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To calculate the area of triangle \( \triangle ACH \), we first need to break down the geometry of the cuboid and triangle. Here's the information we have:

- \( AB = 89 \, \text{m} \) (length of the cuboid)
- \( BC = 77 \, \text{m} \) (width of the cuboid)
- \( AE = 73 \, \text{m} \) (height of the cuboid)

### Step 1: Determine the coordinates of the vertices.
For convenience, let's assign the following coordinates based on the dimensions of the cuboid:

- \( A(0, 0, 0) \)
- \( B(89, 0, 0) \)
- \( C(89, 77, 0) \)
- \( H(89, 77, 73) \) (since \( H \) is directly above \( C \), at a height of 73 m)

Thus, the coordinates of the vertices of triangle \( ACH \) are:
- \( A(0, 0, 0) \)
- \( C(89, 77, 0) \)
- \( H(89, 77, 73) \)

### Step 2: Calculate the area of \( \triangle ACH \).
The area of a triangle can be calculated using the formula:

\[
\text{Area} = \frac{1}{2} \left| \mathbf{AB} \times \mathbf{AC} \right|
\]

We can use the vector cross product approach by treating the points as vectors. The vectors \( \overrightarrow{AC} \) and \( \overrightarrow{AH} \) are:

- \( \overrightarrow{AC} = C - A = (89, 77, 0) \)
- \( \overrightarrow{AH} = H - A = (89, 77, 73) \)

### Step 3: Compute the cross product.
Now we calculate the cross product \( \overrightarrow{AC} \times \overrightarrow{AH} \).The area of triangle \( \triangle ACH \) is approximately \( 4295.5 \, \text{m}^2 \), correct to one decimal place.

Would you like further explanation on any part of the solution? Here are a few related questions that might interest you:

1. How is the cross product used to find the area of a triangle in 3D space?
2. Can you explain more about the vectors involved in the cross product?
3. How would the area change if the cuboid had different dimensions?
4. What are other methods to find the area of a triangle in 3D geometry?
5. How can the Pythagorean theorem be applied to this cuboid?

**Tip:** The cross product of two vectors results in a vector that is perpendicular to the plane formed by those two vectors. Its magnitude is proportional to the area of the parallelogram spanned by the vectors.

Find the area of triangle ACH in a cuboid with dimensions AB = 89 m, BC = 77 m, and AE = 73 m.

Learn how to calculate the area of triangle ACH in a cuboid with dimensions AB = 89 m, BC = 77 m, and AE = 73 m using vector cross product. Step-by-step solution included, suitable for high school students in Grades 10-12.

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