Find composite areas using nonoverlapping parts and holes

Learning goal: Solve area problems involving triangles, quadrilaterals and composite polygons. Decompose into nonoverlapping regions, use perpendicular heights and account for removed regions without changing the defined whole. This lesson covers the area part of the standard; the lesson "Separate composite volume from exposed surface area" covers its volume and surface-area parts.

Area measures the amount of two-dimensional region covered, in square units. A composite shape can often be partitioned into familiar polygons whose areas add. The pieces must fill the desired region without gaps or overlaps. An internal dividing line helps calculation but does not create extra area or change the outside shape.

Consider a rectangular region six meters long and four meters wide with a triangular extension attached along its four-meter side. The triangle's perpendicular distance from that shared side to its outer vertex is three meters. Rectangle area is 6 × 4 = 24 square meters. Triangle area is (1/2) × 4 × 3 = six square meters. The complete region has area thirty square meters.

The shared four-meter segment is a boundary of each calculation piece, but it has no area by itself. Adding the rectangle and triangle does not double-count a strip if the pieces meet only along that line. Their interiors do not overlap. If two proposed pieces overlap in a positive-area region, that overlap must be removed from a simple sum or the decomposition must be changed.

The triangle's height is perpendicular to its chosen base. In this example the base is vertical and the height is horizontal. A triangle does not require a horizontal base to use A = (1/2)bh. The sloping sides from the outer vertex to the shared side's endpoints are not automatically heights. A right-angle mark or geometric construction identifies the correct distance.

Suppose a one-by-two-meter rectangular opening is removed from inside the six-by-four part. Its area is two square meters. The region to cover is now 24 + 6 − 2 = 28 square meters. The enclosing composite outline still bounds thirty square meters, but the opening is excluded from the covering task. State whether the question asks for the entire enclosed outline or only the remaining material.

The diagram outlines every region, including the opening. A blank area is not automatically irrelevant: it may be part of the whole being measured or a specified exclusion. Here the text explicitly removes the opening, so subtraction is justified. In a different problem, unshaded space inside an outlined whole might still count toward total area.

A trapezoid can be handled through a rectangle and triangles or through its formula. With parallel bases five and nine units and perpendicular height four, its area is [(5 + 9)/2] × 4 = 28 square units. The average of the two parallel base lengths gives the width of an equivalent rectangle with the same height. Use the distance between the parallel bases, not a sloping leg, as height.

An L-shaped region offers an add-or-subtract choice. An outer eight-by-six rectangle with a three-by-two corner removed has area 48 − 6 = 42 square units. Another valid partition may split the L into two rectangles with areas that add to 42. Comparing methods is a useful check only if both decompositions use dimensions justified by the diagram.

Missing lengths can be found through relationships before area calculation. If the total horizontal extent is eight and a known top segment is five along the same direction, the remaining aligned segment is three. Do not infer a missing diagonal by subtracting unrelated horizontal and vertical lengths. Explain which segments lie along the same complete dimension.

Units must be consistent. If one dimension is in centimeters and another in meters, convert to a shared length unit before multiplying unless you deliberately carry a mixed-unit area and then convert it. A result such as 28 meters for the covering region uses a length unit instead of an area unit. Estimation can catch scale errors: the small opening should reduce the thirty-square-meter region only slightly, not by twenty or by a factor of two.

Worked alternative partition: For the eight-by-six outer rectangle with a three-by-two upper corner removed, one partition is a full eight-by-four lower rectangle plus a five-by-two upper rectangle. Areas are 32 and ten, totaling 42. This agrees with 48 minus six and shows the pieces exactly fill the L-shaped region.

Worked height error: A triangular extension has a four-unit base, three-unit perpendicular height and a longer sloping side. Its area uses four and three, regardless of the slope's length. Substituting the sloping side for height would produce a different and unjustified area.

Worked covering cost: If the 28-square-meter remaining region uses a hypothetical material costing 2.5 budget units per square meter with no waste or other fee, cost is 70 units. A real purchase might require a waste allowance or whole packages, but this mathematical model must state such additions before including them.

Decomposition workshop: Draw the six-by-four region, its triangular extension and the one-by-two opening on measured grid paper. Label the shared base and perpendicular height. Color calculation pieces differently while leaving every outline visible. Explain why their interiors do not overlap and why the opening is subtracted.

Independent performance: Create and solve one rectangle-plus-triangle area, one trapezoid, one L-shaped region and one composite with a removed opening. For two problems, use two different valid decompositions. State the desired region and all dimensions, show area formulas with square units and verify that no part is omitted or counted twice. A partner checks every inferred length against the complete aligned dimension from which it came.