Why the area formulas work (cut and move, halve and double), then area problems; volume as layers of the base; the parts of a circle, π, and the circumference; then Pythagoras' theorem for right-angled triangles; for Year 8, composite and irregular shapes, capacity and prism problems, the area of a circle, and time zones.
Parallelograms are rectangles cut and moved; triangles are half a parallelogram; area problems with the right base, height and units; then (Year 8) composite and irregular shapes.
Cut a triangle off one end of a parallelogram and move it to the other end: it becomes a rectangle with the same base and height. So the area of a parallelogram is base × perpendicular height.
Two copies of any triangle, one turned round, make a parallelogram with the same base and height. So a triangle's area is half of base × height: A = ½bh.
Pick the base and the height at a right angle to it, put both lengths in the same unit, and split a shape into rectangles and triangles. Area is in square units: m², cm².
A composite shape is made of shapes you know. Cut it into pieces and add their areas, or see it as a big rectangle with pieces taken away; the area is the same either way. The perimeter is the distance round the outside.
A lake or a leaf has no formula. Lay a grid over it: count the full squares, add half the part squares, and multiply by the area of one square. A finer grid gives a better estimate; straight pieces round the edge estimate the perimeter.
Volume as layers: the area of the base times the height, for rectangular and triangular prisms, and capacity in millilitres; then (Year 8) litres and cubic metres, and prism problems.
Fill a box with unit cubes layer by layer. Each layer has as many cubes as the base has squares, so the volume is the area of the base times the height: V = A × h. Volume is in cubic units, cm³.
Any right prism has the same shape all the way along: its cross-section. Volume = area of the cross-section × length, for triangles too (halve first). Capacity: 1 cm³ holds 1 mL, and 1000 cm³ is a litre.
Volume measures space in cubic units; capacity measures how much a container holds. 1 cm³ holds 1 mL, 1000 cm³ holds 1 L, and 1 m³ holds 1000 L. A cubic metre is 100 × 100 × 100 = 1 000 000 cm³, not 100.
The volume of any right prism is the area of its end (the cross-section) times its length: V = Ah. Work backwards to find a missing length: divide the volume by the area of the end. Give the answer in the units the problem needs.
Centre, radius, diameter and circumference; rolling circles out to find π; and C = πd; then (Year 8) A = πr² and circle problems.
Every point of a circle is the same distance from its centre: the radius. The diameter goes right across through the centre, two radii long (d = 2r). A chord joins any two points; the circumference is the distance all the way round.
Roll any circle out flat and its circumference is just over 3 of its diameters. C ÷ d is the same number for every circle: π (pi), about 3.14. π's digits never end: 3.14 and 22/7 are close, not exact.
Because C ÷ d = π, the circumference is π × diameter: C = πd, or C = 2πr from the radius. Going backwards, d = C ÷ π.
Cut a circle into thin slices and lay them top to tail: they make a shape close to a rectangle, πr long and r high. So the area of a circle is A = πr². It is between 2r² and 4r², about 3.14 times r².
Choose the formula by what you need: C = 2πr for a length round the edge, A = πr² for the space inside. Work backwards from an area with r = √(A ÷ π). Half and quarter circles, rings and tracks are parts of circles added to or taken from other shapes.
Squares on the sides of a right-angled triangle, finding any side, and testing whether a triangle is right-angled.
Draw a square on each side of a right-angled triangle. The two smaller squares together have exactly the same area as the biggest: a² + b² = c², where c is the hypotenuse, the side opposite the right angle.
Pythagoras finds any side. For the hypotenuse, add the squares: c² = a² + b². For a shorter side, subtract from the hypotenuse's square: b² = c² − a². Then take the square root, and check the hypotenuse is still the longest.
Pythagoras works backwards: square the two shorter sides and add, then compare with the longest side squared. Equal: right-angled. The longest square is bigger: obtuse. Smaller: acute.
Changing times between time zones, and when flights land.
A city with a bigger UTC offset is ahead; a smaller offset is behind. Find the difference between the two offsets, then add or take it away; the day changes when you cross midnight. A flight's arrival time is the departure time plus the flight, changed to the local time where it lands.
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For teachers: Teacher guide (notes, answers, solutions, slides and worksheets)