Sec 3 G3 Maths Bukit Timah | Enlargement and Scale

Secondary 3 G3 Mathematics Tutor Bukit Timah | Enlargement and Reduction: Centres, Scale Factors and Areas — Did you know? Doubling every side length of a rectangle makes its area four times as large, not twice as large. A drawing can preserve all its angles yet become much bigger or smaller, and the centre of enlargement explains how every point has moved. Once students distinguish linear scale, area scale and centre position, the diagrams become predictable rather than mysterious.

Quick answer: the enlargement rules

An enlargement with positive length scale factor k keeps corresponding angles equal and multiplies every corresponding side length by k.
Factors greater than 1 enlarge; positive factors between 0 and 1 reduce. Factor 1 preserves the original size.
Perimeter and other corresponding lengths scale by k; areas of similar plane figures scale by k².
For similar solids, corresponding volumes scale by k³, a different three-dimensional rule.
With the origin as centre, positive enlargement maps (x,y) to (kx,ky).
With any centre O, image P′ satisfies the vector relationship OP′=k·OP; thus P′=O+k(P−O) in coordinates.
With positive k, O, P and P′ lie on the same ray from O, and OP′/OP=k when P is not the centre.
Translation is different: it shifts every point by the same vector but does not change lengths.
Before calculating, name the direction: original-to-image or image-to-original. The reverse length factor is 1/k.

The 2027 SEC G3 Mathematics syllabus K310 covers enlargement and reduction of plane figures, scale drawings, proportional sides, and area and volume ratios of similar figures under G2. Coordinate examples here illustrate those relationships; the syllabus does not necessarily require every coordinate-centre formula as a separate memorised rule. Schools may teach the content in different Secondary 3 or Secondary 4 terms.

Jump to: Understand scale · 40 worked examples · Common mistakes · Study plan · FAQs.

Understand enlargement as movement from a centre

A scale factor is a ratio of corresponding lengths

If one corresponding side grows from 5 cm to 15 cm, the length factor from the smaller to the larger figure is 15/5=3. Every other corresponding side should follow the same ratio for the two figures to be similar. The angles remain equal.

A student should first label the original and the image. If the question runs backwards from the larger figure to the smaller one, the factor is 1/3 instead. A ratio without a direction can make both technically correct numbers look contradictory.

Positive enlargement keeps points on a ray from the centre

With centre O and positive scale factor k, an original point P lies on a ray starting at O, and the image P′ lies on the same ray. The distance from the centre becomes OP′=k×OP. Thus for k=2, the image is twice as far from O as the original.

A centre is a fixed point of the enlargement. When P=O, the image is also O. If a drawn image lies on the wrong ray, or its distance has been measured from the wrong centre, the construction is not the intended positive enlargement.

The origin makes coordinate enlargement simple

If O=(0,0), a point P=(3,−2) under factor 2 becomes P′=(6,−4). Each coordinate is multiplied by the same factor because the directed movement from the origin has doubled. The y-value remains negative, preserving direction along the corresponding ray.

This is a particularly accessible starting example. The reverse reduction factor 1/2 maps P′ back to P. To check, compute the original distance from the origin and verify that its image distance doubled. A sketch should show both points aligned with the origin.

A non-origin centre changes the coordinate arithmetic

If O=(2,1) and P=(5,3), then movement OP=(3,2). An enlargement by factor 2 makes movement OP′=(6,4), so P′=(8,5). Simply multiplying P’s coordinates by two would give (10,6), which enlarges about the origin instead of the stated centre.

The centre-based expression P′=O+k(P−O) describes the procedure clearly. Subtract the centre, scale the displacement, then add the centre back. It is not a separate trick: it is the geometry of relative position written with coordinates.

A reduction is an enlargement with a smaller positive factor

With scale factor 1/2 about the origin, P=(8,6) becomes P′=(4,3). The image lies halfway from the centre to the original. Its corresponding lengths, such as the distance between two image vertices, also become half as long.

The area reduction is not one-half. Because two independent dimensions are halved, the image area becomes one-quarter. A student should use the correct dimension of the requested quantity rather than automatically applying the same scale factor to everything.

Why the area ratio uses k squared

Imagine a 3-by-4 rectangle. Doubling the length and width produces a 6-by-8 rectangle. The original area is 12 square units; the image area is 48, exactly four times as much. Each dimension contributes one factor of two, creating 2²=4.

For a general positive scale factor k, area becomes k² times the original. This applies to all similar plane figures, not only rectangles, because a plane figure’s area changes with two-dimensional scaling. Ask for the units: square units are a reminder of the exponent.

Perimeter still scales with the first power

When every side length doubles, the sum of those lengths doubles as well. A triangle with perimeter 30 cm becomes a similar triangle with perimeter 60 cm under factor two. Its area becomes four times the original area, however.

A learner who applies k² to perimeter has confused the number of dimensions being measured. A useful independent check is to list three original side lengths and their scaled versions, then add them. The result confirms the linear rule without memorisation.

Similar solid volume uses a third power

A cube with edge 2 cm has volume 8 cm³. Doubling its edge to 4 cm gives volume 64 cm³, eight times as much. The factor 2³ appears because three perpendicular directions are enlarged.

The 2027 K310 syllabus includes volume ratios of similar solids alongside area ratios. This is a useful extension, but a question about a plane figure’s area still uses k², not k³. Classify the quantity as one-, two- or three-dimensional first.

A photograph can preserve shape without proving scale conditions

A printed diagram may show two objects whose outlines appear similar, but a mathematical question must give or justify consistent length ratios and equal corresponding angles. Visual resemblance alone is not proof. Also, the centre of an enlargement cannot be found reliably from an impression of how two pictures are placed.

Use corresponding vertices and draw joining lines from P through P′ when the figures are known to form an enlargement. For noncoincident corresponding points, their lines can help locate the centre. Then check that the same factor works for multiple vertex pairs.

Enlargement is different from translation and reflection

Translation adds the same movement vector to every point, keeping all distances and angles unchanged. Positive enlargement about a centre changes distances proportionally while preserving corresponding angles. Reflection reverses orientation about a mirror line without changing lengths.

If a student sees a larger figure, translation alone cannot explain it. Conversely a figure shifted right by five units without changing size should not be called an enlargement of factor five. The numerical operation should match the geometric behaviour.

Recovering a factor from area is a square-root problem

Suppose the new figure has nine times the area of the original and is similar to it. The positive length factor satisfies k²=9, so k=3. Taking the area ratio itself as the length factor would wrongly claim every side grew ninefold.

For reductions, an area ratio of 1/16 corresponds to a positive length factor of 1/4. The reciprocal direction matters: image-to-original length factor is then four. State which figure is being used as the reference before taking the square root.

Scale drawings connect Mathematics to useful plans

A scale drawing, classroom seating plan or floor plan represents actual distances through a fixed linear ratio. If a model is enlarged uniformly, corresponding dimensions remain proportional; area changes by the square of that ratio. The same reasoning supports plans and maps, though real-world accuracy depends on the stated measurements.

The goal for a Secondary 3 G3 learner is to move confidently among diagram, length ratio, area ratio and verbal explanation. A tutor can ask whether a result makes the image larger or smaller as intended and then validate it using a second corresponding measurement.

40 worked examples with original and image checks

These are original teaching exercises, not official examination-paper items. Ten topics appear in four numerical versions each. Sketch the original and image, write the scale direction and check how a representative length or area changes. Later mix the types so that the first operation is selected from the problem, not from a worksheet heading.

Positive enlargement about the origin

This set develops positive enlargement about the origin. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 1. Question: Point P=(3, -4) is enlarged about O=(0,0) by factor 2. Find P′. Working: Scale both origin-based coordinates: P′=(6, -8). The position vector OP′=2OP. Check: Dividing the image coordinates by 2 returns P=(3, -4). Common mistake: The negative y-direction is preserved; do not change its sign during positive enlargement.

Independent challenge: Label the original and image before writing a ratio. The direction from one to the other determines whether to multiply or divide. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 2. Question: Point P=(4, -5) is enlarged about O=(0,0) by factor 3. Find P′. Working: Scale both origin-based coordinates: P′=(12, -15). The position vector OP′=3OP. Check: Dividing the image coordinates by 3 returns P=(4, -5). Common mistake: The negative y-direction is preserved; do not change its sign during positive enlargement.

Independent challenge: Mark the centre O and compare the directed movements OP and OP′ rather than guessing from page position. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 3. Question: Point P=(5, -6) is enlarged about O=(0,0) by factor 4. Find P′. Working: Scale both origin-based coordinates: P′=(20, -24). The position vector OP′=4OP. Check: Dividing the image coordinates by 4 returns P=(5, -6). Common mistake: The negative y-direction is preserved; do not change its sign during positive enlargement.

Independent challenge: If the centre is not the origin, subtract its coordinates first, scale the resulting displacement, then add the centre back. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 4. Question: Point P=(6, -7) is enlarged about O=(0,0) by factor 5. Find P′. Working: Scale both origin-based coordinates: P′=(30, -35). The position vector OP′=5OP. Check: Dividing the image coordinates by 5 returns P=(6, -7). Common mistake: The negative y-direction is preserved; do not change its sign during positive enlargement.

Independent challenge: Predict whether the image lies closer to or farther from the centre under the stated positive scale factor. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Reduction about the origin

This set develops reduction about the origin. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 5. Question: Reduce point P=(4, 6) by positive factor 1/2 about O=(0,0). Find P′. Working: Halve both coordinates to obtain P′=(2, 3). The point lies halfway along the ray from O to P. Check: Doubling the image coordinates returns (4, 6). Common mistake: The area scale is 1/4, but coordinates and other lengths use the linear factor 1/2.

Independent challenge: Check whether the requested quantity is a length, perimeter, area or volume before choosing k, k² or k³. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 6. Question: Reduce point P=(8, 12) by positive factor 1/2 about O=(0,0). Find P′. Working: Halve both coordinates to obtain P′=(4, 6). The point lies halfway along the ray from O to P. Check: Doubling the image coordinates returns (8, 12). Common mistake: The area scale is 1/4, but coordinates and other lengths use the linear factor 1/2.

Independent challenge: Use the reverse length scale factor to verify the original point or dimension can be recovered. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 7. Question: Reduce point P=(12, 18) by positive factor 1/2 about O=(0,0). Find P′. Working: Halve both coordinates to obtain P′=(6, 9). The point lies halfway along the ray from O to P. Check: Doubling the image coordinates returns (12, 18). Common mistake: The area scale is 1/4, but coordinates and other lengths use the linear factor 1/2.

Independent challenge: Look for a second corresponding side to confirm that the same ratio holds throughout a similar figure. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 8. Question: Reduce point P=(16, 24) by positive factor 1/2 about O=(0,0). Find P′. Working: Halve both coordinates to obtain P′=(8, 12). The point lies halfway along the ray from O to P. Check: Doubling the image coordinates returns (16, 24). Common mistake: The area scale is 1/4, but coordinates and other lengths use the linear factor 1/2.

Independent challenge: Distinguish positive enlargement from translation by checking whether any distances changed. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Enlargement around a non-origin centre

This set develops enlargement around a non-origin centre. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 9. Question: Centre O=(1, 2), point P=(4, 4) and scale factor 2 are given. Find image P′. Working: Movement OP=(3, 2). Double that displacement to (6, 4) and add centre O, giving P′=(7, 6). Check: Subtract O from P′: its components are twice the original OP components. Common mistake: Multiplying P coordinates alone by 2 would incorrectly use the origin as centre.

Independent challenge: Write the physical units for the original and image measurements; scale ratios themselves have no unit when corresponding quantities use the same unit. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 10. Question: Centre O=(2, 3), point P=(6, 6) and scale factor 2 are given. Find image P′. Working: Movement OP=(4, 3). Double that displacement to (8, 6) and add centre O, giving P′=(10, 9). Check: Subtract O from P′: its components are twice the original OP components. Common mistake: Multiplying P coordinates alone by 2 would incorrectly use the origin as centre.

Independent challenge: Return to a new diagram after two days and see if the method still works without seeing the centre-based formula. Reverse the problem from image to original and compare the resulting factor. If the figure is translated before being enlarged, explain which centre or position information now matters. The student should retain one original and one later unprompted attempt for comparison.

Worked example 11. Question: Centre O=(3, 4), point P=(8, 8) and scale factor 2 are given. Find image P′. Working: Movement OP=(5, 4). Double that displacement to (10, 8) and add centre O, giving P′=(13, 12). Check: Subtract O from P′: its components are twice the original OP components. Common mistake: Multiplying P coordinates alone by 2 would incorrectly use the origin as centre.

Worked example 12. Question: Centre O=(4, 5), point P=(10, 10) and scale factor 2 are given. Find image P′. Working: Movement OP=(6, 5). Double that displacement to (12, 10) and add centre O, giving P′=(16, 15). Check: Subtract O from P′: its components are twice the original OP components. Common mistake: Multiplying P coordinates alone by 2 would incorrectly use the origin as centre.

Recover a centre when factor two is known

This set develops recover a centre when factor two is known. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 13. Question: An enlargement has factor 2 and maps P=(5, 3) to P′=(8, 5). Find the centre O. Working: For factor 2, P′=2P−O, so O=2P−P′. Therefore O=(2, 1). Check: Using this centre, OP=(3, 2) and OP′=(6, 4), confirming factor 2. Common mistake: Do not assume the centre is the midpoint between P and P′; that is not generally true for factor two.

Worked example 14. Question: An enlargement has factor 2 and maps P=(7, 5) to P′=(11, 8). Find the centre O. Working: For factor 2, P′=2P−O, so O=2P−P′. Therefore O=(3, 2). Check: Using this centre, OP=(4, 3) and OP′=(8, 6), confirming factor 2. Common mistake: Do not assume the centre is the midpoint between P and P′; that is not generally true for factor two.

Worked example 15. Question: An enlargement has factor 2 and maps P=(9, 7) to P′=(14, 11). Find the centre O. Working: For factor 2, P′=2P−O, so O=2P−P′. Therefore O=(4, 3). Check: Using this centre, OP=(5, 4) and OP′=(10, 8), confirming factor 2. Common mistake: Do not assume the centre is the midpoint between P and P′; that is not generally true for factor two.

Worked example 16. Question: An enlargement has factor 2 and maps P=(11, 9) to P′=(17, 14). Find the centre O. Working: For factor 2, P′=2P−O, so O=2P−P′. Therefore O=(5, 4). Check: Using this centre, OP=(6, 5) and OP′=(12, 10), confirming factor 2. Common mistake: Do not assume the centre is the midpoint between P and P′; that is not generally true for factor two.

Corresponding lengths in an enlargement

This set develops corresponding lengths in an enlargement. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 17. Question: A small figure has a side 4 cm. Its image is similar with length scale factor 2 from small to large. Find the corresponding large side. Working: Multiply corresponding length by factor: 4×2=8 cm. Check: Dividing the image side 8 cm by 2 gives the original 4 cm. Common mistake: For a forward enlargement the image side should not be shorter when f is greater than one.

Worked example 18. Question: A small figure has a side 5 cm. Its image is similar with length scale factor 3 from small to large. Find the corresponding large side. Working: Multiply corresponding length by factor: 5×3=15 cm. Check: Dividing the image side 15 cm by 3 gives the original 5 cm. Common mistake: For a forward enlargement the image side should not be shorter when f is greater than one.

Worked example 19. Question: A small figure has a side 6 cm. Its image is similar with length scale factor 4 from small to large. Find the corresponding large side. Working: Multiply corresponding length by factor: 6×4=24 cm. Check: Dividing the image side 24 cm by 4 gives the original 6 cm. Common mistake: For a forward enlargement the image side should not be shorter when f is greater than one.

Worked example 20. Question: A small figure has a side 7 cm. Its image is similar with length scale factor 5 from small to large. Find the corresponding large side. Working: Multiply corresponding length by factor: 7×5=35 cm. Check: Dividing the image side 35 cm by 5 gives the original 7 cm. Common mistake: For a forward enlargement the image side should not be shorter when f is greater than one.

Area of a larger similar plane figure

This set develops area of a larger similar plane figure. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 21. Question: A plane figure has area 6 cm². A similar figure is enlarged by length factor 2. Find its area. Working: Area factor=2²=4. Therefore enlarged area=6×4=24 cm². Check: The area ratio is 24:6=4:1, the square of the length ratio. Common mistake: Multiplying area only by 2 would ignore the second dimension.

Worked example 22. Question: A plane figure has area 9 cm². A similar figure is enlarged by length factor 3. Find its area. Working: Area factor=3²=9. Therefore enlarged area=9×9=81 cm². Check: The area ratio is 81:9=9:1, the square of the length ratio. Common mistake: Multiplying area only by 3 would ignore the second dimension.

Worked example 23. Question: A plane figure has area 12 cm². A similar figure is enlarged by length factor 4. Find its area. Working: Area factor=4²=16. Therefore enlarged area=12×16=192 cm². Check: The area ratio is 192:12=16:1, the square of the length ratio. Common mistake: Multiplying area only by 4 would ignore the second dimension.

Worked example 24. Question: A plane figure has area 15 cm². A similar figure is enlarged by length factor 5. Find its area. Working: Area factor=5²=25. Therefore enlarged area=15×25=375 cm². Check: The area ratio is 375:15=25:1, the square of the length ratio. Common mistake: Multiplying area only by 5 would ignore the second dimension.

Perimeter under uniform enlargement

This set develops perimeter under uniform enlargement. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 25. Question: A triangle’s perimeter is 20 cm. Enlarge it by positive length factor 2. What is the new perimeter? Working: Every side becomes 2 times as long, so their sum becomes 20×2=40 cm. Check: Dividing the enlarged perimeter by 2 restores the original perimeter 20 cm. Common mistake: The perimeter scales with f, not f².

Worked example 26. Question: A triangle’s perimeter is 25 cm. Enlarge it by positive length factor 3. What is the new perimeter? Working: Every side becomes 3 times as long, so their sum becomes 25×3=75 cm. Check: Dividing the enlarged perimeter by 3 restores the original perimeter 25 cm. Common mistake: The perimeter scales with f, not f².

Worked example 27. Question: A triangle’s perimeter is 30 cm. Enlarge it by positive length factor 4. What is the new perimeter? Working: Every side becomes 4 times as long, so their sum becomes 30×4=120 cm. Check: Dividing the enlarged perimeter by 4 restores the original perimeter 30 cm. Common mistake: The perimeter scales with f, not f².

Worked example 28. Question: A triangle’s perimeter is 35 cm. Enlarge it by positive length factor 5. What is the new perimeter? Working: Every side becomes 5 times as long, so their sum becomes 35×5=175 cm. Check: Dividing the enlarged perimeter by 5 restores the original perimeter 35 cm. Common mistake: The perimeter scales with f, not f².

Volume factor for similar solids

This set develops volume factor for similar solids. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 29. Question: A cube has side 2 cm. Another similar cube has twice the side length. Compare their volumes. Working: Original volume=2³=8 cm³. Larger volume=4³=64 cm³, so the volume factor is 8=2³. Check: The volume ratio 64:8 is eight to one, as a three-dimensional scale requires. Common mistake: Do not apply the two-dimensional area factor four when the question asks for volume.

Worked example 30. Question: A cube has side 3 cm. Another similar cube has twice the side length. Compare their volumes. Working: Original volume=3³=27 cm³. Larger volume=6³=216 cm³, so the volume factor is 8=2³. Check: The volume ratio 216:27 is eight to one, as a three-dimensional scale requires. Common mistake: Do not apply the two-dimensional area factor four when the question asks for volume.

Worked example 31. Question: A cube has side 4 cm. Another similar cube has twice the side length. Compare their volumes. Working: Original volume=4³=64 cm³. Larger volume=8³=512 cm³, so the volume factor is 8=2³. Check: The volume ratio 512:64 is eight to one, as a three-dimensional scale requires. Common mistake: Do not apply the two-dimensional area factor four when the question asks for volume.

Worked example 32. Question: A cube has side 5 cm. Another similar cube has twice the side length. Compare their volumes. Working: Original volume=5³=125 cm³. Larger volume=10³=1000 cm³, so the volume factor is 8=2³. Check: The volume ratio 1000:125 is eight to one, as a three-dimensional scale requires. Common mistake: Do not apply the two-dimensional area factor four when the question asks for volume.

Recover length scale from area scale

This set develops recover length scale from area scale. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 33. Question: A larger similar plane figure has 4 times the area of a smaller one. The smaller has a corresponding side 4 cm. Find the larger side. Working: The positive length factor satisfies f²=4, so f=2. The larger corresponding side=4×2=8 cm. Check: Squaring 2 gives the stated area factor 4; both size and direction are consistent. Common mistake: The area multiplier 4 is not itself the side-length multiplier.

Worked example 34. Question: A larger similar plane figure has 9 times the area of a smaller one. The smaller has a corresponding side 5 cm. Find the larger side. Working: The positive length factor satisfies f²=9, so f=3. The larger corresponding side=5×3=15 cm. Check: Squaring 3 gives the stated area factor 9; both size and direction are consistent. Common mistake: The area multiplier 9 is not itself the side-length multiplier.

Worked example 35. Question: A larger similar plane figure has 16 times the area of a smaller one. The smaller has a corresponding side 6 cm. Find the larger side. Working: The positive length factor satisfies f²=16, so f=4. The larger corresponding side=6×4=24 cm. Check: Squaring 4 gives the stated area factor 16; both size and direction are consistent. Common mistake: The area multiplier 16 is not itself the side-length multiplier.

Worked example 36. Question: A larger similar plane figure has 25 times the area of a smaller one. The smaller has a corresponding side 7 cm. Find the larger side. Working: The positive length factor satisfies f²=25, so f=5. The larger corresponding side=7×5=35 cm. Check: Squaring 5 gives the stated area factor 25; both size and direction are consistent. Common mistake: The area multiplier 25 is not itself the side-length multiplier.

Compare perimeter and area under reduction

This set develops compare perimeter and area under reduction. Name the centre, the scaling direction and the dimensional quantity before calculating. Four variations allow the learner to practise the same correct reasoning when numbers or coordinates change.

Worked example 37. Question: A rectangle 3 cm by 4 cm is reduced by positive factor 1/2. Find new perimeter and new area. Working: New dimensions are 1.5 cm and 2 cm. Perimeter = 14/2=7 cm. Area = (3/2)×(4/2)=3 cm², one-quarter of the original 12 cm². Check: The length factor 1/2 reduces perimeter by 1/2 and area by (1/2)²=1/4. Common mistake: Do not apply the same area and perimeter factor to different dimensional quantities.

Worked example 38. Question: A rectangle 6 cm by 8 cm is reduced by positive factor 1/2. Find new perimeter and new area. Working: New dimensions are 3 cm and 4 cm. Perimeter = 28/2=14 cm. Area = (6/2)×(8/2)=12 cm², one-quarter of the original 48 cm². Check: The length factor 1/2 reduces perimeter by 1/2 and area by (1/2)²=1/4. Common mistake: Do not apply the same area and perimeter factor to different dimensional quantities.

Worked example 39. Question: A rectangle 9 cm by 12 cm is reduced by positive factor 1/2. Find new perimeter and new area. Working: New dimensions are 4.5 cm and 6 cm. Perimeter = 42/2=21 cm. Area = (9/2)×(12/2)=27 cm², one-quarter of the original 108 cm². Check: The length factor 1/2 reduces perimeter by 1/2 and area by (1/2)²=1/4. Common mistake: Do not apply the same area and perimeter factor to different dimensional quantities.

Worked example 40. Question: A rectangle 12 cm by 16 cm is reduced by positive factor 1/2. Find new perimeter and new area. Working: New dimensions are 6 cm and 8 cm. Perimeter = 56/2=28 cm. Area = (12/2)×(16/2)=48 cm², one-quarter of the original 192 cm². Check: The length factor 1/2 reduces perimeter by 1/2 and area by (1/2)²=1/4. Common mistake: Do not apply the same area and perimeter factor to different dimensional quantities.

Six enlargement mistakes to repair

Using the wrong direction for the length ratio

Name original and image first. Enlarging from small to large multiplies by k; reversing it divides by k. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

Multiplying point coordinates without checking the centre

Coordinate multiplication by k works directly when the centre is the origin. Other centres require subtracting and re-adding the centre. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

Using linear factor k for area

Area scales by k², because two dimensions change. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

Using area factor k² for perimeter

Perimeter is a length sum and scales by k. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

Treating a translated figure as an enlarged one

Translation preserves side lengths, whereas a non-unit enlargement changes corresponding distances. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

Mistaking a drawn resemblance for verified similarity

Check that corresponding angles match and side ratios are consistently proportional. The correction should identify the first unjustified assumption and test a new diagram. Avoid merely replacing the final number; ask the learner to explain the relationship and direction of scale.

A Bukit Timah Mathematics tutor can teach scale through movement

A focused Secondary 3 G3 Mathematics tutor in Bukit Timah can give students a transparent grid with an original point, its centre and a factor. Each pupil draws the image point, explains why it lies on the same ray and checks the distance ratio. The grid makes the scale decision visible.

The tutor can then enlarge a small rectangle and compare perimeter and area. When students discover that side lengths double but area quadruples, the k² relationship becomes a consequence of geometry. Later the lesson can introduce an arbitrary centre and compare the result with a translation. Delayed mixed practice tests independent choice.

The unchanged Clementi Secondary 1 Mathematics small-group reference shows a foundation-stage teaching approach. For the upper-secondary year context see the Secondary 3 G3 Mathematics Bukit Timah overview. Confirm the school scheme of work and current class arrangements separately.

A six-week scale-factor practice plan

Week 1 — fix corresponding points and side pairs

Label the original and image, compare matching lengths and recover a scale factor from a diagram. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Week 2 — enlarge about the origin

Scale positive and negative coordinate components and verify points along the correct rays. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Week 3 — use a non-origin centre

Subtract centre coordinates, scale the displacement and re-add the centre. Check with one second image point. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Week 4 — connect lengths, perimeter and area

Use k and k² on the same figure, explaining why the factors differ. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Week 5 — include reduction and similar-solid volumes

Work with factors below one, reverse scale ratios and contrast 2D area with 3D volume. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Week 6 — contrast enlargement with translation

Read fresh grid diagrams without a heading and choose the correct operation from the geometric information. Save one original independent solution and a later different-scale example. A meaningful improvement is choosing the centre and dimension correctly without hints.

Frequently asked questions

What does a positive scale factor of three do?

It triples all corresponding lengths while preserving corresponding angles. Test this with a new set of corresponding points and ask the student to justify the first step.

What does factor one-half mean?

It is a reduction: corresponding lengths become half as long. Test this with a new set of corresponding points and ask the student to justify the first step.

What happens to area when lengths double?

For similar plane figures, area is multiplied by 2²=4. Test this with a new set of corresponding points and ask the student to justify the first step.

What happens to volume when lengths double?

For similar solids, volume is multiplied by 2³=8. Test this with a new set of corresponding points and ask the student to justify the first step.

Does perimeter scale like area?

No. Perimeter scales with k, while area scales with k². Test this with a new set of corresponding points and ask the student to justify the first step.

Can I multiply every coordinate by k?

That is the direct rule when the centre is (0,0). For another centre, scale relative displacement from that centre. Test this with a new set of corresponding points and ask the student to justify the first step.

How do I find a centre with a positive enlargement?

Corresponding original and image points lie on rays from the centre; joining corresponding point pairs helps locate it. Test this with a new set of corresponding points and ask the student to justify the first step.

Can an enlargement change angles?

An enlargement of the kind considered here preserves corresponding angles and produces a similar figure. Test this with a new set of corresponding points and ask the student to justify the first step.

Is a translation the same as an enlargement?

No. A translation shifts every point by the same vector without changing size; an enlargement with non-unit factor changes size. Test this with a new set of corresponding points and ask the student to justify the first step.

Is enlargement covered in the 2027 G3 syllabus?

Yes. K310 G2 includes enlargement and reduction of plane figures and related similarity ratios. Test this with a new set of corresponding points and ask the student to justify the first step.

Official syllabus reference: 2027 SEC G3 Mathematics K310, G2 Congruence and Similarity. These calculations are original educational practice and do not reproduce examination questions.

The takeaway

Did you know? Enlarging a figure is a systematic change measured from a centre, not just drawing a bigger-looking shape. A careful student names the original and image, measures corresponding lengths, checks k for length and k² for area, and uses the centre to find new points. Those decisions make scale drawings and similarity easier to reason about.

Secondary 3 G3 Mathematics Tutor Bukit Timah | Enlargement and Reduction: Centres, Scale Factors and Areas