How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit rather than leaving it all to the end. No task is longer than one lesson; D2 has a part that may be finished at home.
Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: exact values and surds | A | After lesson 3 | 20 minutes | Individual, written, in class. 16 marks. |
| A2: End-of-unit test | A | End of unit, after lesson 7 | 50 minutes | Individual, written, in class, one lesson. 29 marks in three sections. |
| B1: Polygons inside a circle | B | Mid-unit, after lesson 4 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: Mali's proof | C | After lesson 5 | 35 minutes | Individual, written, in class or as homework. One page of words and diagrams. |
| D1: Survey boat and the fuel | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class. A written solution with explanations and a short note to the harbour office. |
| D2: Turf and edging for a flower bed | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class, one lesson; question 6 may be finished at home. A written solution with explanations and a short note to the caretaker. |
D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts: D1 is a journey and bearings (lesson 1), D2 is a circular segment (lesson 4). Set one, not both.
The unit's inquiry
- Key concept. Relationships: lengths, angles and areas in a circle or a right-angled triangle are tied together, and one measurement fixes the rest.
- Related concepts. Measurement, Justification, Space
- Global context. Orientation in space and time: finding where things are, how high, how far and which way, from a few measurements.
- Statement of inquiry. Relationships between lengths and angles in right-angled triangles and circles let us find what we cannot measure directly, and prove that what we find is always true.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: exact values and surds
Why set it. A short check of lesson 3 while there is still time to reteach: exact values of sine, cosine and tangent at 30°, 45° and 60° in right-angled triangles, and surd arithmetic. Questions 6 and 7 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 20 minutes. Format: Individual, written, in class. 16 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 20 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
No calculator: every answer is exact, so leave surds as surds. Show your working: a correct method earns credit even if the final answer slips. Questions 6 and 7 are meant to make you think, so try something rather than leaving them blank.
Part 1. Exact values
- Find the exact value of sin 60° cos 30° − sin 30° cos 60°. (2 marks)
- A right-angled triangle has a hypotenuse of 18 cm and an angle of 30°. Find the exact lengths of the other two sides. (2 marks)
- Write 4 ÷ (√3 + 1) with a rational denominator, in its simplest form. (2 marks)
Part 2. Special triangles
- A right-angled isosceles triangle has a hypotenuse of 8 cm. Find the exact length of each short side and the exact area. (2 marks)
- An equilateral triangle has sides of 8 cm. Find its exact height and its exact area. (3 marks)
Part 3. Think it through
- Noor says: “tan 30° + tan 30° = tan 60°, because 30 + 30 = 60.” Use exact values to decide whether she is right. (2 marks)
- A rectangle has a shorter side of 5 cm. Its diagonal makes an angle of 30° with the longer side. Find the exact length of the diagonal and the exact area of the rectangle. (3 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You give some of the exact values correctly, for example the sides in question 2, but struggle to carry them through a calculation. |
| 3–4 | You answer most of questions 1 to 5 correctly using exact values, with some slips in surds or in rationalizing. |
| 5–6 | You answer questions 1 to 5 correctly with exact values and clear working, and make a good start on question 6 or 7. |
| 7–8 | You answer questions 1 to 5 correctly and solve question 6 or 7 with reasoning that shows your answer works. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | 1 ÷ 2. (√3 ÷ 2)(√3 ÷ 2) − (1 ÷ 2)(1 ÷ 2) = 3 ÷ 4 − 1 ÷ 4. 1 mark for the four exact values, 1 for the answer. Accept the observation that it is sin(60° − 30°) = sin 30°. |
| 2 | 9 cm and 9√3 cm. The side opposite 30° is 18 sin 30° = 9; the other is 18 cos 30° = 9√3. 1 mark each. 15.6 earns the mark only if the exact form is also given. |
| 3 | 2√3 − 2. 4(√3 − 1) ÷ ((√3 + 1)(√3 − 1)) = 4(√3 − 1) ÷ 2. 1 mark for multiplying by the conjugate, 1 for the simplified answer. (4√3 − 4) ÷ 2 left unsimplified earns 1; 4√3 − 4, with the denominator 2 lost, earns only the first mark. |
| 4 | 4√2 cm each; area 16 cm². 8 cos 45° = 8 ÷ √2 = 4√2, and the area is ½ × 4√2 × 4√2. 1 mark for the side, 1 for the area. 5.66 alone earns nothing for exactness. |
| 5 | Height 4√3 cm; area 16√3 cm². Half the base is 4, so the height is 8 sin 60° (or √(64 − 16)). 2 marks for the height, 1 for ½ × 8 × 4√3. A height of 8 × √3 ÷ 2 written as 4√3 is the same thing; a height of 4 × 2 = 8 is not. |
| 6 | Noor is wrong. tan 30° + tan 30° = 2 ÷ √3 = 2√3 ÷ 3, which is about 1.15, but tan 60° = √3 = 3√3 ÷ 3, which is about 1.73. 1 mark for the sum, 1 for the comparison. Saying “the tangent of a sum is not the sum of the tangents” without values earns 1. |
| 7 | Diagonal 10 cm; area 25√3 cm². The shorter side is opposite 30°, so the diagonal is 5 ÷ sin 30° = 10 (1 mark). The longer side is 5 ÷ tan 30° = 5√3 (1 mark) and the area is 5 × 5√3 (1 mark). A student who takes 5 as the side next to the 30° angle is drawing a different rectangle: look for the diagram. |
| Marks to a level | Total 16. 1 to 4 marks: 1–2. 5 to 8: 3–4. 9 to 12: 5–6. 13 to 16, with question 6 or 7 fully correct: 7–8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3–4 profile | Question 1 right (2); question 2 right (2); question 3 written as 4 ÷ √3 + 4 (0); question 4 gives 4√2 but the area as 8√2 (1); question 5 height 4√3, area 8 × 4 ÷ 2 = 16 (1); nothing on 6 or 7. 6 marks, and the profile matches 3–4: some exact values, slips with surds, nothing unfamiliar. |
| Level 7–8 answer to question 7 | “Draw the rectangle, shorter side 5 opposite the 30° angle. sin 30° = 5 ÷ diagonal, so the diagonal = 5 ÷ (1 ÷ 2) = 10. The longer side is next to the angle: tan 30° = 5 ÷ long, so long = 5 ÷ (1 ÷ √3) = 5√3. Area = 5 × 5√3 = 25√3 cm². Check: 25 + 75 = 100 = 102.” |
| Level 7–8 answer to question 6 | “tan 30° = 1 ÷ √3, so tan 30° + tan 30° = 2 ÷ √3 = 2√3 ÷ 3. But tan 60° = √3 = 3√3 ÷ 3, which is bigger. Noor is wrong: the tangent does not add when the angles add.” |
Differentiation
A way in for anyone stuck
- Give the two special triangles from lesson 3 as a reference card, with their sides labelled but not the values of the ratios.
- Let students sketch each triangle before using it; marks are for the exact value, not for the sketch.
- Read question 7 aloud on request and suggest a labelled diagram first. Reading is not what is being assessed.
Stretch
- Find the exact angle between the diagonal and the shorter side in question 7, and use it to check the area. (60°: 10 sin 60° × 10 cos 60° = 25√3.)
- Find the exact value of sin 15° cos 15°. (1 ÷ 4: it is half of sin 30°.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| exact value | an answer written with surds or fractions, not rounded |
| rationalize | remove the surd from the denominator of a fraction |
| conjugate | the same two terms with the sign between them changed |
| hypotenuse | the longest side of a right-angled triangle, opposite the right angle |
| equilateral | with all three sides the same length |
Sentence starters for task A1:
- The side opposite 30° is … because …
- To remove the surd I multiply by …
- Noor is wrong because …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7–8 comes from. Use the parallel version on this page if the sheet has been seen.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, written, in class, one lesson. 29 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. Give lengths to 3 significant figures and angles to 1 decimal place unless the question asks for an exact value.
Section 1. Familiar questions
- A tower stands on level ground. Looking up at its top from a point A, the elevation is 22°. From a point B, 60 m nearer along the same line, the elevation is 38°. Find the height of the tower. (3 marks)
- A sector of a circle has radius 9 cm and angle 140°. Find the arc length, the area and the perimeter of the sector. (3 marks)
- O is the centre of a circle and A, B, C and D are on it. Angle AOB is 118°. C is on the major arc AB and D is on the minor arc AB. Find angle ACB and angle ADB, giving a reason for each. (3 marks)
Section 2. More complex questions
- A cuboid has a base 7 cm by 24 cm and a height of 10 cm. Find the angle between its space diagonal and the base. (3 marks)
- Find the exact value of (sin 60° − cos 60°)2, giving your answer in the form a + b√3 with a and b fractions. (3 marks)
- Chords AB and CD of a circle cross at P, with AP = 6, PB = 4 and CP = 3. Find PD. Separately, a tangent from an outside point T has length 15 and a secant from T meets the circle at A then B, with TA = 9. Find AB. (4 marks)
- A chord subtends an angle of 90° at the centre of a circle of radius 10 cm. Find the exact area of the minor segment cut off by the chord, then give it to 3 significant figures. (3 marks)
Section 3. Unfamiliar questions
- A circle has radius 25 cm. Two parallel chords of lengths 48 cm and 14 cm lie on opposite sides of the centre. How far apart are they? Is there another possible distance if the chords are allowed on the same side? (4 marks)
- A ladder 5 m long leans against a vertical wall on level ground, and its foot slides slowly away from the wall. Its midpoint M is marked. Show that M stays the same distance from the corner where the wall meets the ground, and say what that distance is. (3 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some Section 1 questions correctly, for example one length or one angle, with the method for finding the others unclear. |
| 3–4 | You answer most of Section 1 correctly and make progress in Section 2, for example finding the space diagonal or an arc. |
| 5–6 | You answer Section 1 and most of Section 2 correctly, with clear methods and exact values where asked, and make a start in Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve most of Section 3, explaining why your method works. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | 50.2 m. With h the height, B is h ÷ tan 38° from the tower and A is h ÷ tan 22°, so 60 = h(1 ÷ tan 22° − 1 ÷ tan 38°) and h = 60 ÷ 1.195. 1 mark for the two distances, 1 for the equation for 60, 1 for 50.2. Using 60 ÷ (tan 38° − tan 22°) = 159.0 is the usual slip. |
| 2 | Arc 22.0 cm, area 99.0 cm², perimeter 40.0 cm. Arc = (140 ÷ 360) × 2π × 9 = 7π = 21.99; area = (140 ÷ 360) × π × 81 = 31.5π = 98.96; perimeter = 21.99 + 18 = 39.99. 1 mark each. The perimeter without the two radii (22.0) is the usual slip. |
| 3 | ACB = 59°, ADB = 121°. The angle at the circumference is half the angle at the centre (118 ÷ 2), and opposite angles of a cyclic quadrilateral add to 180° (180 − 59). 1 mark each for the angles, 1 for both reasons. |
| 4 | 21.8°. The diagonal of the base is √(49 + 576) = 25, so tan(angle) = 10 ÷ 25 and the angle is 21.80°. 1 mark for 25, 1 for tan = 10 ÷ 25, 1 for the angle. Using the space diagonal 26.93 and sin(angle) = 10 ÷ 26.93 also gives 21.8°: accept. |
| 5 | 1 − √3 ÷ 2, that is a = 1 and b = −1 ÷ 2. Expanding: 3 ÷ 4 − 2 × (√3 ÷ 2)(1 ÷ 2) + 1 ÷ 4. 1 mark for the exact values, 1 for the cross term, 1 for the answer. An answer of 1 is the slip of dropping the cross term. |
| 6 | PD = 8; AB = 16. AP × PB = CP × PD gives 24 = 3 PD. TA × TB = 152 gives TB = 225 ÷ 9 = 25, so AB = 25 − 9 = 16. 2 marks each part (1 for the correct product statement). Giving TB = 25 as AB loses the last mark. |
| 7 | 25π − 50 cm² = 28.5 cm². Sector = (1 ÷ 4) × π × 100 = 25π (78.54); triangle = ½ × 10 × 10 = 50. 1 mark each for the sector and the triangle, 1 for the subtraction and the exact form. |
| 8 | 31 cm apart; or 17 cm apart on the same side. Half the chords are 24 and 7. The distances from the centre are √(625 − 576) = 7 and √(625 − 49) = 24. 1 mark for each distance from the centre, 1 for 7 + 24 = 31, 1 for 24 − 7 = 17. A student who gives only 31 has earned 3 of the 4 marks. |
| 9 | M is always 2.5 m from the corner. The wall, ground and ladder make a right-angled triangle, and a right angle is the angle in a semicircle: the corner lies on the circle whose diameter is the ladder, so M is the centre and the distance to the corner is a radius, 5 ÷ 2. 1 mark for the right angle in a semicircle idea, 1 for M as the centre, 1 for 2.5 m. A student who tries two or three positions and sees 2.5 m each time earns 2 for the evidence but not the reason. |
| Marks to a level | Total 29. 1 to 6 marks: 1–2. 7 to 13: 3–4. 14 to 22: 5–6. 23 to 29, with at least 5 of Section 3's 7 marks: 7–8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 5–6 profile | Section 1 all correct (9 marks). Section 2: space diagonal 26.93 and 21.8° (3); question 5 expanded with the cross term dropped (1); question 6 right (4); question 7 right (3): 11 marks. Section 3: question 8, with 7 and 24 found and 31 stated (3); question 9 blank. 23 marks, but Section 3 gives 3 of 7, below 5, so the level is 5–6. |
| Level 7–8 answer to question 9 | “The wall and the ground are at right angles, so the ladder is the hypotenuse of a right-angled triangle with its right angle at the corner. The angle in a semicircle is 90°, so the corner is on a circle whose diameter is the ladder. The centre of that circle is the midpoint M, and its radius is 5 ÷ 2 = 2.5 m. So M is always 2.5 m from the corner. Check: foot 3 m out, top 4 m up, M at (1.5, 2), and the distance from the corner is √(2.25 + 4) = 2.5.” |
Differentiation
A way in for anyone stuck
- Read Sections 2 and 3 aloud on request: reading is not what is being assessed.
- Offer a sketch template for question 1 (two right-angled triangles on one line) and a cuboid outline for question 4.
- For question 9, give a ladder drawn at one position and ask students to mark the midpoint, then draw a second position themselves.
Stretch
- In question 8, could two parallel chords of lengths 48 cm and 14 cm be 35 cm apart? Explain. (No: the greatest separation is 7 + 24 = 31 cm.)
- In question 9, what shape does the midpoint trace as the ladder slides from upright to flat? (A quarter of a circle of radius 2.5 m.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| angle of elevation | the angle above the horizontal when you look up at something |
| chord | a straight line joining two points on a circle |
| tangent | a straight line that touches a circle at exactly one point |
| segment | the part of a circle between a chord and its arc |
| space diagonal | the longest straight line inside a box, from one corner to the opposite corner |
Sentence starters for task A2:
- I drew two right-angled triangles, and …
- The angle at the centre is …, so the angle at the circumference is …
- The ladder is a diameter because …
Task B1: Polygons inside a circle
Why set it. A pattern hunt built on lesson 4: the area of a triangle from two sides and the angle between them (used there for segments), and the idea of a sector. The sheet never names the pattern: students choose how to measure and compare, find the pattern, describe it as a rule, test it and justify it.
When: Mid-unit, after lesson 4. Time: 50 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 5, support you give (a table outline, a reminder of the area of a triangle from two sides and the angle between them) is allowed at 1–2 and 3–4: note it on the student's work.
When: Mid-unit, after lesson 4. Time: 50 minutes. Calculator: scientific calculator. Criterion: B, investigating patterns.
A regular polygon with n sides is drawn so that all of its corners lie on a circle of radius 1 cm. Joining each corner to the centre cuts the polygon into n equal triangles. This task is about what happens to the polygon as n gets larger.
Part 1. Collect results
- Find a formula for the area of the polygon, and one for its perimeter, in terms of n. Each triangle has two sides of 1 cm. Show how you got them.
- Use your formulas to find the area and the perimeter for n = 3, 4, 6, 8 and 12. Record them, to 4 decimal places, in a table you design yourself.
Part 2. Describe what you find
- Compare each polygon with the circle it sits in. Choose your own way to measure how different they are, and add that measure to your table.
- Describe what you notice, as a rule another student could use. Say what you would expect for n = 24.
Part 3. Test your rule
- Find the area and the perimeter for n = 24 and n = 48. Did your rule predict them? Show how close you were.
- Use your rule, not the formulas, to predict the area for n = 96. Then check with the formula.
Part 4. Explain
- Is the area of the polygon ever larger than the area of the circle? Explain, using the diagram, why or why not.
- Does your rule work for the perimeter as well? Give evidence, and say how you decided what to test.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you work out some areas or perimeters, record them, and say something you notice about the numbers. |
| 3–4 | With help, you complete the table for the area and the perimeter and describe a pattern that fits most of it. |
| 5–6 | You organize your results in a clear table, describe a rule that fits the area and the perimeter, and use it to predict n = 24 and check it. |
| 7–8 | You organize your results, describe a general rule that fits all your findings, use it to make an accurate prediction for n = 96, and justify that the rule holds for the perimeter and why the area stays below the circle. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | For n = 6, 12, 24, 48 the area falls short of the circle's by 0.5435, 0.1416, 0.0358, 0.0090: each is about a quarter of the one before (ratios 3.84, 3.96, 3.99). The perimeter falls short of 2π by 0.2832, 0.0715, 0.0179, 0.0045 with the same ratios (3.96, 3.99, 4.00). Both rise with n but stay below. Students may use other measures (a percentage, or area ÷ π): credit any that reveal the pattern. |
| 1 | A = (n ÷ 2) sin(360° ÷ n), from n triangles each of area ½ × 1 × 1 × sin(360° ÷ n). P = 2n sin(180° ÷ n), from n chords of length 2 sin(180° ÷ n), by splitting each triangle in two. |
| 2 | n = 3: A 1.2990, P 5.1962. n = 4: 2.0000, 5.6569. n = 6: 2.5981, 6.0000. n = 8: 2.8284, 6.1229. n = 12: 3.0000, 6.2117. |
| 3 | Any measure that compares with the circle (area 3.1416, circumference 6.2832): the difference, a percentage or a ratio. Credit the measure only if it is used consistently. |
| 4 | A description of the numbers rising and of the differences shrinking. A rule for the area only is evidence for 3–4; a rule precise enough to predict n = 24 is 5–6. |
| 5 | n = 24: A 3.1058, P 6.2653. n = 48: A 3.1326, P 6.2787. The differences from the circle are 0.0358 and 0.0090 (area), 0.0179 and 0.0045 (perimeter). |
| 6 | n = 96: A = 3.1394 (3.1416 − 0.0090 ÷ 4 = 3.1394; the formula gives 3.1394). A student who has the quarter pattern can predict this from n = 48 without the formula. |
| 7 | No. The polygon lies entirely inside the circle: its corners are on the circle and its sides are chords, so every point of the polygon is within the circle. At 7–8 expect this reason, or the comparison of each triangle with the sector of the same angle. |
| 8 | Yes. Test the perimeter beside the area: the ratios are about 4 again. Credit evidence from a second value of n and a statement of why that case was chosen. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3–4 | “n = 6: 2.5981 and 6.0000. n = 12: 3.0000 and 6.2117. The areas get bigger and go up by less each time. It will be about 3.1 for n = 24.” Both tables right and a description that fits, but no measure against the circle and no precise rule or test. |
| Level 7–8 | “A = (n ÷ 2) sin(360° ÷ n). The circle's area is 3.1416. Differences: n = 6, 0.5435; n = 12, 0.1416; n = 24, 0.0358; n = 48, 0.0090. Each is about a quarter of the one before (3.84, 3.96, 3.99). Prediction for n = 96: 0.0090 ÷ 4 = 0.0022, so A = 3.1394; formula 3.1394. The perimeter differences from 6.2832 are 0.2832, 0.0715, 0.0179, 0.0045: also about a quarter each time. The polygon is inside the circle, so the area is always less.” |
Differentiation
A way in for anyone stuck
- Give a table outline with columns n, area and perimeter and let the student add the comparison columns.
- Remind students of the area of a triangle from two sides and the angle between them, and of how a triangle from the centre splits in two.
- Prompt with a question, never the rule: “What is the circle’s own area?”
Stretch
- Do the same with a polygon drawn outside the circle, touching it at the midpoint of each side. What do its area and perimeter do? (They fall towards the circle's values from above.)
- Why is the difference about a quarter when n doubles? (For small angles sin θ is close to θ − θ3 ÷ 6, and the error depends on 1 ÷ n2.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| regular polygon | a flat shape with straight sides of equal length and equal angles |
| corner | a point where two sides meet |
| perimeter | the total length around a shape |
| predict | say what will happen before you check |
| rule | a sentence that tells you what happens every time |
Sentence starters for task B1:
- The area of the polygon is … and the circle's is …
- My rule is: …
- I tested my rule on … and it worked / did not work because …
Task C1: Mali's proof
Why set it. Criterion C on its own: notation, a diagram and a line of reasoning a classmate can follow, built around a circular proof and a confusion between opposite and adjacent angles from lesson 5.
When: After lesson 5. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words and diagrams.
The student sheet, with its task-specific clarification
When: After lesson 5. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Mali missed lesson 5. She wrote her own working for two questions and brings it to you.
Mali’s working
(a) Prove that the angle at the centre of a circle is twice the angle at the circumference. Angle ACB = x. Angle AOB must be double angle ACB, so angle AOB = 2x. Triangle AOB is isosceles, so its other two angles are (180 − 2x) ÷ 2 = 90 − x. This is proved.
(b) PQRS is a cyclic quadrilateral. Angle P = 70°. Find angle Q. The angles of a cyclic quadrilateral add to 180°, so angle Q = 110°.
Write a one-page explanation for Mali. She should be able to follow it on her own, without asking anyone.
Part 1. Put it right
- Find what is wrong with each of Mali’s answers. For each, say what she did, why it does not work and what she should have done.
Part 2. Teach the idea
- Write a correct proof for the case where the centre O is inside triangle ABC. Draw and label a diagram, and give a reason for every line.
- Say what changes in your diagram and your proof when O is outside triangle ABC.
Part 3. Check your explanation
- Write one question about a cyclic quadrilateral that would catch out a student who makes Mali’s mistake in (b). Give its correct answer.
- Read your page as if you were Mali. Check that every angle you name has three letters or a clear label, that each step has its reason, and that you use the words centre, circumference, chord, isosceles and opposite correctly.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your explanation can be followed, but a step or a word is missing, and the diagram or the angle names are hard to follow. |
| 3–4 | You explain both of Mali’s answers with some correct words and angle names; your steps are in a sensible order, so Mali could follow most of it. |
| 5–6 | Each step of your proof has a reason and a clear angle name; your diagram matches your words, and Mali could follow all of it with little effort. |
| 7–8 | Your proof reads as a chain of reasons with no gaps, the diagram is labelled to match every line, the notation is always correct, and the page is ordered so Mali could learn from it alone. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistake in (a) | She uses the result she was asked to prove (“angle AOB must be double angle ACB”), so nothing has been proved. A correct proof joins C to O and extends it to D: OA = OC and OB = OC (radii) make two isosceles triangles, and each exterior angle at O is the sum of the two equal base angles, so angle AOD = 2 × angle ACO and angle BOD = 2 × angle BCO. Adding, AOB = 2 × ACB. |
| The mistake in (b) | It is the opposite angles that add to 180°: P and R, so angle R = 110°. Angle Q is not fixed by P alone (Q + S = 180°). Q = 110° happens to be true in some quadrilaterals, for example when arc PQ = 80°, QR = 60°, RS = 80° and SP = 140°, where Q = R. |
| Case with O outside | The line CO extended still gives two isosceles triangles, but now one angle at C is the difference of the other two: AOB = AOD − BOD = 2 ACO − 2 BCO = 2 ACB. Credit the subtraction and a diagram. |
| Own question (Part 3) | Any question where the opposite angle is asked for and the answer differs from the adjacent one. Example: “PQRS is cyclic. Angle P = 70° and angle S = 95°. Find angles Q and R.” (R = 110°, Q = 85°; a student making Mali’s mistake gives Q = 110°.) |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation, a labelled diagram and a reason for every line decide between 5–6 and 7–8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Mali has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3–4 extract | “Mali, in (a) you used the answer to prove the answer. Draw the line from C through O. The triangles have two same sides so the angles are same. Then angle AOB is two times. In (b) the opposite angles are 180, so R is 110 not Q.” Both mistakes are found, but the proof has no labelled diagram, one reason is missing and two angles are not named. |
| Level 7–8 extract | “Join C to O and extend it to meet the circle at D. OA = OC (radii), so triangle OAC is isosceles and angle OAC = angle OCA = p. The exterior angle AOD of triangle OAC is the sum of the two angles opposite it, so angle AOD = 2p. In the same way angle BOD = 2q, where q = angle OCB. Adding, angle AOB = 2p + 2q = 2(p + q) = 2 × angle ACB.” Each line has its reason, the diagram is labelled with the same letters, and the equals signs join equal values. |
Differentiation
A way in for anyone stuck
- Give the diagram of lesson 5 with the letters and the line CD already drawn, so the student writes the reasoning rather than the figure.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Mali wrote …. This does not work because …. She should write ….”
Stretch
- Is Mali’s answer to (b) ever right, and for which cyclic quadrilaterals? (When Q = R, that is when PQ = RS: an isosceles trapezium with QR parallel to PS.)
- Write a second version of the page for a student who has only seen the case where O is inside. What changes?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| prove | show that something is always true, with a reason for each step |
| isosceles | with two equal sides and two equal angles |
| opposite angles | angles at corners that are not next to each other |
| radius | a straight line from the centre to the circle |
| exterior angle | the angle between a side of a triangle and the extension of the next side |
Sentence starters for task C1:
- Mali wrote …, but …
- This step is true because …
- When O is outside the triangle, …
Task D1: Survey boat and the fuel
Why set it. Journey components, distance and bearing, inside one real decision. An alternative to D2: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class. A written solution with explanations and a short note to the harbour office.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A survey boat leaves a harbour H and sails 12 km on a bearing of 060°. It then sails 9 km on a bearing of 150° to a buoy B. Its fuel tank held 32 litres when it left the harbour, and the boat uses 0.9 litres for each kilometre.
Part 1. Where is the boat?
- Draw the journey, and find how far east and how far north of the harbour the boat is when it reaches the buoy.
- Find the straight-line distance from the harbour to the buoy, and the bearing of the buoy from the harbour.
- State the bearing the boat would sail on, from the buoy, to go straight back to the harbour.
Part 2. Fuel
- The boat now sails straight back to the harbour. How many litres of fuel will the whole trip use, and will 32 litres be enough?
Part 3. Does it make sense?
- The 0.9 litres per kilometre is a rough figure: it could be anything from 0.85 to 0.95. Does your answer to question 4 still hold across that range?
- Write a short note to the harbour office saying what you recommend and why, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example how far east the boat is, but do not reach the distance or the fuel. |
| 3–4 | You find the distance and bearing back to the harbour and say whether 32 litres is enough, and whether your answer makes sense. |
| 5–6 | You choose suitable methods (components, Pythagoras, inverse tangent) for every part, reach correct answers, and discuss how the fuel range affects the decision. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, explain how the range in fuel use changes the decision and why, and judge whether your recommendation is sensible for a real boat. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the distance or the bearing. |
| 3–4 | Your working for the components and the bearing can be followed, with some correct notation, bearings in three figures and units. |
| 5–6 | Your working is clear and complete, with bearings in three figures and units (km, litres), and your note to the harbour office is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, bearings and units are always correct, and your note to the harbour office makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | East 14.9 km, north −1.8 km (1.8 km south). Leg 1: east 12 sin 60° = 10.39, north 12 cos 60° = 6.00. Leg 2: east 9 sin 150° = 4.50, north 9 cos 150° = −7.79. Totals 14.89 east and −1.79 north. 1 mark for each leg correctly resolved, 1 for the totals. |
| 2 | 15.0 km on a bearing of 097°. √(14.892 + 1.792) = 15.0 (the legs are at 90° to each other, 150 − 60, so it is also √(122 + 92)). The bearing is 90° + tan−1(1.79 ÷ 14.89) = 96.9°, which is 097° in three figures. 1 mark for the distance, 1 for the angle, 1 for the bearing. 083° comes from taking the angle north of east instead of south of east. |
| 3 | 277°. 097° + 180°. 1 mark. Repeating 097° is the usual slip. |
| 4 | 36 km, 32.4 litres: not enough, 0.4 litres short. The boat has already sailed 12 + 9 = 21 km; with 15 km to go it will have covered 36 km. 36 × 0.9 = 32.4, which is more than 32. 1 mark for the 36 km, 1 for 32.4 litres, 1 for the conclusion. |
| 5 | Not across the whole range. At 0.85 litres per km: 30.6 litres, enough with 1.4 litres to spare. At 0.95: 34.2 litres, 2.2 litres short. So the decision depends on the value: the boat might make it, but with no reserve it should not rely on it. Credit a statement of both ends with units. |
| 6 (criterion C) | The note gives the distance and bearing of the buoy, the fuel needed against the fuel available, and a recommendation (for example: carry spare fuel, or refuel at the buoy), with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3–4 (D) | “East: 10.4 + 4.5 = 14.9. North: 6 − 7.8 = −1.8. Distance 15 km. Bearing 097. Fuel 36 × 0.9 = 32.4, not enough.” Correct numbers, but no bearing back, no range tested and no reason that the decision depends on the fuel rate. |
| Level 7–8 (D and C) | Components shown in a table; 15.0 km also checked as √(144 + 81); 097° and 277°; “with 0.85 the boat needs 30.6 litres and arrives with 1.4 litres to spare, with 0.95 it needs 34.2 and is 2.2 litres short. I recommend taking a spare 5 litre can.” The note to the harbour office is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give a table with columns for distance, bearing, east and north for each leg.
- Offer a compass outline for the bearings, marking north, and the reminder that bearings are measured clockwise from north.
- For Part 3, ask: “What if the boat used a little more fuel for each kilometre?”
Stretch
- Show that the buoy is 15 km from the harbour whatever the bearing of the first leg, as long as the second leg is at 90° to the first. (The legs form a right-angled triangle.)
- How much fuel would a trip from the harbour straight to the buoy and straight back have used, and how much less is that? (30 km, 27.0 litres, 5.4 litres less.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| bearing | the direction, as an angle from north, measured clockwise, in three figures |
| components | the east part and the north part of a journey |
| buoy | a floating marker in the sea |
| litre | a unit for the volume of a liquid |
| recommend | say what someone should do, and why |
Sentence starters for task D1:
- From the harbour the buoy is … km east and … km …
- The bearing back is … because …
- I recommend … because …
Task D2: Turf and edging for a flower bed
Why set it. Segment area, arc length and trigonometry inside one real decision, with a tolerance on a measurement. An alternative to D1: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class, one lesson; question 6 may be finished at home. A written solution with explanations and a short note to the caretaker.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A community centre has a circular flower bed with a radius of 6.0 m. A straight path will be laid across it along a chord of length 8.0 m, and the small segment between the chord and the circle will be covered in turf. The curved edge of the segment needs a strip of edging. Turf is sold in rolls that cover 1.5 m2 each, and edging comes in lengths of 2.5 m.
Part 1. The segment
- Draw the bed and the chord, and find the angle the chord makes at the centre of the circle.
- Find the area of the segment. Show the sector and the triangle you used.
- Find the length of the curved edge.
Part 2. Buying
- How many rolls of turf and how many lengths of edging must be bought? Explain how you decided each.
Part 3. Does it make sense?
- The chord was measured to the nearest 0.1 m. Between which values could it have been? Does the number of rolls and of lengths you chose still hold across that range?
- Write a short note to the caretaker with what to buy and how sure you are, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the angle at the centre, but do not reach the area or the numbers to buy. |
| 3–4 | You find the area of the segment and the edge length, work out what to buy, and say whether your answer makes sense. |
| 5–6 | You choose suitable methods (inverse sine, sector minus triangle, arc length) for every part, reach correct answers, and discuss whether the purchase still holds across the measuring range. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, explain how the measuring range affects the area and the edge, and why the purchase does or does not change. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you found the angle or the area. |
| 3–4 | Your working for the angle and the areas can be followed, with some correct notation and units. |
| 5–6 | Your working is clear and complete, usually with correct units (m, m²) and notation, and your note to the caretaker is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your note to the caretaker makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | 83.6°. Half the chord is 4.0, so sin(half-angle) = 4 ÷ 6 and the half-angle is 41.8°. 1 mark for the right-angled triangle idea, 1 for 83.6°. The cosine rule, cos = (36 + 36 − 64) ÷ 72, gives the same. 41.8° is the half-angle, not the angle at the centre. |
| 2 | 8.4 m². Sector = (83.62 ÷ 360) × π × 36 = 26.27; triangle = ½ × 36 × sin 83.62° = 17.89 (or ½ × 8 × √20); 26.27 − 17.89 = 8.38. 1 mark each for the sector and the triangle, 1 for the subtraction. |
| 3 | 8.8 m. (83.62 ÷ 360) × 2π × 6 = 8.76. 1 mark; 8.0 (the chord) is the usual slip. |
| 4 | 6 rolls and 4 lengths. 8.38 ÷ 1.5 = 5.6, so 6 rolls; 8.76 ÷ 2.5 = 3.5, so 4 lengths. 1 mark each. Both must be rounded up, not to the nearest whole number: part of a roll cannot be bought. |
| 5 | Yes: still 6 rolls and 4 lengths. The chord is between 7.95 and 8.05 m. The segment is then from 8.20 to 8.56 m², needing 5.5 to 5.7 rolls, and the edge from 8.69 to 8.82 m, needing 3.5 lengths. 1 mark for the range of the chord, 1 for the areas, 1 for the conclusion. |
| 6 (criterion C) | The note gives the numbers to buy with units, says that they hold across the measuring range, and says what is left over (about 0.6 m² of turf, 1.2 m of edging) so that a caretaker could use it. Credit a stated waste or risk. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3–4 (D) | “The angle is 83.6. Area of segment is 8.4. Edge is 8.8. 8.4 ÷ 1.5 = 5.6 so 6 rolls. 8.8 ÷ 2.5 = 3.5 so 4.” Correct numbers and purchase, but the chord's range is ignored, no sector or triangle is shown and units are missing. |
| Level 7–8 (D and C) | “The chord could be 7.95 to 8.05. At 7.95 the angle is 83.0° and the segment 8.2 m²; at 8.05 the angle is 84.3° and the segment 8.6 m². Both need 6 rolls (5.5 to 5.7). The edge is 8.7 to 8.8 m, so 4 lengths in both cases. Buy 6 rolls and 4 lengths; the 6 rolls leave about 0.6 m².” Every answer correct, units throughout, range tested at both ends. |
Differentiation
A way in for anyone stuck
- Give a labelled diagram of the bed with the chord, the centre and the two radii, so the student writes the working rather than the figure.
- For question 1, ask: “What right-angled triangle can you see if you draw the line from the centre to the middle of the chord?”
- Let students round the angle to the nearest degree and say how that changes the area.
Stretch
- What would the area of the segment be if the chord were a diameter of the bed? (18π = 56.5 m², half the circle.)
- Show that the segment with a 60° angle at the centre has area (π ÷ 6 − √3 ÷ 4) r2, and find it for r = 6. (36 × 0.0906 = 3.26 m².)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| flower bed | a piece of ground where flowers are planted |
| edging | a strip used along the edge of a bed |
| turf | grass with its soil, sold in rolls |
| caretaker | the person who looks after a building and its grounds |
Sentence starters for task D2:
- The angle at the centre is … because …
- The segment is the sector minus …
- I need to buy … because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page. Because the sheets are public, a student may have seen the A tasks: use the parallel A2 below when you need an unseen test.
To print a sheet, open it on the unit page, then print the page. A1 and C1 are done without a calculator (exact values, proof). B1, A2, D1 and D2 use a scientific calculator; a GDC may be used where students have one, and every number can be found without one.
Parallel A2 (same marks, same markscheme notes): Q1 elevations 30° and 45° from two points 40 m apart (54.6 m). Q2 sector of radius 12 cm and angle 75° (arc 5π = 15.7 cm; area 30π = 94.2 cm²; perimeter 39.7 cm). Q3 angle at the centre 136° (68°; 112°). Q4 cuboid with base 15 by 8 and height 6: angle between diagonal and base (19.4°). Q5 (sin 30° + cos 30°)2 (1 + √3 ÷ 2). Q6 AP = 8, PB = 3, CP = 4, so PD = 6; tangent length 10 and TA = 4, so TB = 25 and AB = 21. Q7 radius 12 and centre angle 60°, segment area 24π − 36√3 = 13.0 cm². Q8 radius 17 and parallel chords 30 and 16 (15 and 8 from the centre: 23 apart, or 7 on the same side). Q9 ladder 13 m (midpoint always 6.5 m from the corner). For A1, change the numbers the same way and rework the markscheme.
The Year 5 descriptor bands on this page are paraphrases written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.