4MA1: Geometry and trigonometry. Version: Issue 2, November 2017; first assessment June 2018; linear Mathematics A.
Foundation teaching and Higher additions are labelled below. This reference packages the existing native-lesson crosswalk. It does not certify unreviewed specification rows or a whole qualification. Original diagnostics are separate and are not reproduced.
Angles, lengths and area · Foundation
Use angle facts with a stated reason. Similar shapes have equal corresponding angles and proportional corresponding lengths. Areas of rectangles and triangles come from their dimensions; compound shapes can be split into simpler parts.
A rectangle of length 8 cm and width 5 cm has area A=LW=40 cm². A triangle on the same base and height has area A=bh/2=20 cm². For a pentagon, the interior-angle sum is (5-2)×180=540°.
Equal angles alone establish similarity, not equal size. Use corresponding lengths in the same order. Convert linear units before calculating area or volume, or square/cube the conversion factor correctly.
Use a labelled sketch and appropriate units. This Foundation/Core lesson does not test area/volume scale factors or advanced circle-theorem proofs.
Angle reasoning, similarity and mensuration · Higher
For similar shapes with length scale factor k, areas scale by k² and volumes by k³. State angle reasons explicitly. A circle's tangent is perpendicular to the radius at the contact point.
If model-to-real length factor is 3, a model area of 12 cm² gives 12×3²=108 cm² and a model volume of 8 cm³ gives 8×3³=216 cm³. A cylinder with r=3,h=5 has volume πr²h=45π.
Equal angles alone establish similarity, not equal size. Use corresponding lengths in the same order. Convert linear units before calculating area or volume, or square/cube the conversion factor correctly.
A geometric proof should name the relevant theorem, identify the equal angle or ratio, and draw the conclusion. A scale drawing is evidence only when the task permits measurement.
Right triangles and non-right triangles · Foundation
Use Pythagoras in a right triangle and use sine, cosine or tangent with the sides labelled relative to the chosen angle.
The ladder length is c=√(3²+4²)=5 m. Its angle to the ground satisfies tanθ=4/3, so θ≈53.1°. A right triangle with legs 6 and 8 has area 6×8/2=24.
Label sides relative to the chosen angle. Pythagoras needs a right angle. A calculator angle mode error can produce a plausible but wrong result. Keep unrounded values for later steps.
This Foundation/Core lesson uses right-angled triangles only. Sine and cosine rules for non-right triangles belong to the advanced-tier lesson.
Right triangles and non-right triangles · Higher
In a right triangle a²+b²=c²; sinθ=opposite/hypotenuse, cosθ=adjacent/hypotenuse and tanθ=opposite/adjacent. For other triangles, use the sine or cosine rule, or area=ab sin C/2.
The ladder length is c=√(3²+4²)=5 m. Its angle to the ground satisfies tanθ=4/3, so θ≈53.1°. With two sides 6 and 8 enclosing 60°, c²=6²+8²-2×6×8 cos60°=52.
Label sides relative to the chosen angle. Pythagoras needs a right angle. A calculator angle mode error can produce a plausible but wrong result. Keep unrounded values for later steps.
Use a plan or elevation for a three-dimensional problem before applying a triangle rule. Explain why the chosen triangle contains the required length or angle.
Circle theorems and reasoned proofs · Higher
The angle at the centre is twice the angle at the circumference on the same arc. Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral sum to 180°. A radius is perpendicular to a tangent at contact.
If a central angle is 100°, the corresponding angle at the circumference is 50°. In a cyclic quadrilateral with one angle 112°, its opposite angle is 180-112=68°. A radius meeting a tangent gives 90°, even if the drawing looks oblique.
Identify the same chord and the correct arc before using a theorem. Two visible right angles do not prove a quadrilateral cyclic without a valid converse argument. A diagram need not be to scale.
Write one reason alongside each angle calculation. For the alternate-segment theorem, name the tangent and chord, then identify the angle in the opposite segment. Use auxiliary radii only when they help the proof.
Constructions, loci and geometric conditions · Foundation
Points equally distant from A and B lie on the perpendicular bisector of AB. Points at fixed distance r from C lie on a circle. Points equally distant from two intersecting lines lie on their angle bisectors.
For A=(0,0) and B=(6,0), the perpendicular bisector is x=3. Points also 5 units from A satisfy x²+y²=25. Substituting x=3 gives y²=16, so (3,4) and (3,-4) satisfy both conditions.
The perpendicular bisector concerns distance to two points; the angle bisector concerns distance to two lines. A sketch is not a ruler-and-compass construction: preserve arcs as evidence of the method.
Translate each condition into a locus before finding intersections. For a region closer to A than B, choose the correct side of the perpendicular bisector and show whether a boundary is allowed.
