Compare angles and lengths under enlargement
Compare every corresponding side and ordinary or reflex angle under positive enlargement, distinguishing length scaling from angle preservation and congruence at factor 1.
Exam scope and assessment notes
These twelve shared lessons apply to Foundation and inherited Higher for F5.2A–M; no extra Higher row is added. Keep named corners, signed turns, equivalent mirror equations, exact fractions, positive factors and the freely chosen-centre and identity exceptions. General-angle coordinates are final3dp approximations of unrounded geometry. Written explanations need human review; coordinate agreement does not certify proof or physical construction. Linear November2026/June2027 entry and regional-centre rules remain separate from modular awards. Book2 Shape and Space7 tags and Hubs66–71 are navigation only: map223–248 differs from proof222/summary247/next248, with chapter/Hub bodies unavailable.
Before you begin
- Calculate exact lengths or squared lengths from coordinates.
- Identify ordinary and reflex polygon interior angles.
- Use positive scale factors and the meaning of congruence.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If a side length is L and the positive factor is k, which is its image length?
What you will learn
- Compare all named side lengths using one positive factor.
- Preserve corresponding interior angles, including reflex ones.
- Explain why a nonunit enlargement changes size while the factor 1 identity remains congruent.
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, enlarge about (1,0) with positive factor 2. Construct the named image and compare every corresponding side length and angle. Decide whether the figures remain congruent and explain the identity-factor exception where relevant.
Describe your operation, construct every named image point, and record your actual directed traces. Only the question supplies the givens. The board does not transform a shape or recover a missing parameter for you.
Both Cartesian directions have equal drawing scale. Blue is supplied geometry, amber is my actual work, purple is dirty retained work, and green is a separately gated reference. Screen size is not physical instrument calibration.
Actual boundaries retain their entered order, closure, concavity and wrong or incomplete points. No reflected corner, rotation endpoint, mirror foot or target polygon is filled automatically. Construction rays continue beyond their anchors; mirror lines continue both ways; joining segments keep their original endpoints.
Supplied positive enlargement factor 2 about (1, 0). This description does not generate a target boundary.
This finite fixed-case comparison is not proof marking, marks, mastery, physical construction calibration or specialist/formal assessment approval. General-angle points stay unrounded; a separately requested three-decimal written table is an approximation.
Actual operation 1
No actual construction is recorded for this action.
Choose a point slot before adding a keyboard coordinate or picking on the board. Named picks retain exact given or current clean recorded coordinates. Free picks retain their unrounded actual planar coordinates; they do not snap to an expected result.
My written coordinates, quantities and reasoning
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Choose Yes or No after checking your current construction.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Choose Yes or No after checking your current construction.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Choose Yes or No after checking your current construction.
Enter your exact fraction or finite decimal, using the stated conditions.
Choose Yes or No after checking your current construction.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
This sequence is optional. If you add operations, complete every chosen parameter or remove the sequence. The tool generates no inverse for you.
These original cases support mathematical self-checking. They do not award marks, mastery, proof approval, physical construction competence, specialist signoff or formal assessment readiness. Positive scale factors only; unchanged images do not uniquely recover a centre. Keep general-angle construction coordinates unrounded and separately requested written coordinates at three decimal places.
Lengths scale by the same positive factor
For any two object points P,Q under one enlargement, P′−Q′=k(P−Q). Thus every corresponding side length is multiplied by the positive factor k, regardless of where the centre lies. Compare corresponding sides in the named boundary order rather than selecting unrelated equal-looking segments.
If a side has exact length L, its image length is kL. Keep radicals and fractional multipliers exact unless an approximation is explicitly requested. Squared distances scale by k², so they can check a coordinate construction without prematurely rounding square roots; the required length comparison still uses k, not k².
Angles are preserved even in a concave boundary
Positive enlargement scales both arms at each vertex without changing their directions relative to one another, so every corresponding interior angle remains equal. This applies to an ordinary angle and to a reflex angle inside a concave polygon. Keep the actual interior side of the boundary when identifying the angle.
The image need not have the same position or size to have the same angles. Equal angles describe part of similarity, while congruence additionally requires the same size. Do not scale an angle measure by k or use an apparent change in screen width as evidence that an angle changed.
Keep the identity exception precise
For a nondegenerate polygon and k≠1, at least one nonzero side changes length, so the complete image is not congruent to the object. At k=1, every side and angle is unchanged and the image is the identity, so it is congruent. This is the mathematical exception to the usual statement that enlargement does not preserve lengths.
Explain the actual factor and all corresponding features in the task. The inverse factor 1/k restores the original size about the same centre, but it does not turn the first nonunit image into a congruent copy of the original. Fixed checks of a coordinate example do not approve an unexamined general proof or a practical instrument construction.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Transformation geometry Foundation printed23/physical29 and Higher inheritance printed38/physical44
- Pearson 2026/2027 International GCSE entry manual · centre regions physical3 and Mathematics A physical25
- Pearson literal second-edition book/Hub map · Book2 Shape and Space7 physical12 · navigation only
Pause and explain
For a nondegenerate polygon, which positive factor makes an enlargement congruent to its object?
Worked example
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, enlarge about (1,0) with positive factor 2. Construct the named image and compare every corresponding side length and angle. Decide whether the figures remain congruent and explain the identity-factor exception where relevant.
Show the worked solution
- Construct the full image using enlargement about (1,0) with positive factor 2: A′=(1,2), B′=(7,2), C′=(3,6). Preserve the given names and boundary order rather than comparing unrelated sides.
- Compare all corresponding side lengths exactly: AB: 3 → 6; BC: 2√2 → 4√2; CA: √5 → 2√5, in coordinate units. The common length factor is 2; squared distances therefore have factor 4.
- All corresponding interior angles agree: ∠A′=∠A, ∠B′=∠B, ∠C′=∠C. The same rule covers reflex interior angles whenever a boundary is concave; the boundary keeps its corresponding interior sides. Because the nonunit factor 2 changes nonzero side lengths, equal angles alone do not make this image congruent.
- The inverse is enlargement about (1,0) with positive factor 1/2. The identity factor 1 is a valid positive exception to length change, but does not make a nonunit enlargement congruent. Numerical agreement here does not approve a general proof or practical construction.
Answer A′=(1,2), B′=(7,2), C′=(3,6). AB: 3 → 6; BC: 2√2 → 4√2; CA: √5 → 2√5. ∠A′=∠A, ∠B′=∠B, ∠C′=∠C. Not congruent: angles are unchanged but corresponding lengths scale by 2.
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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.
Polygon ABCD has A(0,0), B(3,0), C(4,2), D(1,3). In the given order, enlarge about (-1,2) with positive factor 1/2. Construct the named image and compare every corresponding side length and angle. Decide whether the figures remain congruent and explain the identity-factor exception where relevant.
Give me a hint
Use factor k for side lengths, equal angles for every named corner and the identity exception at k=1.
Compare my reasoning
- Construct the full image using enlargement about (-1,2) with positive factor 1/2: A′=(-1/2,1), B′=(1,1), C′=(3/2,2), D′=(0,5/2). Preserve the given names and boundary order rather than comparing unrelated sides.
- Compare all corresponding side lengths exactly: AB: 3 → 3/2; BC: √5 → √(5/4); CD: √10 → √(5/2); DA: √10 → √(5/2), in coordinate units. The common length factor is 1/2; squared distances therefore have factor 1/4.
- All corresponding interior angles agree: ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D. The same rule covers reflex interior angles whenever a boundary is concave; the boundary keeps its corresponding interior sides. Because the nonunit factor 1/2 changes nonzero side lengths, equal angles alone do not make this image congruent.
- The inverse is enlargement about (-1,2) with positive factor 2. The identity factor 1 is a valid positive exception to length change, but does not make a nonunit enlargement congruent. Numerical agreement here does not approve a general proof or practical construction.
A′=(-1/2,1), B′=(1,1), C′=(3/2,2), D′=(0,5/2). AB: 3 → 3/2; BC: √5 → √(5/4); CD: √10 → √(5/2); DA: √10 → √(5/2). ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D. Not congruent: angles are unchanged but corresponding lengths scale by 1/2.
Look for these in your work
- Compare all named side lengths using one positive factor.
- Preserve corresponding interior angles, including reflex ones.
- Explain why a nonunit enlargement changes size while the factor 1 identity remains congruent.
Polygon ABCDEF has A(0,0), B(4,0), C(4,1), D(1,1), E(1,3), F(0,3). In the given order, enlarge about (2,-1) with positive factor 3/2. Construct the named image and compare every corresponding side length and angle. Decide whether the figures remain congruent and explain the identity-factor exception where relevant.
Give me a hint
Use factor k for side lengths, equal angles for every named corner and the identity exception at k=1.
Compare my reasoning
- Construct the full image using enlargement about (2,-1) with positive factor 3/2: A′=(-1,1/2), B′=(5,1/2), C′=(5,2), D′=(1/2,2), E′=(1/2,5), F′=(-1,5). Preserve the given names and boundary order rather than comparing unrelated sides.
- Compare all corresponding side lengths exactly: AB: 4 → 6; BC: 1 → 3/2; CD: 3 → 9/2; DE: 2 → 3; EF: 1 → 3/2; FA: 3 → 9/2, in coordinate units. The common length factor is 3/2; squared distances therefore have factor 9/4.
- All corresponding interior angles agree: ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D, ∠E′=∠E, ∠F′=∠F. The same rule covers reflex interior angles whenever a boundary is concave; the boundary keeps its corresponding interior sides. Because the nonunit factor 3/2 changes nonzero side lengths, equal angles alone do not make this image congruent.
- The inverse is enlargement about (2,-1) with positive factor 2/3. The identity factor 1 is a valid positive exception to length change, but does not make a nonunit enlargement congruent. Numerical agreement here does not approve a general proof or practical construction.
A′=(-1,1/2), B′=(5,1/2), C′=(5,2), D′=(1/2,2), E′=(1/2,5), F′=(-1,5). AB: 4 → 6; BC: 1 → 3/2; CD: 3 → 9/2; DE: 2 → 3; EF: 1 → 3/2; FA: 3 → 9/2. ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D, ∠E′=∠E, ∠F′=∠F. Not congruent: angles are unchanged but corresponding lengths scale by 3/2.
Look for these in your work
- Compare all named side lengths using one positive factor.
- Preserve corresponding interior angles, including reflex ones.
- Explain why a nonunit enlargement changes size while the factor 1 identity remains congruent.
Polygon ABCD has A(-2,1), B(1,1), C(2,4), D(-1,3). In the given order, enlarge about (3,4) with positive factor 1. Construct the named image and compare every corresponding side length and angle. Decide whether the figures remain congruent and explain the identity-factor exception where relevant.
Give me a hint
Use factor k for side lengths, equal angles for every named corner and the identity exception at k=1.
Compare my reasoning
- Construct the full image using enlargement about (3,4) with positive factor 1: A′=(-2,1), B′=(1,1), C′=(2,4), D′=(-1,3). Preserve the given names and boundary order rather than comparing unrelated sides.
- Compare all corresponding side lengths exactly: AB: 3 → 3; BC: √10 → √10; CD: √10 → √10; DA: √5 → √5, in coordinate units. The common length factor is 1; squared distances therefore have factor 1.
- All corresponding interior angles agree: ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D. The same rule covers reflex interior angles whenever a boundary is concave; the boundary keeps its corresponding interior sides. Because k=1, size is unchanged and the image is congruent to the object.
- The inverse is enlargement about (3,4) with positive factor 1. The identity factor 1 is a valid positive exception to length change, but does not make a nonunit enlargement congruent. Numerical agreement here does not approve a general proof or practical construction.
A′=(-2,1), B′=(1,1), C′=(2,4), D′=(-1,3). AB: 3 → 3; BC: √10 → √10; CD: √10 → √10; DA: √5 → √5. ∠A′=∠A, ∠B′=∠B, ∠C′=∠C, ∠D′=∠D. Congruent: all corresponding lengths and angles are unchanged.
Look for these in your work
- Compare all named side lengths using one positive factor.
- Preserve corresponding interior angles, including reflex ones.
- Explain why a nonunit enlargement changes size while the factor 1 identity remains congruent.
Common mistakes
- Scaling angle measures by the enlargement factor.
- Using k² as the length factor.
- Forgetting that k=1 preserves lengths and congruence.
What can you explain now?
A side length scales by k=5/3. Does its interior angle also become 5/3 as large?
Compare with the explanation
No. The corresponding interior angle remains equal.
Both vertex arms keep their relative directions; the positive factor changes lengths rather than the angle between those arms.
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Make it stick
State and justify the length and angle comparisons, distinguish k from k² and explain the factor 1 congruence exception.
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.