Compound areas and outside perimeters
Split or subtract areas, then trace only the exposed boundary.
Section Paper 1 · Core · Non-calculator
Core and Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use rectangle, triangle and circle measures.
- Identify sector arcs and straight boundaries.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When two shapes share a joining edge, is that edge on the outside boundary?
What you will learn
- Find area of joined shapes, cut-outs and rings.
- Find perimeter without counting internal edges.
Partition without overlaps or gaps
A compound area can be split into familiar non-overlapping shapes, or found by subtracting a missing region from a larger enclosing shape. A 10 by 8 cm rectangle with a 4 by 3 cm corner removed has area 80 − 12 = 68 cm². Show the split or cut-out so each region is counted once.
A circular ring of outer radius R and inner radius r has area π(R² − r²). It is not π(R − r)²: the ring's thickness is not a radius of a replacement circle. If pieces overlap, adding their whole areas double-counts the overlap; subtract that shared region or partition it separately.
Area bookkeeping does not give perimeter
Perimeter follows the exposed outer boundary, plus any inner boundary explicitly requested. Joining a semicircle to a rectangle along its diameter removes that shared diameter from the outside. A rectangle 8 cm long and 6 cm wide with a radius-3 semicircle on one short end has perimeter 8 + 8 + 6 + 3π = 22 + 3π cm.
Removing a corner can leave the same total perimeter: in the 10 by 8 rectangle above, removed boundary lengths 4 and 3 are replaced by notch edges of those same lengths, so perimeter remains 36 cm. A hole creates a new inner boundary, while a filled join hides an old boundary. Trace the specified edges instead of subtracting a cut-out's whole perimeter.
Pause and explain
A circle ring has outer radius 5 cm and inner radius 3 cm. Find its area.
Worked example
A 10 by 8 cm rectangle has a 4 by 3 cm rectangular corner removed. Find the remaining area and perimeter.
Show the worked solution
- Remaining area is 10 × 8 − 4 × 3 = 68 cm².
- Two removed outside lengths total seven centimetres; two new notch edges also total seven.
- The outside perimeter stays 2(10 + 8) = 36 cm.
Answer Area 68 cm²; perimeter 36 cm.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A rectangle 8 by 6 cm has a radius-3 cm semicircle joined along one 6 cm end. Find total area and outside perimeter.
Give me a hint
Add the filled areas but omit the shared diameter from perimeter.
Compare my reasoning
- Area is 48 + (1/2)π × 9 = 48 + 9π/2 cm².
- Three exposed rectangle sides total 8 + 8 + 6 = 22 cm.
- The semicircle arc adds 3π cm, giving perimeter 22 + 3π cm.
Area (48 + 9π/2) cm²; perimeter (22 + 3π) cm.
Look for these in your work
- Counted each region once.
- Omitted the joined diameter.
A ring has outer radius 5 cm and inner radius 3 cm. Find area and the combined length of both circular edges.
Give me a hint
Both edges count only because the question explicitly asks for both.
Compare my reasoning
- Area is π(25 − 9) = 16π cm².
- The outer edge is 10π cm and inner edge is 6π cm.
- Combined edge length is 16π cm, which has length units despite the same coefficient.
Area 16π cm²; combined edge length 16π cm.
Look for these in your work
- Subtracted the hole area.
- Included both specified boundaries.
A stadium-shaped track surrounds a 10 by 6 m rectangle with a semicircle at each short end. Find enclosed area and outside perimeter.
Give me a hint
Two semicircles make one circle of radius three metres.
Compare my reasoning
- Area is 60 + 9π m².
- The two straight exposed sides total 20 m.
- The two curved ends total 6π m, giving perimeter 20 + 6π m.
Area (60 + 9π) m²; perimeter (20 + 6π) m.
Look for these in your work
- Combined the semicircles correctly.
- Excluded both internal diameters.
Common mistakes
- Counting internal joins in outside perimeter.
- Adding a hole's area.
- Squaring the difference of ring radii instead of subtracting their squares.
What can you explain now?
Can a removed corner leave perimeter unchanged?
Compare with the explanation
Yes.
New notch edges can replace removed boundary lengths exactly, so trace the boundary rather than assuming subtraction.
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Make it stick
Tomorrow, draw a joined rectangle and semicircle. Colour area pieces and trace exposed edges separately.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.