Answer
Flashcards · J560 · Topic 10
Mensuration flashcards
28 cards on Topic 10, subtopics 10.01 to 10.05. Pick a subtopic or study them all.
Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.
Choose your cards
Session complete
Worth another look:
All 28 cards
For a quick read-through or printing. Cover the right-hand side to test yourself. Cards marked Higher are Higher tier content only.
10.01 Units and measurement
- Area and volume conversions from 1 m = 100 cm
- 1 m² = 100² = 10 000 cm². 1 m³ = 100³ = 1 000 000 cm³. 1 litre = 1000 cm³; 1 m³ = 1000 litres.
- Convert 4 h 45 min to hours, and 1.4 h to hours and minutes
- 45 ÷ 60 = 0.75, so 4.75 h. 0.4 × 60 = 24, so 1 h 24 min.
- Convert 8 m/s to km/h
- × 3.6: 8 × 3.6 = 28.8 km/h. (m/s to km/h is × 3.6; km/h to m/s is ÷ 3.6.)
- Density and pressure formulae and unit links
- Density = mass ÷ volume; pressure = force ÷ area. 1 g/cm³ = 1000 kg/m³. Pressure in N/m² and N/cm² differ by a factor of 10 000.
- A map has scale 1 : 2000. What real length is 8 cm on the map?
- 8 × 2000 = 16 000 cm = 160 m.
- Bearings and back bearings
- Measured from north, clockwise, with three figures (such as 047°). Back bearing: add or subtract 180°. A bearing of 305° from X to Y gives 125° from Y to X.
10.02 Perimeter calculations
- Circumference of a circle
- C = πd = 2πr
- Arc length for angle θ and radius r
- Arc length = (θ/360) × 2πr
- Perimeter of a sector and of a semicircle
- Sector: arc length + 2r. Semicircle: πr + 2r.
- Perimeter of a rectangle with a corner notch cut out
- The same as the outer rectangle: the two inner edges of the notch replace the two missing outer lengths.
10.03 Area calculations
- Area of a triangle, parallelogram and trapezium
- Triangle: ½ × base × perpendicular height. Parallelogram: base × perpendicular height. Trapezium: ½ × (a + b) × h, where a and b are the parallel sides.
- Sector area, and the area of a sector of radius 5 cm and angle 288°
- Sector area = (θ/360) × πr². 288/360 = 4/5, so 4/5 × 25π = 20π ≈ 62.8 cm².
- Area of a ring with outer radius 7 cm and inner radius 3 cm
- π × 7² − π × 3² = 49π − 9π = 40π ≈ 126 cm². It is not π(7 − 3)².
- A trapezium has area 66 cm² and parallel sides 8 cm and 14 cm. Find its height.
- 66 = ½ × (8 + 14) × h = 11h, so h = 6 cm.
- Area of a triangle from two sides and the included angleHigher
- Area = ½ab sin C, where C is the angle between sides a and b. Example: 7 cm and 12 cm with 30°: ½ × 7 × 12 × 0.5 = 21 cm².
10.04 Volume and surface area calculations
- Volume and surface area of a prism
- Volume = area of cross-section × length. Surface area = 2 × cross-section + perimeter of cross-section × length.
- Cylinder of radius r and height h
- Volume = πr²h. Curved surface = 2πrh; total surface area = 2πr² + 2πrh.
- Cone: volume, curved surface and slant height
- Volume = ⅓πr²h (perpendicular height). Curved surface = πrl (slant height). l = √(r² + h²). Example: r = 6, h = 8 gives l = 10 and curved area 60π cm².
- Sphere and solid hemisphere
- Sphere: V = (4/3)πr³, surface area 4πr². Solid hemisphere: V = (2/3)πr³, total surface area 3πr² (curved part 2πr²).
- Volume of a pyramid
- ⅓ × base area × perpendicular height. Example: 6 cm by 5 cm base, height 9 cm: ⅓ × 30 × 9 = 90 cm³.
- Volume units: 1 cm³ in ml, and 1000 cm³
- 1 cm³ = 1 ml; 1000 cm³ = 1 litre. Convert all lengths to one unit before multiplying.
10.05 Triangle mensuration
- Pythagoras: finding the hypotenuse compared with a shorter side
- a² + b² = c², c the hypotenuse (opposite the right angle). Hypotenuse: add the squares, then √ (5 and 12 give 13). Shorter side: subtract the squares, then √ (10 and 6 give 8).
- Right-angled trigonometry ratios
- sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj. Unknown side on top: multiply; on the bottom: divide. For an angle use sin⁻¹, cos⁻¹ or tan⁻¹.
- Exact values of sin, cos and tan for 0°, 30°, 45°, 60°, 90°
- sin: 0, 1/2, √2/2, √3/2, 1 cos: 1, √3/2, √2/2, 1/2, 0 tan (0° to 60°): 0, √3/3, 1, √3
- Angles of elevation and depression
- Both are measured from the horizontal: elevation looking up, depression looking down.
- Space diagonal of an 8 × 6 × 5 cuboidHigher
- Base diagonal: √(64 + 36) = 10. Space diagonal: √(10² + 5²) = √125 ≈ 11.2. Keep exact values until the end.
- Sine ruleHigher
- a/sin A = b/sin B = c/sin C (or sin A/a = … for angles). Use it when you know a side and its opposite angle. For an angle, check whether 180° minus the answer also fits.
- Cosine ruleHigher
- a² = b² + c² − 2bc cos A, for two sides and the included angle. cos A = (b² + c² − a²)/(2bc), for an angle from three sides. A negative cosine means an obtuse angle.
Revise Mensuration further
- Study guideOCR GCSE Mathematics: Area: triangles, parallelograms, trapezia, circles, sectors and composite shapes (J560)
- Revision notesOCR GCSE Mathematics: Area: triangles, parallelograms, trapezia, circles, sectors and composite shapes (J560) – Revision Notes
- Practice questionsOCR GCSE Mathematics: Area: triangles, parallelograms, trapezia, circles, sectors and composite shapes (J560) – Practice Questions