IB MYP Mathematics: Geometry, area, volume and circles (MYP Mathematics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set tests the geometry strand of the spatial reasoning branch of the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022. Questions run from MYP 1-3 skills through the standard and extended MYP 4-5 content, and enrichment items are labelled. MYP has no SL/HL split. Round to 3 significant figures unless told otherwise.

Revise the content in the study guide and the revision notes. The course hub is IB MYP Mathematics and the printable checklist lists each skill.

Questions

1. Each exterior angle of a regular polygon is 18°. Find (a) the number of sides, (b) each interior angle, (c) the sum of the interior angles. [3]

2. Two parallel lines are crossed by a transversal. A pair of alternate angles measure (4x − 15)° and (2x + 39)°. Find x, the size of each alternate angle, and the size of the co-interior angle to one of them. [4]

3. A garden bed is a rectangle 9 m long and 4 m wide with a semicircle of diameter 4 m attached to one short end. Find (a) its area and (b) the length of its outer boundary. [5]

4. (a) Convert 3.2 m² to cm². [1] (b) Convert 450 000 cm³ to m³. [1] (c) (Extended) A tank measures 1.5 m by 0.8 m by 0.6 m. Find its capacity in litres. [2]

5. A tin is a cylinder of radius 3.5 cm and height 12 cm.

(a) Find its volume. [2] (b) A paper label covers the curved surface exactly. Find the area of the label. [2] (c) The label is drawn as a net. Find the length of the rectangle’s edge that wraps around the tin. [1]

6. A cone has base radius 9 cm and slant height 15 cm.

(a) Find its vertical height. [2] (b) Find its volume. [2] (c) Find the area of its curved surface. [1]

7. A candle holder is a cube of side 6 cm with a square-based pyramid of height 8 cm fixed on top. The pyramid’s base is the top face of the cube.

(a) Find the total volume. [2] (b) Find the exposed surface area, including the underside of the cube. [4]

8. A circle has radius 8 cm. A sector has an angle of 60° at the centre. The chord joining the ends of the bounding arc forms a segment.

(a) Calculate the length of the arc. [2] (b) Calculate the sector’s area. [2] (c) The triangle formed by the two radii and the chord is equilateral. Find the area of the segment. [2] (d) Find the perimeter of the segment. [1]

9. (a) A chord of length 30 cm is in a circle of radius 17 cm. Find the distance from the centre to the chord. [3] (b) A tangent PT of length 40 cm touches a circle of radius 9 cm at T. The centre is O. Find OP. [3]

10. A, B, C and D lie in order on a circle with centre O. Angle AOC is 140°, where this angle stands on the minor arc AC and D is on that arc. B is on the major arc.

(a) Find angle ABC, giving a reason. [2] (b) Find angle ADC and state the rule you used. [2] (c) A tangent is drawn at A. Find the angle between this tangent and chord AC on the side of D, using the alternate segment theorem. [3]

11. (Enrichment) (a) A sector has radius 9 cm and angle π/3 radians. Find the arc length and the sector area. [2] (b) A square is inscribed in a circle of radius 5 cm. Find the side of the square and the percentage of the circle it fills. [3]

12. (Enrichment) A fractal curve starts as an equilateral triangle of perimeter 18 cm. At each stage every side is replaced by four sides, each one third as long.

(a) Find the perimeter at stages 1, 2 and 3. [2] (b) Find the first stage at which the perimeter exceeds 100 cm. [2]

Answers

1. (a) 360 ÷ 18 = 20 sides [1] (b) Interior = 180 − 18 = 162° [1] (c) (20 − 2) × 180 = 3240° [1] Examiner insight: Each part is a separate line of working. Writing only the final sum, without showing how you found n, leaves nothing to credit if the value of n is wrong.

2. Alternate angles are equal: 4x − 15 = 2x + 39 [1] 2x = 54, so x = 27 [1] Angle = 4(27) − 15 = 93° [1] Co-interior angle = 180 − 93 = 87° [1] Examiner insight: Name the rule (alternate angles, parallel lines) next to the equation. Mixing up alternate and co-interior would give the equation 6x + 24 = 180, a different value of x, so check the answer against the diagram.

3. (a) Rectangle 9 × 4 = 36 [1]; semicircle ½ × π × 2² = 2π = 6.28 [1]; area = 42.3 m² [1] (b) Boundary = 9 + 9 + 4 + arc π × 2 [1] = 22 + 6.28 = 28.3 m [1] Examiner insight: The boundary leaves out the diameter where the semicircle joins the rectangle. Including 4 m for that join is the usual slip on compound perimeters.

4. (a) 3.2 × 10 000 = 32 000 cm² [1] (b) 450 000 ÷ 1 000 000 = 0.45 m³ [1] (c) Volume = 1.5 × 0.8 × 0.6 = 0.72 m³ [1]; capacity = 0.72 × 1000 = 720 L [1] Examiner insight: Using the length factor (100) for area, or 100 for volume, is the typical error. Write the factor you apply, for example 100² = 10 000.

5. (a) V = π × 3.5² × 12 [1] = 462 cm³ [1] (b) Curved area = 2π × 3.5 × 12 [1] = 264 cm² [1] (c) Circumference = 2π × 3.5 = 22.0 cm [1] Examiner insight: Part (b) uses 2πrh, not the total surface area. Reading what the label covers tells you which faces to include.

6. (a) h² = 15² − 9² = 144 [1]; h = 12 cm [1] (b) V = ⅓ × π × 81 × 12 [1] = 324π = 1020 cm³ [1] (c) Curved area = π × 9 × 15 = 135π = 424 cm² [1] Examiner insight: Volume uses vertical height and curved area uses slant height. Using 15 in the volume formula is a common slip, so label which length you are using.

7. (a) Cube 6³ = 216 [1]; pyramid ⅓ × 36 × 8 = 96; total 312 cm³ [1] (b) Pyramid face height = √(8² + 3²) = √73 = 8.544 [1] Four triangles: 4 × ½ × 6 × 8.544 = 102.5 [1] Cube exposed: four sides 4 × 36 = 144 plus underside 36 = 180 [1] Total = 180 + 102.5 = 283 cm² [1] Examiner insight: The top face of the cube is joined to the pyramid, so it is hidden. A good solution says which faces are counted before adding them.

8. (a) Arc = (60/360) × 2π × 8 [1] = 8π/3 = 8.38 cm [1] (b) Sector = (60/360) × π × 64 [1] = 33.5 cm² [1] (c) Triangle = ¼ × √3 × 64 = 16√3 = 27.7 [1]; segment = 33.5 − 27.7 = 5.80 cm² [1] (d) Perimeter = arc + chord = 8.38 + 8 = 16.4 cm [1] Examiner insight: A segment is sector minus triangle, not sector minus a rectangle or a single radius. Keep unrounded values until the final subtraction, since the two areas are close in size.

9. (a) Half the chord = 15 cm, as the perpendicular from the centre bisects it [1] d² = 17² − 15² = 64 [1]; d = 8 cm [1] (b) The radius OT is perpendicular to the tangent PT, so triangle OTP is right-angled at T [1] OP² = 9² + 40² = 1681 [1]; OP = 41 cm [1] Examiner insight: State the theorem that justifies the right angle. Without it, the use of Pythagoras is not supported by a geometric reason.

10. (a) Angle at circumference = ½ × 140 = 70°, since the central angle on arc AC is double the angle at B on that arc [2] (b) Opposite angles of a cyclic quadrilateral add to 180°, so ADC = 180 − 70 = 110° [2] (c) Triangle OAC is isosceles, so angle OAC = (180 − 140) ÷ 2 = 20° [1] The tangent at A is perpendicular to OA, so the angle is 90 − 20 = 70° [1] This matches angle ABC in the alternate segment, which confirms the result [1] Examiner insight: Each answer needs its reason, not only the number. Check which arc your angle stands on before using the centre theorem: B on the major arc gives half of 140°, not half of the reflex angle.

11. (a) Arc = 9 × π/3 = 3π = 9.42 cm; sector = ½ × 81 × π/3 = 13.5π = 42.4 cm² [2] (b) The square’s diagonal is the diameter, 10 cm. Side = 10 ÷ √2 = 7.07 cm [1]; area = 50 cm² [1] Circle area = 25π = 78.5; fraction = 50 ÷ 78.5 = 63.7% [1] Examiner insight: In radians the angle must stay in radians in the formulas rθ and ½r²θ. Mixing it with the degree formula gives a wrong answer.

12. (a) Each stage multiplies the perimeter by 4/3: stage 1 = 24 cm, stage 2 = 32 cm, stage 3 = 42.7 cm [2] (b) Stage 5 = 75.9, stage 6 = 18 × (4/3)⁶ = 101.1 [1]; so stage 6 [1] Examiner insight: Show the multiplying factor and the repeated pattern. A list of stages with no stated rule makes it hard to see your reasoning.

Where marks are usually lost

• Using diameter where a formula needs the radius.
• Including the join between two shapes in a compound perimeter.
• Counting hidden faces in the surface area of a compound solid.
• Converting area or volume with the length factor.
• Using slant height for volume or vertical height for curved surface area.
• Giving an angle with no theorem named.
• Mixing alternate and co-interior angle rules.
• Rounding in the middle of a multi-step solution.
• Omitting units, or writing cm² for a volume.

Next steps

• Review the revision notes for formulas and methods.
• Reread the study guide for the worked examples.
• Check the IB MYP Mathematics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

IB Mathematics guide (for use from September 2020/January 2021), MYP framework, spatial reasoning: Geometry. International Baccalaureate Organization, first assessment 2022.