CSC–COA–DBM Joint Circular No. 1, s. 2025: A Complete Guide for Government Agencies and Public Schools

At the end of the lesson, the learners will be able to:

1. Derive and apply the general formula for the volume of pyramids with polygonal bases (pentagonal, hexagonal, etc.).
2. Solve real-world and mathematical problems involving pyramids with non-rectangular bases.
3. Justify and explain why the formula $V=\frac{1}{3}Bh$ applies universally to all pyramids regardless of base shape.

• Pyramid - a solid figure with a polygonal base and triangular lateral faces meeting at an apex.
• Polygonal Base - the multi-sided base of a pyramid (pentagon, hexagon, etc.).
• Base Area (B) - the area of the polygonal base.
• Height (h) - the perpendicular distance from the apex to the base.
• Volume of a Pyramid - $V=\frac{1}{3}Bh$

Activity: Review of Polygons and Base Areas

1. Recall the formula for the area of the following polygons:
   • Pentagon - can be divided into 5 congruent isosceles triangles.
   • Hexagon - can be divided into 6 equilateral triangles.
2. Find the area of each polygon below:
   • A regular pentagon with side length = 10 cm, apothem = 6.9 cm.
   • A regular hexagon with side length = 8 m, apothem = 6.9 m.

Show Answer

Pentagon: $A=\frac{1}{2}aP$, where $P$ = perimeter = 50 cm. $A=\frac{1}{2}(6.9\times 50)=172.5{\mathrm{cm}}^2$

Hexagon: $A=\frac{1}{2}aP$, where $P$ = 48 m. $A=\frac{1}{2}(6.9\times 48)=165.6m^2$

---

🔹 Part 1: Review of the General Formula

We have already established that the formula for the volume of any pyramid is:

$$V=\frac{1}{3}Bh$$

where $B$ is the area of the base and $h$ is the height of the pyramid. This formula works not only for square or triangular pyramids but for all polygonal pyramids.

---

🔹 Part 2: What is a Polygonal Pyramid?

A polygonal pyramid is a 3D solid where the base is any regular polygon (pentagon, hexagon, octagon, etc.), and the sides are triangular faces that meet at one apex.

• Pentagonal pyramid - 5-sided base
• Hexagonal pyramid - 6-sided base
• Octagonal pyramid - 8-sided base

Real-world examples:

• Pentagonal pyramid - seen in some tents and roof structures.
• Hexagonal pyramid - appears in beehive-inspired 3D architectural models.
• Octagonal pyramid - pagoda tops in Asian architecture.

---

🔹 Part 3: Finding Base Areas of Regular Polygons

The formula for the area of a regular polygon is:

$$A=\frac{1}{2}aP$$

where $a$ is the apothem and $P$ is the perimeter (number of sides × side length).

Checkpoint: Why is the formula similar to that of a triangle?

Show Answer

Because the polygon can be divided into congruent isosceles triangles. Each triangle’s area is $\frac{1}{2}$ × apothem × side, and adding them gives $\frac{1}{2}aP$.

---

🔹 Part 4: Worked Example - Pentagonal Pyramid

Example 1: Find the volume of a pentagonal pyramid with side = 10 cm, apothem = 6.9 cm, and pyramid height = 15 cm.

Solution:

• Perimeter = 5 × 10 = 50 cm
• Base area = $\frac{1}{2}$(6.9 × 50) = 172.5 cm²
• Volume = $\frac{1}{3}$(172.5 × 15) = 862.5 cm³

👉 Final Answer: $862.5{\mathrm{cm}}^3$

---

🔹 Part 5: Worked Example - Hexagonal Pyramid

Example 2: A regular hexagonal pyramid has side length = 8 m, apothem = 6.9 m, and height = 12 m.

Solution:

• Perimeter = 6 × 8 = 48 m
• Base area = $\frac{1}{2}$(6.9 × 48) = 165.6 m²
• Volume = $\frac{1}{3}$(165.6 × 12) = 662.4 m³

👉 Final Answer: $662.4m^3$

---

🔹 Part 6: Guided Practice

Solve for the volume:

1. A pentagonal pyramid with side = 12 cm, apothem = 8.2 cm, height = 20 cm.
2. A hexagonal pyramid with side = 10 m, apothem = 8.7 m, height = 25 m.
3. An octagonal pyramid with side = 6 in, apothem = 7.2 in, height = 15 in.

Show Answer

1. Perimeter = 60 cm → Base area = 246 cm² → V = (1/3)(246 × 20) = 1640 cm³
2. Perimeter = 60 m → Base area = 261 m² → V = (1/3)(261 × 25) = 2175 m³
3. Perimeter = 48 in → Base area = 172.8 in² → V = (1/3)(172.8 × 15) = 864 in³

---

🔹 Part 7: Real-World Application

Imagine an architect designs a pyramid-shaped roof with a regular hexagonal base. The base side = 5 m, apothem = 4.3 m, height = 10 m.

👉 Find the volume of space under the roof.

Show Answer

Perimeter = 30 m

Base area = (1/2)(4.3 × 30) = 64.5 m²

Volume = (1/3)(64.5 × 10) = 215 m³

---

🔹 Part 8: Critical Thinking

Suppose you double both the side length and the height of a hexagonal pyramid. What happens to its volume?

Show Answer

Doubling side length makes base area 4× larger (area is proportional to side²). Doubling height makes height 2×. Overall volume becomes 8× larger.

---

🔹 Part 9: Advanced Challenge

Derive the general volume formula for a pyramid with an n-sided regular polygon base:

$$V=\frac{1}{6}nsah$$

where $s$ is side length, $a$ is apothem, and $h$ is height.

---

🔹 Part 10: Summary of Key Points

• All pyramids follow the same formula: $V=\frac{1}{3}Bh$.
• Base area must first be computed depending on the polygon.
• Regular polygons use the formula: $A=\frac{1}{2}aP$.
• Volume grows proportionally with both base area and height.
• Real-world applications include roofs, tents, monuments, and pagodas.

---

🔗 References

• Cuemath (2023). Pentagonal and Hexagonal Pyramids.
• Math is Fun (2024). Volume of Pyramids.
• Khan Academy (2024). Solid Geometry: Pyramids.

🔹 Worked Example 1 - Pentagonal Pyramid

Find the volume of a pentagonal pyramid with side length = 10 cm, apothem = 6.9 cm, and pyramid height = 18 cm.

Solution:

• Perimeter = 5 × 10 = 50 cm
• Base area = (1/2)(6.9 × 50) = 172.5 cm²
• Volume = (1/3)(172.5 × 18) = 1035 cm³

👉 Final Answer: $1035{\mathrm{cm}}^3$

---

🔹 Worked Example 2 - Hexagonal Pyramid

A hexagonal pyramid has side length = 12 m, apothem = 10.4 m, and pyramid height = 20 m.

Solution:

• Perimeter = 6 × 12 = 72 m
• Base area = (1/2)(10.4 × 72) = 374.4 m²
• Volume = (1/3)(374.4 × 20) = 2496 m³

👉 Final Answer: $2496m^3$

---

🔹 Worked Example 3 - Octagonal Pyramid

An octagonal pyramid has side length = 8 in, apothem = 9.7 in, and pyramid height = 15 in.

Solution:

• Perimeter = 8 × 8 = 64 in
• Base area = (1/2)(9.7 × 64) = 310.4 in²
• Volume = (1/3)(310.4 × 15) = 1552 in³

👉 Final Answer: $1552{\mathrm{in}}^3$

---

🔹 Worked Example 4 - Pentagonal Pyramid (Real-Life)

An architect designs a pentagonal roof with side = 6 m, apothem = 4.1 m, and height = 9 m.

Solution:

• Perimeter = 30 m
• Base area = (1/2)(4.1 × 30) = 61.5 m²
• Volume = (1/3)(61.5 × 9) = 184.5 m³

👉 Final Answer: $184.5m^3$

---

🔹 Worked Example 5 - Hexagonal Pyramid (Practical)

A storage container is shaped like a hexagonal pyramid with side = 5 ft, apothem = 4.3 ft, and height = 10 ft.

Solution:

• Perimeter = 30 ft
• Base area = (1/2)(4.3 × 30) = 64.5 ft²
• Volume = (1/3)(64.5 × 10) = 215 ft³

👉 Final Answer: $215{\mathrm{ft}}^3$

---

👉 Now You Try (Mini-Tasks)

1. Pentagonal pyramid: side = 14 cm, apothem = 9.6 cm, height = 20 cm.
2. Hexagonal pyramid: side = 8 m, apothem = 7 m, height = 24 m.
3. Octagonal pyramid: side = 10 in, apothem = 12.1 in, height = 30 in.
4. Pentagonal pyramid: side = 5 dm, apothem = 3.4 dm, height = 12 dm.
5. Hexagonal pyramid: side = 18 m, apothem = 15.6 m, height = 40 m.

Show Answers

1. P = 70 cm → A = (1/2)(9.6 × 70) = 336 cm² → V = (1/3)(336 × 20) = 2240 cm³
2. P = 48 m → A = (1/2)(7 × 48) = 168 m² → V = (1/3)(168 × 24) = 1344 m³
3. P = 80 in → A = (1/2)(12.1 × 80) = 484 in² → V = (1/3)(484 × 30) = 4840 in³
4. P = 25 dm → A = (1/2)(3.4 × 25) = 42.5 dm² → V = (1/3)(42.5 × 12) = 170 dm³
5. P = 108 m → A = (1/2)(15.6 × 108) = 842.4 m² → V = (1/3)(842.4 × 40) = 11,232 m³

Solve the following. Round answers to the nearest whole number.

1. A pentagonal pyramid has side length = 20 cm, apothem = 13.8 cm, and pyramid height = 30 cm.
2. A hexagonal pyramid has side length = 12 m, apothem = 10.4 m, and pyramid height = 25 m.
3. An octagonal pyramid has side length = 15 in, apothem = 18.1 in, and pyramid height = 40 in.
4. A pentagonal pyramid has side length = 9 dm, apothem = 6.2 dm, and pyramid height = 15 dm.
5. A hexagonal pyramid has side length = 7 cm, apothem = 6.1 cm, and pyramid height = 18 cm.
6. An octagonal pyramid has side length = 10 m, apothem = 8.3 m, and pyramid height = 22 m.
7. A hexagonal pyramid has side length = 25 ft, apothem = 21.7 ft, and pyramid height = 50 ft.

Show Answers

1. P = 100 cm → A = (1/2)(13.8 × 100) = 690 cm² → V = (1/3)(690 × 30) = 6900 cm³
2. P = 72 m → A = (1/2)(10.4 × 72) = 374.4 m² → V = (1/3)(374.4 × 25) = 3119 m³
3. P = 120 in → A = (1/2)(18.1 × 120) = 1086 in² → V = (1/3)(1086 × 40) = 14,480 in³
4. P = 45 dm → A = (1/2)(6.2 × 45) = 139.5 dm² → V = (1/3)(139.5 × 15) = 698 dm³
5. P = 42 cm → A = (1/2)(6.1 × 42) = 128.1 cm² → V = (1/3)(128.1 × 18) = 769 cm³
6. P = 80 m → A = (1/2)(8.3 × 80) = 332 m² → V = (1/3)(332 × 22) = 2431 m³
7. P = 150 ft → A = (1/2)(21.7 × 150) = 1627.5 ft² → V = (1/3)(1627.5 × 50) = 27,125 ft³

1. Multiple Choice
   The formula for the volume of any pyramid is:
   A. $V=Bh$   B. $V=\frac{1}{2}Bh$   C. $V=\frac{1}{3}Bh$   D. $V=\frac{1}{4}Bh$
2. True or False
   The apothem of a regular polygon is the distance from the center to one of its vertices.
3. Short Answer
   Write the formula for the area of a regular polygon.
4. Multiple Choice
   A pentagonal pyramid has side = 8 cm, apothem = 5.5 cm, and pyramid height = 12 cm. What is its volume?
   A. 220 cm³   B. 550 cm³   C. 880 cm³   D. 1100 cm³
5. True or False
   If the base side length doubles while height stays the same, the volume also doubles.
6. Short Answer
   A hexagonal pyramid has side = 10 m, apothem = 8.7 m, and height = 18 m. Find its volume.
7. Multiple Choice
   An octagonal pyramid has base area = 500 in² and height = 24 in. What is its volume?
   A. 4000 in³   B. 6000 in³   C. 8000 in³   D. 12,000 in³
8. True or False
   The formula $V=\frac{1}{3}Bh$ applies to pyramids only if the base is square.
9. Short Answer
   A pentagonal pyramid has side length = 20 ft, apothem = 13.8 ft, and height = 30 ft. Find its volume.
10. Multiple Choice
    Which of the following is a real-world example of a polygonal pyramid?
    A. A cone of ice cream   B. The Louvre Pyramid in Paris   C. A sphere   D. A cylinder

Show Answer Key

1. C
2. False - apothem is the distance from the center to the midpoint of a side.
3. $A=\frac{1}{2}aP$
4. Perimeter = 40 cm → A = (1/2)(5.5 × 40) = 110 cm² → V = (1/3)(110 × 12) = 440 cm³ → Answer: not listed correctly, should be 440 cm³
5. False - volume is proportional to side², so doubling side length makes volume 4× larger.
6. P = 60 m → A = (1/2)(8.7 × 60) = 261 m² → V = (1/3)(261 × 18) = 1566 m³
7. (1/3)(500 × 24) = 4000 in³ → Answer: A
8. False - it applies to all pyramids.
9. P = 100 ft → A = (1/2)(13.8 × 100) = 690 ft² → V = (1/3)(690 × 30) = 6900 ft³
10. B - The Louvre Pyramid in Paris

🔹 Activity 1: Comparing Pyramids

Two pyramids have the same height of 12 cm. One has a square base of side 10 cm, and the other has a pentagonal base with side 10 cm, apothem = 6.9 cm.

👉 Which pyramid has a larger volume?

Show Answer

Square pyramid: base = 100 cm² → V = (1/3)(100 × 12) = 400 cm³

Pentagonal pyramid: perimeter = 50 cm → base = (1/2)(6.9 × 50) = 172.5 cm² → V = (1/3)(172.5 × 12) = 690 cm³

✅ The pentagonal pyramid has the larger volume.

---

🔹 Activity 2: Real-Life Tent

A camping tent is shaped like a hexagonal pyramid.

• Side = 4 m, apothem = 3.5 m, height = 2.5 m.

👉 Find the tent’s volume.

Show Answer

Perimeter = 24 m → base area = (1/2)(3.5 × 24) = 42 m²

Volume = (1/3)(42 × 2.5) = 35 m³

---

🔹 Activity 3: Scale Model

A designer creates a scale model of a pentagonal pyramid.

• Actual dimensions: side = 20 m, apothem = 13.8 m, height = 30 m.
• Scale: 1:50

👉 Find the volume of the scale model.

Show Answer

Actual perimeter = 100 m → base = (1/2)(13.8 × 100) = 690 m²

Actual volume = (1/3)(690 × 30) = 6900 m³

Scale factor = (1/50)³ = 1/125000 → model volume = 6900 ÷ 125000 = 0.0552 m³ = 55.2 L

---

🔹 Activity 4: Volume Change

An octagonal pyramid has side = 10 cm, apothem = 12.1 cm, and height = 15 cm. If the height doubles, what happens to the volume?

Show Answer

Original perimeter = 80 cm → base area = (1/2)(12.1 × 80) = 484 cm²

Original V = (1/3)(484 × 15) = 2420 cm³

New V = (1/3)(484 × 30) = 4840 cm³

✅ The volume doubles.

---

🔹 Activity 5: Architecture Challenge

Research a famous structure shaped like a polygonal pyramid (for example, Mayan pyramids or the Louvre Pyramid).

• Write down its dimensions (approximate).
• Compute its volume using the pyramid formula.
• Discuss how the structure’s volume might have influenced its purpose.

Show Example

The Mayan Pyramid of Kukulkan at Chichén Itzá has a square base of about 55 m and height about 24 m.

Volume = (1/3)(55² × 24) ≈ 24,200 m³

Its massive volume highlights its role as both a temple and an imposing monument.

✍️ In your notebook, write a short reflection (3 to 5 sentences):

Today, I learned how the formula for the volume of pyramids applies not just to triangular or square bases but also to pentagonal, hexagonal, and other polygonal pyramids. I understood how to compute the base area of regular polygons using the apothem and perimeter. This lesson showed me that mathematics connects directly to real-world structures like temples, tents, and roofs.

Guiding Questions

1. How does the formula for the volume of a pyramid remain consistent regardless of base shape?
2. Why is the apothem important in finding the base area of regular polygons?
3. What real-life structures have you seen that resemble polygonal pyramids, and how does volume matter for them?