Derive triangle and parallelogram area from rectangles

Learning goal: Find areas of right and nonright triangles and parallelograms by composing or rearranging rectangles. Identify a base and its perpendicular height. Explain why a slanted side is generally not the height used in the area formula.

A rectangle six centimeters long and four centimeters wide has area twenty-four square centimeters. Draw a diagonal joining opposite vertices. The two triangles are congruent, meaning the same size and shape, and together cover the rectangle without gaps or overlap. Each triangle therefore has area twelve square centimeters, half of six times four.

This gives the triangle relationship A = (1/2)bh, where b is a chosen base length and h is the perpendicular distance from the opposite vertex to the line containing that base. The units multiply to square units. The factor one half arises because two copies of the triangle compose a rectangle or parallelogram with the same base and height. It is not a rule about perimeter, and it does not apply to a side length that is not the matched perpendicular height.

A nonright triangle can also be decomposed using an altitude, a segment drawn from one vertex perpendicular to the line containing the opposite side. Consider a triangle with base six centimeters, height four centimeters and the altitude meeting the base two centimeters from its left endpoint. The altitude splits it into two right triangles with bases two and four centimeters and a shared height four. Their areas are (1/2) × 2 × 4 = 4 and (1/2) × 4 × 4 = 8 square centimeters. Together they give twelve, equal to (1/2) × 6 × 4.

For an obtuse triangle, the altitude to a selected base may meet an extension of the base outside the triangle. The height is still perpendicular distance to the base line. You can compose the triangle with another region into a larger triangle or use a congruent copy to make a parallelogram. The formula remains one half base times perpendicular height; the altitude's location does not change the area relationship.

A parallelogram has two pairs of parallel opposite sides. With base six centimeters and perpendicular height four centimeters, its area is twenty-four square centimeters. When the perpendicular from a top corner lands on the base, as it does in the diagram, cut the triangular piece off that end along the perpendicular line and slide it to the other end. The rearranged pieces form a six-by-four rectangle with the same total area, provided no piece overlaps or is discarded.

A parallelogram that leans farther, so that the perpendicular from the top corner lands beyond the end of the base, cannot be finished with that single cut. Enclose it in a rectangle instead. With base six, height four and the top edge shifted eight centimeters, the enclosing rectangle is fourteen by four, with area fifty-six. The two right triangles outside the parallelogram each have area (1/2) × 8 × 4 = 16, leaving 56 − 32 = 24 square centimeters. The area is still base times perpendicular height.

The diagram shows a parallelogram with its top edge shifted two centimeters relative to its bottom edge. A dashed perpendicular segment identifies the four-centimeter height. The slanted edge is longer than four centimeters, but it is not the perpendicular distance between the parallel bases. Multiplying the base by that slanted length would overestimate area.

Choose another side as base only if you also choose the height perpendicular to that side's line. A base and height are a matched pair. Mixing the original base with a height belonging to another side does not generally preserve the correct product. Marking a right angle helps identify which lengths belong together.

Area and perimeter answer different questions. Two parallelograms can share a base and perpendicular height while having different slants and perimeters. Rearranging a shape without stretching can preserve area while changing its boundary length. A border-material problem requires perimeter; a covering-material problem requires area.

Fractional dimensions work in the same relationships. A triangle with base 3.5 meters and height 2 meters has area (1/2) × 3.5 × 2 = 3.5 square meters. A parallelogram with the same base and height has area seven square meters. The numerical area can happen to match a side's numerical length, but their units and meanings remain different.

Worked missing-height check: A triangle has base eight centimeters and area twenty square centimeters. Its doubled area is forty, so eight times the height equals forty. The perpendicular height is five centimeters. Check (1/2) × 8 × 5 = 20.

Worked correction: A parallelogram has base six centimeters, height four centimeters and a slanted side of about 4.47 centimeters. Its area is 6 × 4 = 24 square centimeters, not 6 × 4.47. The given slanted side measures the boundary, not the separation between parallel bases.

Worked real-world covering problem (hypothetical): A garden bed is shaped like a parallelogram with a 5-meter base, a 3-meter perpendicular height and slanted sides 3.5 meters long. Mulch must cover the whole bed, so the question asks for area: 5 × 3 = 15 square meters. Edging for the border would instead need the perimeter, 5 + 3.5 + 5 + 3.5 = 17 meters. Solving a real-world problem starts by deciding which measurement the situation needs, then applying the matched base and height.

Guided construction: Cut a paper rectangle diagonally and compare the two triangle areas. Draw a nonright triangle and decompose it with an altitude. Cut and rearrange a parallelogram into a rectangle, labeling the original and new base and height.

Independent performance: Find areas of a right triangle, a nonright triangle and a parallelogram, including fractional dimensions. Show a composition or decomposition argument, mark perpendicular heights and state square units. Explain an error caused by using a slanted side as height. Solve one covering problem and identify why perimeter would answer a different question.