Calculate corresponding lengths in similar figures
Use and reverse a directed length factor, then justify the correspondence in a triangle with a parallel cross-section.
Section 4.11 · Similarity
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
- Recognise a justified similarity statement and identify corresponding sides.
- Multiply and divide exact fractions and decimals.
- Use collinear point order and angle facts from parallel lines.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A target length equals k times its original length. Which equation recovers the original?
What you will learn
- Find a directed length factor from a known corresponding pair.
- Multiply to find a target length and divide to recover an original length.
- Use parallel-angle conditions and ordered points to justify a smaller-to-full triangle correspondence.
Two similar figures have corresponding sides 4 cm and 10 cm. A second side of the smaller figure is 7 cm and a third side of the larger figure is 12.5 cm. Find the corresponding missing larger and smaller sides.
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.
Both drawing axes have equal scale in cm. Whole-view turns rotate all geometry together; free points retain actual coordinates without snapping.
Two similar figures have corresponding sides 4 cm and 10 cm. A second side of the smaller figure is 7 cm and a third side of the larger figure is 12.5 cm. Find the corresponding missing larger and smaller sides.
Representative triangles organise the corresponding side data; generic similar-figure questions do not gain an extra triangle hypothesis. The missing vertex is not supplied.
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.
Both drawing axes have equal scale in cm. Whole-view turns rotate all geometry together; free points retain actual coordinates without snapping.
Two similar figures have corresponding sides 4 cm and 10 cm. A second side of the smaller figure is 7 cm and a third side of the larger figure is 12.5 cm. Find the corresponding missing larger and smaller sides.
Representative triangles organise the corresponding side data; generic similar-figure questions do not gain an extra triangle hypothesis. The missing vertex is not supplied.
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 polygon 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 3 and 12. 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.
Write the direction before calculating
When two figures are known to be similar, choose one as the original and the other as the target. If their known corresponding lengths are a and b in that direction, write k=b/a. Every original length is multiplied by k to obtain its matching target length. The factor is dimensionless only after the two given lengths have been expressed in a common unit.
A length factor is multiplicative. A fixed difference between one pair of lengths is not an amount to add to every other side. Keep the factor as an exact fraction when possible, and draw arrows from each original length to its corresponding target. If k is between zero and one, the target is smaller; the same multiplication rule still applies.
Reverse the operation deliberately
If the known length is on the target figure, the equation is still target=k×original. Recover the original by dividing the target by k. Do not change which figure is called original in the middle of a calculation. A small table headed original, factor and target helps when different missing sides occur on different figures in the same question.
Check every recovered length by putting it back into the original equation. Multiplying the recovered original by k should reproduce the given target. This check is especially useful with decimal or fractional factors: dividing by a number smaller than one increases a positive length, which may be exactly what recovering the larger original requires. A zero or negative physical length cannot be rescued by silently taking its absolute value.
Justify a parallel cross-section
In a triangle, a segment parallel to one side can form a smaller triangle with the same vertex. Name both triangles and explain the angle correspondence: the angle at the shared vertex is common, and corresponding or alternate angles come from the stated parallel lines. Only after establishing those conditions should you use a common side ratio.
Distinguish a whole side from a remainder. If D lies between A and B, then AD is measured from the shared vertex while DB is the leftover part. A smaller-to-full factor therefore compares AD with AB, not with DB. Scale corresponding complete sides first, recover any requested full side by division, and subtract only after the required whole and part refer to the same segment.
Sources for this lesson
Pause and explain
D lies between A and B, and ADE is a smaller triangle similar to the full triangle ABC. Which ratio is the smaller-to-full factor?
Worked example
Two similar figures have corresponding sides 4 cm and 10 cm. A second side of the smaller figure is 7 cm and a third side of the larger figure is 12.5 cm. Find the corresponding missing larger and smaller sides.
Show the worked solution
- Choose the smaller figure as original and the larger as target. Their known pair gives k=10/4=5/2.
- The side corresponding to the smaller 7 cm length is 7×5/2=35/2=17.5 cm on the larger figure.
- The larger third side is already known. Reverse target=k×original: its smaller counterpart is 12.5÷(5/2)=5 cm.
- Check both directions: 17.5/7=5/2 and 12.5/5=5/2, matching the original 10/4 ratio.
Answer The larger missing side is 17.5 cm, and the smaller missing third side is 5 cm. The common smaller-to-larger factor is 5/2.
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.
Similar triangles have corresponding lengths 18 and 27 cm. Another side of the smaller triangle is 14 cm and a third side of the larger triangle is 30 cm. Find the corresponding missing larger and smaller sides.
Give me a hint
Write larger=k×smaller, then decide which missing side needs multiplication and which needs division.
Compare my reasoning
- The smaller-to-larger factor is 27/18=3/2. Keep that direction for both missing sides.
- Multiply the smaller 14 cm side by 3/2 to obtain its larger counterpart, 21 cm.
- Recover the smaller side corresponding to 30 cm by division: 30÷(3/2)=20 cm.
- Check 21/14=30/20=27/18=3/2. The smaller side lengths 18,14,20 and their larger counterparts are consistent.
The larger missing side is 21 cm and the smaller missing third side is 20 cm, using the factor 3/2 in the stated direction.
Look for these in your work
- Establish one smaller-to-larger factor.
- Use multiplication and inverse division for their correct sides.
- Substitute both results back into the common ratio.
Similar figures have corresponding sides 9.5 m and 47.5 m. Another side of the smaller figure is 6.2 m and a third side of the larger figure is 38 m. Find the corresponding missing larger and smaller sides and check the ratios.
Give me a hint
Recover one exact directed factor and check both missing sides against it.
Compare my reasoning
- Compare the given corresponding lengths in the same metre unit: k=47.5/9.5=5 from smaller to larger.
- Multiply 6.2 m by 5 to find the larger matching side, 31 m.
- The larger third side is 38 m, so divide by 5 to recover its smaller counterpart, 7.6 m.
- The checks 31/6.2=5 and 38/7.6=5 reproduce the known factor 47.5/9.5=5.
The missing larger side is 31 m and the missing smaller third side is 7.6 m. Every corresponding larger-to-smaller length ratio is 5.
Look for these in your work
- Keep all lengths in metres and use the same direction.
- Recover both a larger and a smaller matching side.
- Check both resulting ratios.
In triangle ABC, AB=28 cm and AC=21 cm. D lies on AB with AD=16 cm, E lies on AC and DE=20 cm is parallel to BC. Find AE,BC,DB and EC, stating the correspondence.
Give me a hint
Use the parallel-angle conditions to justify ADE/ABC, then compare smaller sides with full sides before subtracting remainders.
Compare my reasoning
- Use D on AB and E on AC with DE parallel to BC. The angle at A is common and the corresponding parallel-line angles establish ADE similar to ABC, with A↔A, D↔B and E↔C.
- The smaller-to-full factor is AD/AB=16/28=4/7. Apply it to the corresponding full AC to obtain AE=(4/7)×21=12 cm.
- Here the smaller corresponding side DE=20 cm is given. Recover the full BC by division: BC=20÷(4/7)=35 cm.
- Use the stated interior point order to subtract the remainders: DB=28−16=12 cm and EC=21−12=9 cm. Check AE/AC=DE/BC=4/7.
AE=12 cm, BC=35 cm, DB=12 cm and EC=9 cm. The justified ADE-to-ABC factor is 4/7; it compares smaller sides with full corresponding sides.
Look for these in your work
- Justify the named ADE/ABC correspondence using actual parallel angles.
- Use the smaller-to-full factor to recover both a smaller length and a full length.
- Subtract remainders only after respecting A-D-B and A-E-C.
Common mistakes
- Adding a known difference instead of multiplying by a common factor.
- Multiplying a known target length when the original must be recovered by division.
- Comparing a smaller triangle side with a remainder rather than the corresponding full side.
What can you explain now?
A target is 0.8 times its original and its corresponding side is 4.8 cm. What was the original side, and how can you check it?
Compare with the explanation
6 cm; 0.8×6=4.8 cm.
The target is already known, so divide 4.8 by 0.8. Multiplying the recovered original by the stated factor checks both the arithmetic and the direction.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do you decide between multiplication and division, and why must a triangle remainder stay out of the whole-side ratio?
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.