Higher: Solve problems with similar areas and volumes
Choose the dependency between lengths, corresponding areas and volumes, keeping units, matching regions and practical assumptions explicit.
Exam scope and assessment notes
The first four lessons are shared Foundation content inherited by Higher; the last three are additional Higher area, volume and mixed-similarity work. Match corresponding vertices and keep one positive length factor in a stated direction. Maps and plans use exact stated lengths; screen centimetres, physical ruler accuracy and print calibration are not certified. Similar solids require corresponding dimensions to scale together; total surface area includes the whole stated external boundary. Exact numerical comparisons and text-entry attempts do not approve written proof, marks, mastery or physical construction. Linear 4MA1 Issue 2 is checked against November 2026/June 2027 entries; regional options depend on centre eligibility and modular awards remain separate. Publisher links are navigation only: literal Book2 Shape and Space6 F4.11B, Shape and Space7 F4.11A, and Shape and Space9 H4.11A/B/C tags are retained, with no Book1 tag invented. The map/proof page and topic-location conflicts remain explicit; the relevant chapter/Hub bodies are unavailable. Full-book, specialist and formal-assessment approval remain pending.
Before you begin
- Use the positive square and cube relationships for genuine similarity.
- Convert between metric linear, square and cubic units.
- Recognise a parallel cone section and distinguish full, removed and remaining quantities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A corresponding surface-area factor is R for similar solids. Which expression gives their volume factor?
What you will learn
- Choose a square, cube or positive-root relationship from the quantity supplied.
- Combine unit conversion with corresponding areas and volumes in similar models.
- Justify similar cones from a parallel cut and shared vertex before subtracting volumes.
Two similar closed boxes have total surface areas in the ratio 81:144. The smaller volume is 2700cm³. Find the larger volume.
Construct your own complete figures, directed routes and corner correspondence. Enter every requested quantity, substitution and reason; missing outlines, dimensions and scale factors remain your own work.
Given panel 1 — drawing unit: cm
Blue: original supplied geometry. Amber: my actual submitted work. Each frame uses its stated drawing unit; compare mixed units by conversion, not by apparent screen size. Unlabelled sides and angles stay for calculation.
This is a schematic 3D camera view. Projected angles and distances are not metric measurements. Choose an actual visible named corner or enter all three coordinates; no unique screen-to-3D inverse is used.
Two similar closed boxes have total surface areas in the ratio 81:144. The smaller volume is 2700cm³. Find the larger volume.
A closed cuboid is a representative coordinate witness of a generic similar solid; the paper does not thereby specify cuboid dimensions.
Only literal givens are initially labelled. Every requested scalar, complete missing outline, chosen correspondence and explained reference remains learner work.
Compare other given figures in their stated units
Frames with the same unit and dimension share a drawing scale. Different units use separate labelled metric frames; convert physical lengths rather than comparing apparent screen sizes.
Given panel 2 — drawing unit: cm
Blue: original supplied geometry. Amber: my actual submitted work. Each frame uses its stated drawing unit; compare mixed units by conversion, not by apparent screen size. Unlabelled sides and angles stay for calculation.
This is a schematic 3D camera view. Projected angles and distances are not metric measurements. Choose an actual visible named corner or enter all three coordinates; no unique screen-to-3D inverse is used.
Two similar closed boxes have total surface areas in the ratio 81:144. The smaller volume is 2700cm³. Find the larger volume.
A closed cuboid is a representative coordinate witness of a generic similar solid; the paper does not thereby specify cuboid dimensions.
Only literal givens are initially labelled. Every requested scalar, complete missing outline, chosen correspondence and explained reference remains learner work.
Mark or construct the actual complete cuboid in Given panel 1
No geometry is recorded for this action.
Enter your own count and names in boundary, route or anchor order. A cuboid needs all eight actual corners: base A–B–C–D and corresponding E–F–G–H, with all faces closing. Generic figure and solid drawings are coordinate witnesses, not extra hypotheses for the paper question.
This draft is incomplete: Enter your own integer count between 8 and 8. All entered raw text and points remain visible; no valid replacement figure is invented.
My quantities, equations and reasons
Enter exact supported fractions or decimals. A field asking for a coefficient of π takes the rational coefficient alone. Keep two decimal places where money is requested. Every response or unit edit closes previous feedback and references.
Required for the whole attempt. Give an exact supported scalar.
Required for the whole attempt. Give an exact supported scalar.
Required for the whole attempt. Give an exact supported scalar.
Required for the whole attempt.
Written substitutions and reasons require human review; they are not automatically proved or marked.
Required for the whole attempt.
Written substitutions and reasons require human review; they are not automatically proved or marked.
This tool compares fixed mathematics with actual coordinate constructions. Written reasoning needs human review; it awards no marks, mastery, physical-instrument competence, specialist signoff or formal assessment approval. Mixed physical units require conversion; projected 3D angles and lengths are schematic.
Make a dependency chain
A mixed similarity problem often gives one kind of quantity and asks for another. Write a short chain such as area ratio → positive length factor → volume factor. This prevents an area ratio from being used directly as a volume ratio. Each arrow represents a particular operation: a square root recovers k from k², while cubing k produces k³.
Keep the direction consistent through the chain. If the target is the larger object, form target divided by original for the first ratio and retain that direction for later factors. Check that the result has a sensible size relationship: a positive factor greater than one enlarges lengths, areas and volumes, although by different multipliers. None of those numerical ratios establishes similarity unless that condition is actually given or justified.
Match units and count the whole surface
A model height given in centimetres and a real height given in metres must be converted before finding a dimensionless length factor. After scaling, convert square and cubic quantities separately: centimetres squared and centimetres cubed have different metre conversion factors. Keep the unrounded factor until the end so that avoidable rounding does not affect both subsequent calculations.
In these closed-solid problems, total surface area means the entire external boundary. For a cuboid, count all six faces, including the opposing pairs, and check the given area and volume against its actual dimensions. The paint model coats every external face once at the stated rate. Calculate that full charge before applying a requested money rounding rule; the numerical model does not establish material density or physical paint consumption.
Use an actual similar solid before subtraction
A plane parallel to a cone base can cut off a smaller cone with the same vertex. The shared vertex and parallel section give corresponding cross-sections and a common linear factor. Compare the smaller height with the full height, then cube that factor to find the smaller volume. A ratio involving the remaining height describes a different pair of lengths.
The remaining frustum volume is the full volume minus the removed smaller-cone volume. Check that both volumes use the same unit and that the smaller height lies strictly between zero and the full height. This is a justified application of similar-cone scaling followed by subtraction, not an unexplained general frustum formula or approval of an instrument construction. Explain the geometric conditions as well as the arithmetic.
Sources for this lesson
Pause and explain
A cone is cut parallel to its base. Which facts justify using a height ratio for the removed smaller cone and the full cone?
Worked example
Two similar closed boxes have total surface areas in the ratio 81:144. The smaller volume is 2700cm³. Find the larger volume.
Show the worked solution
- The surface-area ratio is smaller:larger=81:144, so the target-to-original area factor is 144/81=16/9.
- Recover the physical length factor from the area ratio using its positive square root, k=√(16/9)=4/3.
- Use that recovered length factor in the volume dependency by cubing it: k³=(4/3)³=64/27.
- The larger volume is 2700×64/27=6400 cm³. Check that its ratio to 2700 is 64/27, not the surface-area factor 16/9.
Answer The larger box volume is 6400 cm³. The linear and volume factors are 4/3 and 64/27 respectively.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 model cuboid is 18 cm high, 10 cm wide and 4 cm deep; its total surface area is 584cm² and volume 720cm³. A geometrically similar real object is 54 m high. Find the real total surface area and volume in square and cubic metres.
Give me a hint
Convert heights to one unit, then scale all six faces and the volume with different powers.
Compare my reasoning
- Check the supplied cuboid model: 18×10×4=720 cm³, and 2(18×10+18×4+10×4)=584 cm² counts all six faces.
- Convert the real height to the model unit: 54 m=5400 cm. The model-to-real length factor is k=5400/18=300.
- Scale total surface area by k²: 584×300²=52560000 cm². Divide by 10000 to obtain 5256 m².
- Scale volume by k³: 720×300³=19440000000 cm³. Divide by 1000000 to obtain 19440 m³.
- An independent dimension check gives real dimensions 54 m by 30 m by 12 m. Their six-face area is 5256 m² and their product is 19440 m³.
The real total surface area is 5256 m² and the real volume is 19440 m³. The common length factor is 300.
Look for these in your work
- Verify the stated model volume and all six face areas from dimensions.
- Convert height, area and volume units with their correct powers.
- Check the real totals from actual scaled dimensions.
A cone is cut parallel to its base, leaving a smaller cone with the same vertex. The full height is 10 cm, the smaller height 4 cm and the full volume 900cm³. Find the smaller-cone volume and the remaining frustum volume.
Give me a hint
Use the parallel cut and shared vertex to compare the smaller cone with the full cone, not with the remainder.
Compare my reasoning
- The cut is parallel to the cone base and the smaller cone shares the full cone vertex, so the cones are similar in the stated height correspondence.
- The smaller-to-full length factor is k=4/10=2/5, not 4/6 from comparing the smaller cone with the leftover height.
- Cube the factor and scale the full volume: 900×(2/5)³=900×8/125=57.6 cm³ for the smaller cone.
- The remaining frustum volume is 900−57.6=842.4 cm³. These positive volumes add back to the supplied full 900 cm³.
The smaller cone volume is 57.6 cm³ and the remaining frustum volume is 842.4 cm³.
Look for these in your work
- Justify the similar cones using the given cut and vertex.
- Cube the smaller-to-full height ratio.
- Subtract the removed volume and check that the parts reproduce the full volume.
Two similar solid cuboid ornaments have volumes 500cm³ and 864cm³. The smaller measures 10 cm by 10 cm by 5 cm and has total surface area 400cm². Find the larger total surface area and its painting cost at$0.012 per cm² to the nearest cent, assuming every external face is painted once.
Give me a hint
Recover the length factor from the volume ratio, count the whole external surface and round the final charge once.
Compare my reasoning
- The smaller cuboid has volume 10×10×5=500 cm³ and total surface area 2(10×10+10×5+10×5)=400 cm², consistent with the given data.
- The larger-to-smaller volume factor is 864/500=216/125. Its positive cube root is k=6/5.
- Square that factor for total surface area: A=400×(6/5)²=400×36/25=576 cm². The larger dimensions 12 cm by 12 cm by 6 cm also give volume 864 cm³ and total area 576 cm².
- Every external face is painted once at the stated uniform rate, so the exact charge is 576×$0.012=$6.912.
- Round the final charge once to the nearest cent: $6.91. No separate density, wall-thickness or physical paint-consumption claim follows from this numerical cost model.
The larger total surface area is 576 cm². The exact painting charge is $6.912, which rounds to $6.91 to the nearest cent.
Look for these in your work
- Verify the actual smaller and larger cuboid dimensions, total areas and volumes.
- Use the square of a recovered length factor for total surface area.
- Apply the all-faces-once cost model and round only the final charge.
Common mistakes
- Using an area or volume ratio directly as a length factor.
- Scaling with mismatched centimetres and metres, then using the wrong square or cube conversion.
- Comparing a smaller cone with the frustum remainder instead of the full cone.
What can you explain now?
For similar solids, a corresponding area factor is 49/16. What are the positive length and volume factors?
Compare with the explanation
Length factor 7/4; volume factor 343/64.
Take the positive square root of 49/16 to recover the length factor. Cube 7/4 to obtain the volume factor, keeping the same direction of correspondence.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Can you explain each arrow from a given area or volume ratio to the quantity actually requested, including its unit and geometric conditions?
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.