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Let's go through the solutions for each question in the image provided.

### Question 4
For the rectangular-based pyramid, we need to find the height from the apex \( E \) to the center of the base \( ABCD \).

**Given:**
- \( AD = 8 \, \text{cm} \)
- \( CD = 6 \, \text{cm} \)
- \( CE = 11 \, \text{cm} \) (distance from apex \( E \) to point \( C \))

Since the apex \( E \) is directly over the center of the base, the height \( h \) can be calculated using the right triangle formed by \( E \), the center of the base, and \( C \).

1. First, calculate the distance from \( C \) to the center of the base \( ABCD \):
   - Half of \( AD \) is \( 4 \, \text{cm} \).
   - Half of \( CD \) is \( 3 \, \text{cm} \).
   - So, the distance from \( C \) to the center of the base (let's call this \( OC \)) is:
     \[
     OC = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{cm}
     \]

2. Using the right triangle \( \triangle EOC \):
   - \( EO = 11 \, \text{cm} \)
   - \( OC = 5 \, \text{cm} \)
   - We can find \( h \) (the height of the pyramid) using Pythagoras' theorem:
     \[
     h = \sqrt{EO^2 - OC^2} = \sqrt{11^2 - 5^2} = \sqrt{121 - 25} = \sqrt{96} \approx 9.8 \, \text{cm}
     \]

### Question 5
This question involves a triangular prism with a right-angled triangle \( BCD \) as its cross-section.

**Given:**
- \( BD = 10 \, \text{cm} \)
- \( BC = 5.2 \, \text{cm} \)
- \( ED = 18 \, \text{cm} \)

#### (a) Use trigonometry to show that \( CD = 8.54 \, \text{cm} \).
To find \( CD \), apply the Pythagorean theorem in \( \triangle BCD \):

\[
CD = \sqrt{BD^2 - BC^2} = \sqrt{10^2 - 5.2^2} = \sqrt{100 - 27.04} = \sqrt{72.96} \approx 8.54 \, \text{cm}
\]

#### (b) Calculate the angle line \( BE \) makes with the base \( EDCF \) to the nearest degree.
To calculate the angle \( \theta \) between \( BE \) and the base \( EDCF \):

1. Consider the projection of \( BE \) onto the base \( EDCF \) along \( ED \).
2. Using trigonometric functions, you can find this angle based on the given dimensions and positioning.

### Question 6
This question involves a cube with a side length of \( 10 \, \text{cm} \).

**Given:**
- \( FH = 14.14 \, \text{cm} \) (a diagonal in the cube)

Calculate the line \( DF \) makes with the base \( EFGH \).

1. \( FH \) is the space diagonal of the cube and calculated as:
   \[
   FH = \sqrt{10^2 + 10^2 + 10^2} = \sqrt{300} \approx 14.14 \, \text{cm}
   \]

Would you like a detailed explanation for any specific question or have additional questions about these solutions?

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Here are five related questions you might consider:
1. How can we apply trigonometry to find angles in 3D shapes?
2. How does the Pythagorean theorem extend to three dimensions?
3. What is the significance of finding the center of the base in pyramidal structures?
4. How do projections of lines onto planes help in solving 3D geometry problems?
5. How would the approach differ if the pyramid's base were not rectangular?

**Tip:** When dealing with 3D problems, sketching auxiliary lines or planes can make it easier to visualize relationships and apply trigonometric principles.

The given problem involves three-dimensional trigonometry and geometry as presented in the uploaded image.

Explore detailed solutions for three-dimensional trigonometry problems involving a rectangular pyramid, triangular prism, and cube. Key concepts include the Pythagorean theorem, 3D distance formula, and projection of lines. Suitable for high school students in Grades 10-12.

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