Perimeter, area, surface area and volume
Select the right dimensional relationship, decompose a composite figure and rearrange formulas to recover lengths.
What you will tackle
- Calculate perimeter and area of standard and composite plane figures
- Calculate surface area and volume of standard and composite solids
- Choose and rearrange formulas in contexts
- Use length, area and volume scaling correctly
Identify what is being measured
Perimeter is boundary length, area is two-dimensional coverage, surface area adds exposed faces, and volume is enclosed three-dimensional space. Split composite figures into known shapes without overlapping or omitting regions. A cylinder has volume πr²h and total surface area 2πr² + 2πrh; an open container may exclude one circular face.
Dimensions and scale provide checks
For similar objects with length scale k, area scales by k² and volume by k³. Rearranging V = πr²h gives h = V/(πr²), provided r ≠ 0. The height of a triangle or cone is perpendicular to its base; a sloping edge is not automatically that height.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
A closed cylinder has radius 3 cm and height 5 cm. Find its volume and total surface area. What changes for a cylinder open at the top?
Show a starting hint
List each exposed face before adding areas.
Compare with the explained answer
- Volume = π·3²·5 = 45π cm³.
- Closed surface area = 2π·3² + 2π·3·5 = 48π cm².
- Removing the top subtracts 9π cm², leaving 39π cm².
Answer / response criteria: 45π cm³; 48π cm² closed; 39π cm² open.
Guided 2
A solid consists of a cylinder of radius 2 and height 3 joined to a hemisphere of radius 2. Find volume and exposed surface area, including the bottom but excluding the internal joining circle.
Compare with the explained answer
- Cylinder volume is 12π; hemisphere volume is half of (4/3)π·2³ = 16π/3.
- Total volume is 52π/3.
- Exposed areas are cylinder side 12π, hemisphere curved area 8π and bottom 4π, totaling 24π.
Answer / response criteria: Volume 52π/3 cubic units; exposed area 24π square units.
Independent practice
Independent 1
A 10 by 6 rectangle has a 3 by 2 rectangular corner removed. Find its remaining area and perimeter. A similar solid has volume factor 64; find its length and surface-area factors.
Compare with the explained answer
- Area = 60 − 6 = 54 square units.
- The corner removal replaces two outer edge lengths by equal inner lengths, so perimeter remains 2(10 + 6) = 32.
- k³ = 64 gives k = 4 and area factor k² = 16.
Answer / response criteria: Area 54; perimeter 32; length factor 4 and area factor 16.
Independent 2
A 10 by 8 by 6 block has a rectangular 3 by 2 opening passing vertically through its full height 6. Find remaining volume and total exposed area, including the new internal walls.
Compare with the explained answer
- Volume = 10·8·6 − 3·2·6 = 444.
- The original area is 2(80 + 60 + 48) = 376; top and bottom lose 2·6 = 12.
- The four hole walls add 2(3 + 2)·6 = 60; total 376 − 12 + 60 = 424.
Answer / response criteria: Volume 444; exposed area 424, in cubic/square units respectively.
Application practice
Application 1
A tank of radius 2 m must hold 20π m³. Find its cylindrical height and the material area for the sides and bottom only.
Compare with the explained answer
- 20π = π·2²·h gives h = 5 m.
- The curved side has area 2πrh = 20π m².
- The bottom adds πr² = 4π m², giving 24π m²; construction allowances are outside this ideal model.
Answer / response criteria: Height 5 m; sides plus bottom 24π m².
Application 2
A tank is a cylinder of radius 3 m and height 4 m topped by a hemisphere of radius 3 m. Find its ideal internal capacity and external material area including the flat bottom. Explain why the joined circle is not added to either exposed surface.
Compare with the explained answer
- Volumes are 36π and 18π, giving 54π m³.
- External areas are cylinder side 24π, hemisphere curve 18π and bottom 9π, giving 51π m².
- The joining disc lies inside the completed solid and is not exposed.
Answer / response criteria: Capacity 54π m³; external area 51π m², ignoring thickness and joins.
What can you explain now?
A tank is a cylinder of radius 3 m and height 4 m topped by a hemisphere of radius 3 m. Find its ideal internal capacity and external material area including the flat bottom. Explain why the joined circle is not added to either exposed surface.
Compare with the explanation
Capacity 54π m³; external area 51π m², ignoring thickness and joins.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide