Compound shapes and parts of shapes.
Split, subtract and track the exposed boundary, from an L-shape to a frustum.
Calculate compound areas and perimeters · combine solids · account for hidden joins · use similar cones for frustums
Split the area; trace the actual boundary
A compound shape combines simpler shapes whose measurements you already know. Instead of searching for a new formula, draw dividing lines that turn the region into rectangles, triangles or other familiar pieces.
For area, add non-overlapping pieces or subtract a missing piece from a larger shape. For perimeter, ignore the dividing lines inside the shape and follow only the actual exterior boundary. Find unlabelled lengths from matching totals before substituting into formulas.
Add simple pieces or subtract a missing piece from a larger whole. For perimeter, follow every exposed boundary edge once; internal joins are not boundary edges.
Find the area and perimeter of this L-shape.
Area = large rectangle − removed rectangle.
Missing outer lengths are 12 − 5 = 7 cm and 8 − 3 = 5 cm. Trace six boundary edges:
The dashed edges show the removed rectangle; do not count them as part of this shape’s perimeter.
A 10 cm by 6 cm rectangle is topped by a semicircle of diameter 10 cm.
Perimeter = 10 + 6 + 6 + 5π = 22 + 5π cm
The shared 10 cm diameter is inside the shape and is not included in its outside perimeter.
Pause and explainShould an internal dividing line be counted in the compound shape’s perimeter?
No. It is a helper line inside the region, not part of the outside boundary.
Parts of circles and holes
Removing a hole decreases area but can introduce additional boundary. A circular hole in a sheet removes material from the inside while leaving a new circular edge.
State exactly which measurement is requested. For remaining area, subtract the removed region. For the total boundary of the material, include the inner edge if the question includes hole boundaries. Keep circles, semicircles and sectors separate so you remove the correct fraction.
A circular hole of radius 2 cm is cut from a 10 cm square. Find the remaining area and total boundary length.
Total boundary = 40 + 4π cm
The hole removes area but adds an inner boundary. If asked only for the outer perimeter, that remains 40 cm.
A circle has radius 10 cm. Find the area of the minor segment cut off by a 90° central angle.
Subtract the triangle between the two radii from the quarter-circle sector. The radii form a right angle.
= 25π − 50 cm²
Pause and explainDoes cutting a hole always decrease the total length of the material’s boundary?
No. It removes area but introduces an inner boundary, so the total boundary length can increase.
Compound solids: joined faces disappear
For a compound solid, volume describes how much space the pieces occupy together, while surface area describes only the faces exposed to the outside or specified cavity.
Add volumes for joined pieces and subtract volumes for empty spaces. When finding exterior area, exclude touching faces hidden at a join. For a hollow solid, include exposed interior surfaces only when they belong to the requested total surface.
A cylinder of radius 3 cm and height 8 cm is topped by a hemisphere of the same radius. The bottom is closed.
Add the volumes, but count only surfaces visible from outside.
Outside area = 48π + 18π + 9π = 75π cm²
The three exposed parts are the cylinder wall, hemisphere curve and bottom circle. The two matching circular faces at the join are internal.
A hollow cylinder has outer radius 5 cm, inner radius 3 cm and height 8 cm. Find the material volume.
For its entire surface area, with both ends open, include outer wall, inner wall and two annular end faces:
Pause and explainWhy can adding the full surface areas of two joined cuboids overcount?
The touching faces become internal and are not exposed. Their areas must be removed from the exterior total.
Frustums: subtract similar cones
A frustum is the portion left after the top of a cone or pyramid is cut off parallel to the base. The removed small solid is similar to the original complete solid.
Use the similarity ratio to recover a missing full or small height before subtracting volumes. For a conical frustum’s curved area, subtract the small cone’s curved area from the full cone’s. Then add the flat end faces if total surface area is required.
Find volume and total surface area for the frustum shown.
- The removed small cone and original cone have radius ratio 3 : 6 = 1 : 2.
- The frustum height is 4 cm, so the original cone has height 8 cm and the removed cone height 4 cm.
- The frustum slant height is 5 cm. The original and removed slant heights are 10 cm and 5 cm.
= 96π − 12π = 84π cm³
Subtract curved areas, then add both flat circular ends for a closed frustum.
Total area = 45π + 36π + 9π = 90π cm²
Check the frustum slant height: √[4² + (6 − 3)²] = 5 cm. The radius difference, not the full lower radius, forms the horizontal leg.
Pause and explainWhy can the radii of the two cone bases help find their height ratio?
The small removed cone and original cone are similar, so corresponding radii and heights share the same length ratio.
Your turn. Mark the hidden joins.
Draw the pieces or the original-and-removed solids before writing a calculation. Choose whether addition or subtraction gives the required quantity.
For surface area, mark every exposed face and cross out hidden joins. For a frustum, establish similarity first rather than assuming the labelled remaining height is the full cone’s height.
A 9 cm by 7 cm rectangle has a 3 cm by 2 cm corner removed. Find its area.
Reveal worked answer
Find the area of a ring with outer radius 6 cm and inner radius 4 cm.
Reveal worked answer
A closed cylinder of radius 2 cm and height 5 cm is topped by a hemisphere of radius 2 cm. Find its exposed area.
Reveal worked answer
A right-cone frustum has radii 4 cm and 2 cm and height 3 cm. Find its volume by subtracting similar cones.
Reveal worked answer
A semicircle of radius 4 cm is attached along its diameter to an 8 cm by 3 cm rectangle. Find the outside perimeter.