Sometimes tested Papers 1 & 2

Geometry: Angles, Circle Theorems, Pythagoras, Trigonometry, Vectors and Mensuration

Speed in TMUA geometry comes from recognising angle facts, circle theorems and standard triangle results, then reading off an exact answer. Pythagoras in 2D and 3D, the sine and cosine rules, similarity ratios and vectors all appear, always without a calculator and against the clock.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • Angle facts to have instant: on a straight line sum to 180, around a point 360, in a triangle 180, in a quadrilateral 360; interior-angle sum of an n-gon = (n - 2) × 180; each exterior angle of a regular n-gon = 360/n.
  • Parallel-line angles: alternate angles are equal, corresponding angles are equal, and co-interior (allied) angles sum to 180 (they are NOT equal).
  • Similarity scale factor k acts differently on each dimension: lengths scale by k, areas by k2, volumes by k3.
  • Pythagoras: a2 + b2 = c2 in 2D; the space diagonal of a cuboid is d = √(a2 + b2 + c2) in 3D. Recognise triples 3-4-5, 5-12-13, 8-15-17 to avoid surds.
  • Circle theorems: angle at centre = 2 × angle at circumference on the same arc; angle in a semicircle = 90; angles in the same segment are equal; opposite angles of a cyclic quadrilateral sum to 180; a tangent meets the radius at 90; alternate segment theorem.
  • Trigonometry: SOH CAH TOA for right triangles; the sine and cosine rules for any triangle; area of a triangle = (1/2) × a × b × sin(C).
  • Vectors add and subtract componentwise with magnitude |(x, y)| = √(x2 + y2); two vectors are parallel when one is a scalar multiple of the other, which also proves three points are collinear.
  • Mensuration to memorise: circle area = pi × r2, circumference = 2 × pi × r; volume of a prism = cross-section area × length; cone = (1/3) × pi × r2 × h; sphere = (4/3) × pi × r3.

Diagram

4353^2 + 4^2 = 5^2
Pythagoras in a right-angled triangle: a2 + b2 = c2 (the 3, 4, 5 triangle: 9 + 16 = 25).

Formulae

a2 + b2 = c2

Right-angled triangle: c is the hypotenuse (opposite the right angle). Rearrange to find a shorter side.

d = √(a2 + b2 + c2)

Length of the space diagonal of a cuboid with edges a, b, c (3D Pythagoras).

a/sin(A) = b/sin(B) = c/sin(C)

Sine rule: use when you have a matching side-angle pair plus one more side or angle.

c2 = a2 + b2 - 2 × a × b × cos(C)

Cosine rule: use for two sides and the included angle, or all three sides. Note the MINUS sign.

area = (1/2) × a × b × sin(C)

Area of any triangle from two sides a, b and the included angle C.

|v| = √(x2 + y2)

Magnitude (length) of a 2D vector v = (x, y); extend to √(x2 + y2 + z2) in 3D.

Definitions

Congruent figures
Figures that are identical in both shape and size, so one maps exactly onto the other by rotation, reflection or translation. Established by SSS, SAS, ASA or RHS.
Similar figures
Figures with the same shape but possibly different size: corresponding angles are equal and corresponding sides are in a fixed ratio k, the scale factor.
Cyclic quadrilateral
A quadrilateral whose four vertices all lie on a single circle. Its opposite angles sum to 180.
Position vector
The vector from a fixed origin O to a point; the position vector of A is written vector OA, and vector AB = vector OB - vector OA.

Worked examples

1

Points B, C and D lie on a circle and BD is a diameter. The angle DBC = 27 degrees. Find the angle BDC.

  1. 1

    Angle in a semicircle is 90:

    angle BCD = 90
  2. 2

    Angles in a triangle sum to 180:

    angle BDC = 180 - 90 - 27
  3. 3
    angle BDC = 63

Answer: angle BDC = 63 degrees.

2

In triangle ABC, AB = 5, AC = 8 and the included angle A = 60 degrees. Find the exact length BC.

  1. 1

    Cosine rule:

    BC2 = 52 + 82 - 2 × 5 × 8 × cos(60)
  2. 2
    Use cos(60) = 1/2:
    BC2 = 25 + 64 - 80 × (1/2)
  3. 3

    Simplify:

    BC2 = 89 - 40 = 49
  4. 4
    BC = 7

Answer: BC = 7.

Common mistakes

  • ×Confusing area and volume scale factors: for a length scale factor k, areas scale by k2 and volumes by k3, not by k.
  • ×Applying the angle-at-centre theorem when the centre and circumference angles do not subtend the SAME arc; check the arc before doubling or halving.
  • ×Dropping the minus sign in the cosine rule; it is a2 + b2 - 2ab cos(C), and using plus gives a wrong (too large) answer.
  • ×Assuming a triangle is right-angled or isosceles just because the figure looks that way; TMUA diagrams are not drawn to scale, so use only stated facts.
  • ×Treating co-interior (allied) angles as equal; they sum to 180, whereas only alternate and corresponding angles are equal.

No-calculator tips

  • ✓Memorise exact trig values: sin(30) = 1/2, cos(30) = √(3)/2, sin(45) = cos(45) = 1/√(2), tan(30) = 1/√(3), tan(60) = √(3).
  • ✓Know the common Pythagorean triples (3-4-5, 5-12-13, 8-15-17 and their multiples) so you can read off a side without surd arithmetic.
  • ✓Keep surds exact and simplify: √(50) = 5 × √(2), and rationalise denominators, e.g. 1/√(2) = √(2)/2.
  • ✓Eliminate impossible options: the longest side lies opposite the largest angle, every triangle angle is under 180, and an obtuse angle forces a negative cosine.
  • ✓Exploit symmetry: two radii of a circle are equal, so the triangle they form is isosceles and its base angles are equal, often giving the answer in one step.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.Points A and B lie on a circle with centre O. The angle AOB (the non-reflex angle at the centre) is 120 degrees. Point C lies on the major arc AB. What is the size of angle ACB?

  • A. 120 degrees
  • B. 240 degrees
  • C. 60 degrees
  • D. 30 degrees
  • E. 90 degrees
Show answer

Answer: C — 60 degrees

The angle at the centre is twice the angle at the circumference on the same arc, so angle ACB = 120 / 2 = 60 degrees. Choosing 120 forgets to halve; 240 doubles instead of halving.

Q2.A solid cuboid has edge lengths 2, 3 and 6. What is the length of its longest interior (space) diagonal?

  • A. 11
  • B. 7
  • C. √(11)
  • D. 3 × √(5)
Show answer

Answer: B — 7

The space diagonal is √(22 + 32 + 62) = √(4 + 9 + 36) = √(49) = 7. The value 11 comes from adding the edges, √(11) from forgetting to square them, and 3*√(5) = √(32 + 62) from using only two of the three edges.

Q3.In triangle OAB, the vector OA = a and the vector OB = b. M is the midpoint of AB. Which expression gives the vector OM in terms of a and b?

  • A. (1/2)(b - a)
  • B. a + (1/2)*b
  • C. (1/2)*a + b
  • D. (1/2)(a + b)
Show answer

Answer: D — (1/2)(a + b)

OM = OA + AM = a + (1/2)(b - a) = (1/2)(a + b). The option (1/2)(b - a) is the vector AM (from A to M), not OM.

Read this topic in the official UAT-UK TMUA content specification →

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