E5.5 Compound shapes and parts of shapes

Syllabus
0580–2028–2029
Topic
E5.5
Level
Extended

Learning objectives

Solve compound area and perimeter problems

A compound shape is built from familiar pieces or from a familiar shape with a piece removed. Area combines whole regions; perimeter follows only the final outside boundary.

Quantity Reliable construction What not to count
area add non-overlapping component areas, or enclosing area minus removed areas overlaps twice
perimeter trace the outside edge once and add its straight and curved lengths shared or internal edges

First mark missing lengths using totals, equal sides, radii or diameters. Choose a decomposition whose pieces do not overlap. Write every component area before adding or subtracting; for a shaded region, identify clearly which area is removed. For perimeter, start at one point and travel once around the finished shape, recording each boundary segment in order. Use arc length—not circle area—for curved edges, and keep units consistent.

A 1010 cm by 66 cm rectangle has a 44 cm by 22 cm corner removed. The remaining area is 10×64×2=52 cm210\times6-4\times2=52\text{ cm}^2. Tracing its six outside edges gives 10+4+4+2+6+6=3210+4+4+2+6+6=32 cm. The two cut edges must be included because they become exposed, but the removed corner is not part of the area.

Do not add the perimeters of separate pieces: their shared edges would be counted even though they are inside the final shape. Area uses square units; perimeter uses linear units. Three-dimensional solids are handled in the next card.

Solve compound solid and partial-solid problems

For a compound solid, volume combines the space occupied by its components, while surface area counts only surfaces exposed after the components are joined or cut.

Situation Volume Surface area
solids joined add component volumes add only exposed faces; omit both sides of each join
piece removed or empty space whole volume minus removed volume include new cut faces that become exposed
fraction of a solid multiply the full volume by the fraction take the relevant curved fraction and include exposed cut faces
frustum from a parallel cut large cone/pyramid minus the similar small top outer sloping/curved part plus both parallel end faces

Draw up a component ledger before calculating. Keep one unit throughout, use each full-solid formula, and preserve π\pi until the end. A join does not change total volume, but it hides matching surfaces. For a frustum, extend it mentally to the original cone or pyramid; use similarity to find the removed top's dimensions, then subtract its volume. Lengths scale by kk, areas by k2k^2 and volumes by k3k^3.

A cone of height 1212 cm and radius 66 cm is cut parallel to its base so the removed small cone has height 44 cm. Similarity gives its radius as 6×4/12=26\times4/12=2 cm. Frustum volume =13π(6)2(12)13π(2)2(4)=416π3 cm3=\frac13\pi(6)^2(12)-\frac13\pi(2)^2(4)=\frac{416\pi}{3}\text{ cm}^3. For surface area, count the outer curved frustum and both exposed circular ends, not the removed cone's base as an internal join.

For an isolated hemisphere, a flat cut face may be exposed; at a join it is hidden. Never subtract a joined face from volume. A frustum subtraction relies on a cut parallel to the base so the small and large solids are similar.