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To solve the volume of the cuboid (balok) for each case, we need to determine the dimensions: length (\( l \)), width (\( w \)), and height (\( h \)). The volume formula for a cuboid is:

\[
V = l \times w \times h
\]

Let's analyze each part of the problem:

### (a) Given:
- \( AB = 12 \, cm \) (length)
- \( AD = 9 \, cm \) (width)
- \( AE = 5 \, cm \) (height)

**Volume:**
\[
V = 12 \, cm \times 9 \, cm \times 5 \, cm = 540 \, cm^3
\]

### (b) Given:
- \( AB = 16 \, cm \) (length)
- \( AE = 12 \, cm \) (height)
- Area of \( AFGD = 100 \, cm^2 \)

Since \( AFGD \) is the front face, it represents the width (\( AD \)) and height (\( AE \)):
\[
AD \times AE = 100 \, cm^2 \Rightarrow w \times 12 = 100 \Rightarrow w = \frac{100}{12} \approx 8.33 \, cm
\]

**Volume:**
\[
V = 16 \, cm \times 8.33 \, cm \times 12 \, cm \approx 1599.36 \, cm^3
\]

### (c) Given:
- \( AB : BD : AE = 4 : 3 : 5 \)
- Area of \( BCGF = 60 \, cm^2 \)

We can express the dimensions as \( AB = 4x \), \( BD = 3x \), and \( AE = 5x \). Since \( BCGF \) is the side face:
\[
BD \times AE = 60 \Rightarrow 3x \times 5x = 60 \Rightarrow 15x^2 = 60 \Rightarrow x^2 = 4 \Rightarrow x = 2
\]

Therefore:
- \( AB = 4x = 8 \, cm \)
- \( BD = 3x = 6 \, cm \)
- \( AE = 5x = 10 \, cm \)

**Volume:**
\[
V = 8 \, cm \times 6 \, cm \times 10 \, cm = 480 \, cm^3
\]

### (d) Given:
- Area of \( ABCD = 90 \, cm^2 \)
- Area of \( BCGF = 30 \, cm^2 \)
- Area of \( ABFE = 48 \, cm^2 \)

\( ABCD \) represents length \( l \) and width \( w \), so:
\[
l \times w = 90 \Rightarrow l \times w = 90
\]

\( BCGF \) represents width \( w \) and height \( h \), so:
\[
w \times h = 30
\]

\( ABFE \) represents length \( l \) and height \( h \), so:
\[
l \times h = 48
\]

We have three equations:
1. \( l \times w = 90 \) 
2. \( w \times h = 30 \) 
3. \( l \times h = 48 \)

Solving these, we can find \( l = 10 \, cm \), \( w = 9 \, cm \), and \( h = 4 \, cm \).

**Volume:**
\[
V = 10 \, cm \times 9 \, cm \times 4 \, cm = 360 \, cm^3
\]

### (e) Given:
- Perimeter of \( ABCD = 44 \, cm \)
- Perimeter of \( BCGF = 32 \, cm \)
- Perimeter of \( ABFE = 48 \, cm \)

From the perimeter formulas:
1. \( 2(l + w) = 44 \Rightarrow l + w = 22 \)
2. \( 2(w + h) = 32 \Rightarrow w + h = 16 \)
3. \( 2(l + h) = 48 \Rightarrow l + h = 24 \)

We solve these equations:
- From \( l + w = 22 \) and \( l + h = 24 \), we can find:
  \( w = 8 \, cm \), \( h = 14 \, cm \), and \( l = 14 \, cm \).

**Volume:**
\[
V = 14 \, cm \times 8 \, cm \times 14 \, cm = 1568 \, cm^3
\]

### Summary:
1. **(a)** \( 540 \, cm^3 \)
2. **(b)** \( 1599.36 \, cm^3 \)
3. **(c)** \( 480 \, cm^3 \)
4. **(d)** \( 360 \, cm^3 \)
5. **(e)** \( 1568 \, cm^3 \)

Would you like further clarification or details on any of these calculations?

**Questions to Expand:**
1. How can you find volume if only the areas of sides are given?
2. Can volume still be determined if only the ratios of sides are provided?
3. Why is it important to have accurate side dimensions before calculating volume?
4. What are other methods to solve if more details are given for volume problems?
5. How do perimeter and area relate to calculating dimensions of 3D shapes?

**Tip:** Always recheck dimensions when substituting values to prevent calculation errors.

Tentukan volume balok jika diketahui berbagai informasi seperti panjang sisi, luas permukaan, dan perbandingan panjang sisi.

Learn how to calculate the volume of a cuboid (balok) using given side lengths, areas of faces, and side ratios. This problem involves geometry and algebra, and is suitable for students in Grades 9-11.

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