Series route: This guide belongs to the complete PSLE Learner’s Guide volume directory. The original PSLE Learning Guide supplies the wider English, Mathematics and Science route, while PSLE Mathematics provide verified subject-specific continuation.
Find your route through this guide
This volume solves one precise learner problem: the learner sees a line from a vertex to the opposite side and assumes equal areas without checking that the line reaches the midpoint, or assumes equal area means congruent triangles. The aim is not to memorise a slogan. It is to understand the mechanism well enough to recognise the idea when the wording, numbers, diagram or context changes. A Primary 6 learner should be able to explain the decision, perform it under time pressure, and detect a tempting answer that only looks familiar.
- Understand the mechanism before memorising a form.
- Study the worked examples and compare near-misses.
- Repair common misconceptions with targeted tests.
- Complete graded independent practice before reading the answers.
- Use delayed transfer and examination control to make the skill durable.
- Use the parent and tutor notes only after the learner has attempted the work.
The problem beneath the mistake
Draw triangle ABC and mark M as the midpoint of BC. Segment AM is a median because it joins vertex A to the midpoint of the opposite side.
Triangles ABM and ACM have bases BM and MC on the same straight line. Because M is the midpoint, BM equals MC.
Both smaller triangles use the same perpendicular distance from A to line BC. Their heights are equal even if the altitude falls outside an obtuse triangle.
The area formula one half times base times height now gives equal products. Equal base and common height, not visual symmetry, produce the result.
The two smaller triangles need not be congruent. If AB and AC differ, their shapes and perimeters can differ while their areas remain equal.
A line from a vertex to any point on the opposite side is a cevian, but it is a median only when that point is the midpoint. The midpoint evidence is essential.
If the opposite side is split in a ratio of 2:3, the two smaller triangles share a height and their areas are also in the ratio 2:3. The equal-area result is the special 1:1 case.
A midpoint may be stated, shown by matching tick marks, or found from lengths. Do not rely on a drawing that merely looks balanced.
The median property can find an unknown area without knowing the triangle’s height. If one side region is 18 square centimetres, the other is also 18 square centimetres.
Conversely, if two triangles share a vertex and their bases partition the same straight segment, equal areas imply equal base lengths because the common height and one-half factor match.
A triangle has three medians, one from each vertex. Each median halves the area, although the six small triangles formed by all three medians require further reasoning.
The reliable routine is identify the vertex, verify the opposite midpoint, choose the two collinear bases, recognise the shared perpendicular height, compare area products, then separate area claims from shape claims.
Core decision: A median joins a vertex of a triangle to the midpoint of the opposite side. It creates two smaller triangles with equal base lengths on the same line and the same perpendicular height from the shared vertex, so their areas are equal.
A useful rule has three parts: a trigger, an operation and a check. The trigger tells you when the rule applies. The operation tells you what to do. The check tells you whether the result still matches the meaning of the question. Learners who remember only the operation often apply it in the wrong place; learners who remember only an example often freeze when the surface details change.
Mechanism: why the rule works
1. Verify the midpoint
A midpoint divides one segment into two equal lengths and lies on that segment.
Diagnostic question: What evidence shows the two base parts are equal? Reliable response: Use stated lengths, matching marks or a calculation.
2. Name the two bases
For the two small triangles, use the adjacent segments of the opposite side as bases.
Diagnostic question: Which base belongs to each region? Reliable response: Keep both bases on the same straight line.
3. Recognise the common height
The perpendicular distance from the shared vertex to the common base line is identical for both triangles.
Diagnostic question: From which vertex is height measured? Reliable response: Do not use different sloping sides as heights.
4. Compare products
Each area equals one half times its base times the shared height.
Diagnostic question: Which factors cancel or match? Reliable response: Equal bases make equal areas.
5. Use ratio transfer
With a common height, the area ratio equals the base ratio.
Diagnostic question: If the base is not halved, what is its split ratio? Reliable response: Transfer that ratio directly to the two areas.
6. Limit the conclusion
Equal area does not guarantee equal side lengths, angles, perimeters or congruence.
Diagnostic question: What quantity has actually been proved equal? Reliable response: State area only unless more evidence is supplied.
Worked examples: make the decision visible
Do not race through these examples. Before reading each explanation, name the evidence that matters and the information that does not. Then compare the correct route with the tempting route. That comparison is where the transferable learning sits.
Worked example 1: Direct midpoint
Task. M is the midpoint of BC in triangle ABC. Area of ABM is 24 cm². Find area of ACM.
Reasoning. BM equals MC and both triangles share the height from A to BC. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Area of ACM is 24 cm².
Independent check. Add them to obtain total area 48 cm². A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 2: Total area
Task. Triangle PQR has area 70 cm² and PM is a median to QR.
Reasoning. The median divides the total into two equal regions. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Each smaller triangle has area 35 cm².
Independent check. 35 + 35 reconstructs 70. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 3: Length check
Task. BN is 6 cm and NC is 6 cm in triangle ABC.
Reasoning. N is the midpoint of BC, so AN is a median. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. The two areas are equal.
Independent check. Their common altitude makes both products one half × 6 × h. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 4: Not a midpoint
Task. BD is 4 cm and DC is 7 cm.
Reasoning. The bases are unequal though the height is shared. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Areas are in ratio 4:7, not equal.
Independent check. The larger base has the larger area. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 5: Ratio area
Task. BE:EC = 2:5 and area of ABE is 14 cm².
Reasoning. Common height means area ratio 2:5. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Area of AEC is 35 cm².
Independent check. 14 ÷ 2 × 5 = 35. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 6: Unknown base
Task. Two regions from A have equal areas and bases BD and DC on line BC.
Reasoning. The common height and half factor match. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. BD equals DC, so D is the midpoint.
Independent check. If one base were longer, its area would be larger. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 7: Obtuse triangle
Task. The perpendicular from A meets an extension of BC.
Reasoning. Both component triangles still use the same distance to line BC. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. A median still divides the area equally.
Independent check. The altitude need not lie inside the outline. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 8: Different perimeters
Task. AB is much longer than AC, but M is the midpoint of BC.
Reasoning. Unequal sloping sides change shape and perimeter, not the matched base-height products. The controlling idea is equal base segments on one straight line together with the shared perpendicular height from the opposite vertex. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Areas remain equal.
Independent check. Do not claim congruence. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Deepening lab: compare cases, conditions and limits
A method becomes independent when you can say where it applies, where it stops and which change would alter the conclusion. The cases below deliberately vary one structural feature at a time. For each case, pause before the analysis and predict whether the core rule remains available. Then name the exact condition that controls the decision. This is harder than repeating a worked solution, but it is also closer to what an unfamiliar examination item demands.
Deepening case 1: Tick marks
Case. Matching marks on BM and MC show equality.
Analysis. Diagram notation supplies midpoint evidence. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. State the evidence explicitly. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: Visual midpoint only
Case. M looks central but no lengths or marks are given.
Analysis. Appearance is not proof. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Request or derive midpoint evidence. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: One-third point
Case. BD is one third of BC.
Analysis. The areas are in ratio 1:2 because DC is two thirds. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Find both from a total. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: Total from one half
Case. One median region has area 31 cm².
Analysis. The other equals it and total area is 62 cm². Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Write the two-step reasoning. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: Fraction language
Case. A median gives each region one half of the original triangle’s area.
Analysis. The two equal parts exhaust the whole without overlap. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert to percentage. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: Rotated triangle
Case. The opposite side is vertical on the page.
Analysis. Orientation does not alter midpoint or perpendicular distance. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Choose the side by relationship, not position. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: Outside altitude
Case. An obtuse triangle’s altitude falls beyond the base segment.
Analysis. Distance to the base line remains common. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Extend the line mentally. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: Shared vertex condition
Case. Two triangles use different top vertices.
Analysis. Their heights may differ even if bases are equal. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not transfer the rule. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Non-collinear bases
Case. Two equal-looking base segments lie on different lines.
Analysis. A shared vertex does not automatically give equal heights to different lines. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check perpendicular distances. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Same area, different shape
Case. A long narrow triangle and compact triangle can share area.
Analysis. Area equality is weaker than congruence. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. List unproved properties. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Median versus altitude
Case. A median reaches a midpoint; an altitude meets a line at 90 degrees.
Analysis. One segment can be both only in special triangles. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Name each defining condition. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Median versus angle bisector
Case. An angle bisector divides an angle equally, not necessarily the opposite side.
Analysis. Different constructions answer different questions. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Avoid naming by appearance. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: Three medians
Case. Every triangle has three medians.
Analysis. Each one independently halves total area. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Draw all three carefully. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Reverse inference
Case. Equal component areas with a common height imply equal collinear bases.
Analysis. The area ratio equals the base ratio. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Prove the midpoint. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Composite triangle
Case. A median appears inside a larger composite diagram.
Analysis. Ignore unrelated outer lengths until the target triangle is isolated. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Trace its boundary. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Area ratio table
Case. Base ratios 1:1, 2:3 and 4:1 give matching area ratios.
Analysis. The common factor one half × height cancels. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Complete missing entries. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Units
Case. Base centimetres and height centimetres produce square centimetres.
Analysis. Ratio work may cancel units, but final areas need square units. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Attach the correct unit. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Scale drawing
Case. Doubling all lengths multiplies every area by four, but the median still halves the new total.
Analysis. Similarity scaling and equal partition coexist. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check both claims. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Coordinate grid
Case. A midpoint can be found by equal horizontal or vertical intervals on a grid.
Analysis. Once midpoint is secured, the area result follows. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not rely only on counting visual cells. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 20: Final audit
Case. Vertex, midpoint, collinear bases, common height, ratio and conclusion limit complete the proof.
Analysis. Each item protects a logical step. Notice that the answer depends on equal base segments on one straight line together with the shared perpendicular height from the opposite vertex; changing a decorative detail would not be enough to change the reasoning.
Challenge. Audit ten diagrams. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Finish the lab with a blank-page recall. Write the rule, its conditions, one confirming example, one near-miss and one independent check from memory. Compare that page with the article only after finishing. If the rule is correct but the condition is missing, the learning is not yet safe: an overgeneralised rule can produce confident errors. If the condition is correct but no example can be generated, practise construction rather than rereading.
Common misconceptions and targeted repairs
Misconception 1: Any vertex line halves area
What the learner may think. A line from a vertex automatically gives equal regions.
Why it fails. Only a midpoint split gives equal bases. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Check the two base lengths. Use this short test immediately: Is the endpoint a midpoint?
Misconception 2: Symmetry is required
What the learner may think. Only an isosceles triangle can be halved by a median.
Why it fails. Common height and equal base segments work in every triangle. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use the area formula. Use this short test immediately: Do the factors match?
Misconception 3: Median means perpendicular
What the learner may think. The segment must meet at 90 degrees.
Why it fails. Median is defined by midpoint; altitude is defined by perpendicularity. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Name constructions separately. Use this short test immediately: Which condition is marked?
Misconception 4: Equal area means congruent
What the learner may think. All sides and angles must match.
Why it fails. Area alone does not determine shape. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Limit the conclusion to area. Use this short test immediately: What extra data prove congruence?
Misconception 5: Height is the median
What the learner may think. The drawn median is used as height for both small triangles.
Why it fails. Their bases lie along the opposite side, so height is perpendicular from the shared vertex to that line. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Choose compatible bases and height. Use this short test immediately: Is the median perpendicular?
Misconception 6: Drawing proves midpoint
What the learner may think. The line looks centred.
Why it fails. Figures may not be to scale. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use labels, marks or calculation. Use this short test immediately: What explicit evidence is given?
Graded independent practice
Attempt each item without looking ahead. Write one line that identifies the controlling evidence before writing the final answer. Items 1–3 check recognition, 4–7 require explanation, and the later items change the surface form so that memory alone is insufficient.
Practice 1: Recognition
Question. M is midpoint of BC. Compare areas ABM and ACM.
Explained answer. They are equal.
Why this answer earns trust. Equal bases share one height. Transfer variation: State the formula proof.
Practice 2: Total
Question. Total area is 86 cm². Find each median region.
Explained answer. 43 cm².
Why this answer earns trust. The median halves the total. Transfer variation: Check by addition.
Practice 3: Ratio
Question. BD:DC = 3:4. Find the area ratio.
Explained answer. 3:4.
Why this answer earns trust. Common height transfers the base ratio. Transfer variation: Use a total of 49 cm².
Practice 4: Unknown
Question. Area ABD equals area ADC. What follows about BD and DC?
Explained answer. They are equal, provided both bases are collinear and share height from A.
Why this answer earns trust. Area ratio equals base ratio. Transfer variation: Name D’s role.
Practice 5: Counterexample
Question. Does an angle bisector always halve area?
Explained answer. No; an angle bisector need not be a median.
Why this answer earns trust. It need not meet the opposite side at its midpoint. Transfer variation: State when it would.
Practice 6: Orientation
Question. Does rotating the triangle change the result?
Explained answer. No; rotation preserves the geometric relationships.
Why this answer earns trust. Geometric relationships remain. Transfer variation: Relabel the base.
Practice 7: Limit
Question. Can equal area prove equal perimeter?
Explained answer. No; area equality alone cannot determine perimeter.
Why this answer earns trust. Perimeter uses boundary lengths. Transfer variation: Give a scalene example.
Practice 8: Construction
Question. How do you construct a median conceptually?
Explained answer. Find the midpoint of a side and join it to the opposite vertex.
Why this answer earns trust. Both defining parts are named. Transfer variation: Repeat for another side.
Practice 9: Rule recall
Question. State why a median halves area.
Explained answer. It creates equal bases with a shared perpendicular height.
Why this answer earns trust. The mechanism, not a slogan, is recalled. Transfer variation: Recall after one week.
Practice 10: Transfer
Question. Solve ten mixed cevian diagrams.
Explained answer. Every solution first classifies midpoint or ratio evidence.
Why this answer earns trust. This prevents automatic halving. Transfer variation: Include three non-medians.
Delayed transfer: prove that the learning lasts
Transfer test 1: Blank proof
Rebuild the one-half × base × height proof after a delay. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Both equal-base and common-height reasons are explicit. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Ratio extension
Solve splits other than 1:1. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Area ratios match collinear base ratios. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Misleading drawing
Work from labels on a deliberately not-to-scale figure. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. No conclusion depends on appearance. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Examination control
1. Read the job before the story. Underline the command and the noun it governs. “Explain the change,” “state the word,” “find the fraction” and “identify the evidence” are different jobs even when the surrounding context is similar.
2. Separate given information from remembered information. Use only the facts supplied for the current task plus secure subject knowledge. Do not let a previous question, a familiar worksheet or an expected pattern silently become new evidence.
3. Write enough working to expose the decision. A marker or teacher should be able to see the relationship that justifies the answer. Extra sentences are not automatically stronger; the strongest working is concise, relevant and checkable.
4. Reserve a different-route check. For language, read the sentence for meaning and grammar. For mathematics, estimate or reverse the operation. For science, check the claim against the stated observation and a conservation or fair-test principle.
5. Stop when the requested job is complete. Unrequested speculation can contradict a correct answer. Add detail only when it explains the mechanism or satisfies another explicit part of the question.
Under time pressure, use a compact four-part margin routine: trigger → evidence → operation → check. Trigger names the feature that activates the method. Evidence identifies the exact words, values or observations allowed. Operation performs the language, mathematical or scientific step. Check asks whether the final statement answers the requested job, preserves units or referents, and avoids a stronger claim than the evidence supports.
A compact error log
After practice, do not record only “careless mistake.” Name the broken decision. Useful labels include trigger missed, evidence imported, relationship reversed, referent lost, unit dropped, operation correct but job wrong, and check repeated the same route. Add one prevention cue beside the label. A short accurate log is more useful than a long emotional account because it tells the learner what to do on the next item.
Review the log after two days and again after a week. Reattempt only one or two representative items first. If the method transfers, move to a mixed set. If it fails, repair the exact link between trigger and operation. This spacing matters because a solution can feel obvious while it is still visible and disappear when the page is closed.
Notes for parents and tutors
Cut a paper triangle and mark the midpoint of one side with a ruler. Drawing the median makes the equal bases visible before formulas are used.
Ask the learner to distinguish median, altitude and angle bisector by definition. Naming the marked condition prevents diagram guessing.
Use ratio splits after the equal case. Seeing 1:1 as one member of a general base-ratio rule deepens transfer.
The adult’s most useful question is not “Why were you careless?” but “Which decision did the evidence require?” Ask the learner to point to the controlling word, value, relationship or observation. If the learner can explain that link, reduce support. If the learner cannot, model one example and then change the surface details for the next attempt.
Frequently asked questions
What is a triangle median?
A segment joining a vertex to the midpoint of the opposite side.
Why are the two areas equal?
The component triangles have equal bases on one line and share the same perpendicular height.
Must the triangle be isosceles?
No. The result holds for every triangle.
Is a median always an altitude?
No. It is perpendicular only in special cases.
Are the two triangles congruent?
Not necessarily; equal area does not prove congruence.
What if the base split is 2:3?
The component areas are also in the ratio 2:3.
Next routes
Return to the complete volume directory to locate the series, use the original PSLE Learning Guide for the foundation route, and continue through PSLE Mathematics for verified subject guidance. The next useful task is a mixed set in which this idea is sometimes relevant and sometimes irrelevant; that is the point at which a learner must choose, not merely copy.