Use rectangular and triangular faces to find surface area

Learning goal: Interpret nets containing triangles and rectangles, identify every outside face and calculate surface area. Distinguish a triangle's perpendicular face height from a solid's vertical height and from an edge length.

A right triangular prism has congruent right-triangle ends, the same size and shape, with legs three and four centimeters and longest side, called the hypotenuse, five centimeters. Its length between the triangular ends is six centimeters. The triangle dimensions are supplied; no theorem for discovering the hypotenuse is required for this task.

Each triangular end has area (1/2) × 3 × 4 = 6 square centimeters. Together the two ends contribute twelve. The three lateral faces, the side faces joining the two ends, are rectangles with dimensions 6 × 3, 6 × 4 and 6 × 5, contributing eighteen, twenty-four and thirty square centimeters. The full surface area is 12 + 18 + 24 + 30 = 84 square centimeters.

The net arranges those three rectangles in a strip sharing their six-centimeter edges. The strip widths are three, four and five centimeters, matching the three edges of the triangular end. A congruent right triangle attaches to each of the two four-centimeter end edges of the middle rectangle. Folding the strip around the triangular boundary joins the remaining edges and closes both ends.

The longest rectangular face uses the triangle's five-centimeter hypotenuse as its width. Its area is 6 × 5 = 30, not half of that. The factor one half belongs to each triangular face's area calculation. Classify the face shape before choosing a formula.

Now consider a square pyramid with base side four centimeters and a supplied slant height of three centimeters for each triangular face. The base has area sixteen square centimeters. Each of four triangular faces has base four and perpendicular face height three, so each area is six. The total surface area is 16 + 4 × 6 = 40 square centimeters.

In this pyramid, slant height means the distance measured along a triangular face from the midpoint of its base edge to the apex, the top point of the pyramid, perpendicular to that base edge within the face. It is not the vertical distance from the apex down to the center of the square base. The triangular-face area uses the face's own perpendicular height.

The pyramid net contains one square with a congruent triangle attached to each of its four sides. When folded, the triangle tips meet at the apex. A dashed segment within one triangle marks the supplied three-centimeter face height. Measuring the diagonal boundary edge instead would give a different length and an incorrect face-area calculation.

Prisms and pyramids have different face structures. The triangular prism has two congruent parallel triangular ends and three rectangular lateral faces. The square pyramid has one square base and four triangular lateral faces meeting at an apex. A formula remembered from one solid should not be transferred without accounting for the other solid's actual faces.

Count only the outside faces requested by the problem. For a triangular-prism sleeve with both ends open, the three rectangular faces alone total seventy-two square centimeters. For the closed prism, include both six-square-centimeter ends to obtain eighty-four. In a mathematical net, shared fold edges have no area and are not counted as extra faces.

Physical construction provides a useful check. A net should have matching edge lengths where faces meet and a plausible folding arrangement. If two faces collide or an end remains open unexpectedly, revisit the connections. Numerical face totals alone cannot establish that an arbitrary flat arrangement is a valid net.

Worked comparison: The triangular-prism sleeve area is 6(3 + 4 + 5) = 72 square centimeters. Adding two triangular end areas gives 72 + 12 = 84. The grouped expression uses the perimeter of one triangular end times prism length for the three rectangles, then accounts for both ends separately.

Worked height correction: A pyramid face with base four and face height three has area six, even though its sloping boundary edge is longer than three. Use the dashed perpendicular within the face. The solid's vertical height would not directly supply that triangle area.

Worked real-world painting problem (hypothetical): A wooden doorstop has the shape of the closed triangular prism above and rests on its 6 × 4 rectangular face. A crafter paints every outside face except the one on the floor. The painted area is the full surface area minus the bottom face: 84 − 24 = 60 square centimeters. Applying the net to a real-world problem means deciding which faces the situation actually includes.

Guided net study: Label all five prism faces and all five pyramid faces. Calculate their areas separately before summing. Trace which edges meet after folding, and explain why the two solids have different combinations of face shapes.

Independent performance: Construct or sketch a valid triangular-prism net and a square-pyramid net from supplied dimensions. Identify triangle bases and perpendicular face heights, find full surface areas and solve one open-surface variation. Explain which dimensions are supplied and which measurement each formula uses. Keep square units and any omitted faces explicit.