Cone volume, slant height and surface area
Use perpendicular height for volume and slant height for the curved surface.
Core and Extended.
Before you begin
- Use circle area.
- Use Pythagoras in a right triangle.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a right cone's axial triangle, which length is the hypotenuse?
What you will learn
- Calculate cone volume and curved or total area.
- Relate radius, perpendicular height and slant height in a right cone.
Different heights do different jobs
A cone of radius r and perpendicular height h has volume πr²h/3. The slant height l runs along the curved surface from the apex to the circular rim. For a right circular cone, r, h and l form a right triangle and l² = r² + h². With r = 3 and h = 4 cm, l = 5 cm.
Do not insert l where the volume formula asks for h. If a cone has known radius and slant height, find perpendicular height using √(l² − r²). All lengths must be in compatible units, and l must be longer than r for a non-flat right cone.
The curved face is a sector
A cone's curved surface opens into a sector of radius l, whose arc wraps around the base circumference 2πr. Its area is πrl. The total area of a closed cone includes the base circle, giving πrl + πr². For r = 3, h = 4 and l = 5 cm, curved area is 15π cm² and total area 24π cm².
A conical party hat without a base needs only curved area; a closed solid needs the base too. In a compound solid, a base joined to another component becomes internal and is omitted from outside area, even though both component volumes still contribute. Identify the surfaces before adding formulas.
Pause and explain
A cone has r = 3 cm and h = 4 cm. Find its curved area.
Worked example
A right cone has radius 3 cm and height 4 cm. Find slant height, volume and total area.
Show the worked solution
- Pythagoras gives l = √(3² + 4²) = 5 cm.
- Volume is π × 9 × 4/3 = 12π cm³.
- Total area is π × 3 × 5 + π × 9 = 24π cm².
Answer Slant 5 cm; volume 12π cm³; total area 24π cm².
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 right cone has radius 5 cm and height 12 cm. Find slant height and volume.
Give me a hint
Use Pythagoras, then the perpendicular height in volume.
Compare my reasoning
- Slant height is √(25 + 144) = 13 cm.
- Volume is π × 25 × 12/3 = 100π cm³.
- The slant thirteen is not substituted for the volume height twelve.
Slant 13 cm; volume 100π cm³.
Look for these in your work
- Found the hypotenuse correctly.
- Kept solid and slant heights separate.
A right cone has radius 4 cm and slant height 5 cm. Find its perpendicular height and total area.
Give me a hint
Subtract radius squared to recover height squared.
Compare my reasoning
- h = √(25 − 16) = 3 cm.
- Curved area is π × 4 × 5 = 20π cm².
- The base adds 16π cm², giving total 36π cm².
Height 3 cm; total area 36π cm².
Look for these in your work
- Recovered the missing leg by subtraction.
- Included exactly one base.
A base-free conical hat has radius 6 cm and perpendicular height 8 cm. Find the card area, excluding tabs.
Give me a hint
The hat uses the curved surface only.
Compare my reasoning
- Slant height is √(36 + 64) = 10 cm.
- Curved area is π × 6 × 10 = 60π cm².
- The circular base is absent, so no extra 36π is added.
60π cm².
Look for these in your work
- Used slant height for curved area.
- Excluded the absent base and tabs.
Common mistakes
- Using slant height for volume.
- Using perpendicular height for curved area.
- Adding a circular base to a base-free cone.
What can you explain now?
Which height appears in πr²h/3?
Compare with the explanation
The perpendicular height.
The volume factor measures separation from the apex to the base plane, not distance along the curved surface.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, find both heights of a right cone and annotate which is used for volume and surface area.
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources