Volume of Regular Pyramids (2nd) Second Quarter Grade 8 Matatag Revised K-12 Tagalog Math Tutorial

The video introduces the topic of calculating the volume of pyramids, specifically focusing on pyramids with regular polygonal bases other than square and rectangular. The learning competencies include exploring the volume of pyramids inductively, finding the volume of various pyramids, and solving related problems.

A pyramid is defined as a three-dimensional solid with a polygonal base and triangular faces meeting at an apex. Key parts of a pyramid are reviewed, along with formulas for the area of rectangles and squares, and the volume of a prism (length x width x height or base area x height).

The video demonstrates how the formula for the volume of a pyramid (1/3 * base area * height) is derived by showing that three identical pyramids can form a rectangular prism. This relationship illustrates that a pyramid's volume is one-third the volume of a prism with the same base and height.

The formulas for the volume of rectangular pyramids (1/3 * length * width * height or 1/3 * base area * height) and square pyramids (1/3 * side * side * height or 1/3 * side squared * height or 1/3 * base area * height) are summarized.

The main lesson begins, focusing on pyramids with regular polygonal bases like triangles, pentagons, and hexagons. The general formula for the volume of any pyramid (1/3 * base area * height) holds true, with the primary difference being how to calculate the base area of the specific polygon.

For a triangular pyramid, the base is a triangle. The area of the triangular base is calculated using the formula: 1/2 * base * height, where 'base' and 'height' refer to the dimensions of the triangular base itself.

To find the area of a regular pentagonal base, it can be divided into five congruent triangles. The area of one such triangle is 1/2 * side * apothem (where apothem is the height of the triangle from the center to the midpoint of a side). Therefore, the total base area is 5 * (1/2 * side * apothem) or (5/2) * side * apothem.

Similarly, for a regular hexagonal base, it can be divided into six congruent triangles. The area of one triangle is 1/2 * side * apothem. Thus, the total base area is 6 * (1/2 * side * apothem) which simplifies to 3 * side * apothem.

An example problem is solved step-by-step: A triangular pyramid has a base with a base of 8 cm and a height of 6 cm. The pyramid's height is 12 cm. The base area is calculated first (1/2 * 8 cm * 6 cm = 24 cm²), then the volume (1/3 * 24 cm² * 12 cm = 96 cm³).

A problem involving a pentagonal pyramid is presented: base side of 10 in, apothem of 7 in, and pyramid height of 15 in. The base area is found using the formula (5/2 * 10 in * 7 in = 175 in²), and then the volume (1/3 * 175 in² * 15 in = 875 in³).

The final example calculates the volume of a hexagonal pyramid: base side of 12 m, apothem of 10 m, and pyramid height of 20 m. The base area is calculated (3 * 12 m * 10 m = 360 m²), followed by the volume (1/3 * 360 m² * 20 m = 2400 m³).

A practice problem for a triangular pyramid is given: base 9 cm, height of triangle 5 cm, pyramid height 14 cm. The base area is 22.5 cm², and the volume is 105 cm³.

A practice problem for a pentagonal pyramid is presented: base side 8 in, apothem 6 in, pyramid height 18 in. The base area is 120 in², and the volume is 720 in³.

A practice problem for a hexagonal pyramid is provided: base side 1.2 m, apothem 1 m, pyramid height 2.4 m. The base area is 3.6 m², and the volume is 2.88 m³.

The video concludes by encouraging viewers to practice and mentions that the next lesson will cover the volume of pyramids with irregular bases.