is rac{5}{2} ext{sqrt}(6) cm.

Conclusion: The Answer

Therefore, the correct answer is d. rac{5}{2} ext{sqrt}(6).

I hope this explanation was helpful, guys! Remember, the key is to visualize, break down the problem, and use the right formulas. Geometry can be fun, and with practice, you'll become a pro at these problems! Keep practicing, and you'll ace these tests in no time. If you have any questions, feel free to ask!