Geometry Exam Practice

Having reached the end of the geometry course, use the seven Edexcel-style questions below to consolidate every topic from lessons 1–10: angles and polygons, 2D and 3D shape properties, perimeter/area/volume, transformations, congruence and similarity, Pythagoras and trigonometry, constructions and loci, and vectors. Each question is presented in the mark allocation and style of an Edexcel GCSE Mathematics (1MA1) question, with a full mark-scheme-style solution and an examiner note.

Attempt these under timed conditions: target 40 minutes for the full set (total 40 marks, at 1 minute per mark). Then self-mark and use the examiner notes to identify where you are losing marks.

Question G1: Angles in Polygons (4 marks, Foundation)

The interior angle of a regular polygon is 162°. Work out the number of sides of the polygon.

Mark-scheme solution

Exterior angle = 180° − 162° = 18°. (M1)

Number of sides = 360° ÷ exterior angle = 360 ÷ 18. (M1)

= 20 sides. (A1)(A1 for a fully correct method with answer)

Examiner note. Always work with the exterior angle when finding the number of sides: the sum of exterior angles of any polygon is 360°, so sides = 360 ÷ exterior angle. Candidates who try to use the interior angle formula (n − 2) × 180 ÷ n = 162 must solve algebraically — valid, but more work and more places to slip. The exterior-angle approach is a two-line solution. Common error: computing 360 ÷ 162 = 2.2… and rounding — always check whether an integer answer is expected, and if not, re-examine your method.

Question G2: Area of a Compound Shape (4 marks, Foundation)

A shape is made from a rectangle of width 12 cm and height 8 cm with a semicircle of diameter 8 cm attached to one of the short sides. Work out the total area of the shape, giving your answer to 1 decimal place. Use π = 3.142.

Mark-scheme solution

Area of rectangle = 12 × 8 = 96 cm². (M1)

Radius of semicircle = 8 ÷ 2 = 4 cm. Area of semicircle = (1/2) × π × 4² = (1/2) × π × 16 = 8π. (M1)

Using π = 3.142: 8 × 3.142 = 25.136 cm². (A1)

Total area = 96 + 25.136 = 121.136 ≈ 121.1 cm² (1 d.p.). (A1)

Examiner note. On a compound shape, split it into simple parts, compute each area, then add. Common errors: (i) using the diameter 8 as the radius, giving a circle area four times too big; (ii) forgetting to halve for the semicircle; (iii) rounding π to 3.14 when the question specifies 3.142 — the final answer changes and you lose the A1. Write the working line-by-line so an examiner can follow and award method marks even if your arithmetic slips.

Question G3: Pythagoras' Theorem in 3D [H] (4 marks, Higher)

A cuboid has dimensions 6 cm × 4 cm × 3 cm. Calculate the length of the longest diagonal of the cuboid, giving your answer to 2 decimal places.

Mark-scheme solution

Diagonal of the base = $\sqrt{6^{2} + 4^{2}}$62+42​ = $\sqrt{36 + 16}$36+16​ = $\sqrt{52}$52​. (M1)

Longest diagonal = $\sqrt{diagonal_base^{2} + height^{2}}$diagonalb​ase2+height2​ = $\sqrt{52 + 3^{2}}$52+32​ = $\sqrt{52 + 9}$52+9​ = $\sqrt{61}$61​. (M1)

= 7.8102… (M1)

= 7.81 cm (2 d.p.). (A1)

Examiner note. The 3D Pythagoras shortcut: diagonal = $\sqrt{a^{2} + b^{2} + c^{2}}$a2+b2+c2​ — here $\sqrt{36 + 16 + 9}$36+16+9​ = $\sqrt{61}$61​ = 7.81 cm. Candidates who apply Pythagoras only once (for example $\sqrt{6^{2} + 3^{2}}$62+32​) miss that the diagonal goes from corner to opposite corner through the cuboid, involving all three dimensions. Writing it out in two 2D steps (diagonal of the base, then use that with the height) is the safest method and scores full marks. Always check the answer is bigger than the longest single edge (here, bigger than 6) — a quick sanity check.

Question G4: Trigonometry — SOH CAH TOA (5 marks, Higher)

A ladder of length 5.2 m leans against a vertical wall. The foot of the ladder is 1.6 m away from the base of the wall.

(a) Calculate the angle the ladder makes with the ground. Give your answer to 1 decimal place. (3)

(b) Calculate how high up the wall the ladder reaches. Give your answer to 3 significant figures. (2)

Mark-scheme solution

(a) The horizontal distance (1.6 m) is adjacent to the angle; the ladder (5.2 m) is the hypotenuse. So use cos.

cos θ = 1.6 / 5.2 = 0.30769… (M1)

θ = $\cos^{-1}(0.30769…)$cos−1(0.30769…) = 72.079…° (M1)

= 72.1° (1 d.p.). (A1)

(b) Height up the wall = opposite to the angle. Use sin, or Pythagoras.

By Pythagoras: height = $\sqrt{5.2^{2} − 1.6^{2}}$5.22−1.62​ = $\sqrt{27.04 − 2.56}$27.04−2.56​ = $\sqrt{24.48}$24.48​ = 4.9477… (M1)

= 4.95 m (3 s.f.). (A1)

Examiner note. SOH CAH TOA is the go-to mnemonic: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent. Identify which two sides you have (adjacent + hypotenuse here), and that tells you which ratio. Candidates who pick the wrong ratio lose the M1 even if the arithmetic is correct. In part (b), either sin(72.1°) × 5.2 or Pythagoras works — Pythagoras is often safer because it avoids accumulating rounding error from part (a). Store the full angle value (not the rounded 72.1°) in your calculator if you do use trig.

Question G5: Similar Shapes and Scale Factors (5 marks, Higher)

Two similar cones have surface areas in the ratio 9 : 25.

(a) Find the ratio of their heights. (2)

(b) The volume of the smaller cone is 108 cm³. Find the volume of the larger cone. (3)

Mark-scheme solution

(a) Surface areas scale by the square of the length scale factor. If lengths are in ratio k : 1, areas are in ratio k² : 1.

So k² : 1 corresponds to 9 : 25, meaning the length ratio (smaller : larger) is $\sqrt{9}$9​ : $\sqrt{25}$25​ = 3 : 5. (M1)(A1)

(b) Volumes scale by the cube of the length scale factor.

Volume ratio = 3³ : 5³ = 27 : 125. (M1)

Setting up: 27 parts = 108 cm³, so 1 part = 4 cm³. (M1)

Larger volume = 125 × 4 = 500 cm³. (A1)

Examiner note. The length : area : volume rule — 1 : k, 1 : k², 1 : k³ — is one of the highest-leverage facts in GCSE geometry. Candidates who forget and cube the area ratio (getting 729 : 15,625) or treat it as linear get part (b) completely wrong. A quick sanity-check: larger cone is bigger, so larger volume should exceed 108 — and 500 > 108 ✓. Edexcel examiners specifically look for the square-root step in (a) and the cubing step in (b); if you can state both clearly, you bank all 5 marks even with minor numerical slips.

Question G6: Sine Rule [H] (6 marks, Higher — Calculator)

In triangle ABC, angle A = 48°, angle B = 72° and the side opposite to angle A (side a = BC) is 7.5 cm.

(a) Find the length of the side opposite to angle B (side b = AC). Give your answer to 2 decimal places. (3)

(b) Work out the area of the triangle. Give your answer to 1 decimal place. (3)

Mark-scheme solution

(a) Sine rule: a / sin A = b / sin B.

7.5 / $\sin 48°$sin48° = b / $\sin 72°$sin72°. (M1)

b = 7.5 × $\sin 72°$sin72° / $\sin 48°$sin48° = 7.5 × 0.95106 / 0.74314 = 7.5 × 1.27981… = 9.5986… (M1)

= 9.60 cm (2 d.p.). (A1)

(b) Angle C = 180° − 48° − 72° = 60°. (M1)

Area = (1/2) × a × b × sin C = (1/2) × 7.5 × 9.5986 × $\sin 60°$sin60°. (M1)

= 0.5 × 7.5 × 9.5986 × 0.86603 = 31.2 cm² (1 d.p.). (A1)

Examiner note. The sine rule and the area formula (1/2)ab sin C are on the Edexcel formula sheet at both tiers — so you do not need to memorise them, but you must know when to apply each. Use the sine rule when you have a matched pair (a side and its opposite angle) plus one more piece of information. Use the cosine rule when you have either two sides and the included angle, or all three sides. In part (b), use the unrounded value of b from part (a) for maximum accuracy — if you substitute 9.60 you may lose a decimal place in the final answer. A common student error is to apply (1/2) × base × height without realising the triangle is not right-angled; the (1/2)ab sin C formula generalises this to any triangle.

Question G7: Vector Geometry Proof [H] (6 marks, Higher)

In triangle OAB, vector OA = a and vector OB = b. M is the midpoint of AB. The point C is on OA such that OC = (2/3)a. The point D is on OM produced such that OD = (3/2) OM.

(a) Express vector OM in terms of a and b. (2)

(b) Show that B, C and D are collinear. (4)

Mark-scheme solution