Geometry: Exam Practice

This lesson pulls the whole Geometry and Measures strand of OCR GCSE Mathematics (J560) together with exam-style questions and full worked solutions. The questions range across both tiers: Foundation-level items (grades $1$1–$5$5) and Higher-only items (grades $4$4–$9$9, marked [Higher]). For each question the marks and tier are shown, and the solution explains where the method and accuracy marks fall. Work each one before reading the solution, then map your performance to the grade-band guidance. The skills assessed span AO1 (technique), AO2 (reasoning and communication) and AO3 (problem solving).

How Marks Are Awarded

It is worth understanding how an OCR mark scheme is built before you attempt the questions. Marks fall into three broad types. Method marks (M) are given for using a correct process — for example, setting up the right trig ratio, or splitting a composite shape sensibly — even if a later arithmetic slip spoils the final number. Accuracy marks (A) are given for a correct value, but usually only once the method mark has been earned. Communication or reasoning marks reward a clearly stated geometric reason (naming an angle fact or a circle theorem), a correct unit, or a properly drawn construction with its arcs. The single most important consequence in geometry is this: name your reasons and show your working at every stage. A correct angle written with no reason loses the reasoning mark even though the number is right, and a perfectly accurate construction with no visible arcs loses the method marks because there is no evidence of the method used. The same principle applies to units: an answer of "$36$36" where the mark scheme expects "$36$36 cm$^{2}$2" can drop the final mark. As you read each solution below, notice exactly where each method, accuracy, reasoning and unit mark is described, and aim to reproduce that level of detail in your own answers.

Exam-Style Questions

Question 1 — Angle reasoning (2 marks, Foundation/Higher, grades 2–3)

Two angles on a straight line are $x$x and $47^{\circ}$47∘. Work out $x$x, giving a reason.

Solution: Angles on a straight line sum to $180^{\circ}$180∘ (the reason mark), so $x = 180^{\circ} - 47^{\circ} = 133^{\circ}$x=180∘−47∘=133∘ (the accuracy mark). A common slip is to subtract from $360^{\circ}$360∘ by confusing "on a line" with "at a point".

Question 2 — Polygon angles (3 marks, Foundation/Higher, grades 3–4)

Work out the size of each interior angle of a regular decagon ($10$10 sides).

Solution: The exterior angle of a regular polygon is $\dfrac{360^{\circ}}{n} = \dfrac{360^{\circ}}{10} = 36^{\circ}$n360∘​=10360∘​=36∘ (method). Each interior angle is $180^{\circ} - 36^{\circ} = 144^{\circ}$180∘−36∘=144∘ (accuracy). The alternative route via $(10-2)\times 180^{\circ} \div 10$(10−2)×180∘÷10 gives the same $144^{\circ}$144∘ and would also earn full marks.

Question 3 — Area of a composite shape (3 marks, Foundation/Higher, grades 3–4)

A shape is made of a rectangle $10$10 cm by $4$4 cm with a triangle of base $10$10 cm and height $3$3 cm sitting on top. Work out its total area.

Solution: Rectangle area $= 10 \times 4 = 40$=10×4=40 cm$^{2}$2 (method) and triangle area $= \dfrac{1}{2} \times 10 \times 3 = 15$=21​×10×3=15 cm$^{2}$2 (method). Total $= 40 + 15 = 55$=40+15=55 cm$^{2}$2 (accuracy). A frequent error is forgetting the $\dfrac{1}{2}$21​ in the triangle area.

Question 4 — Circle (3 marks, Foundation/Higher, grades 4–5)

A circle has diameter $20$20 cm. Work out its area, giving your answer in terms of $\pi$π.

Solution: First halve the diameter: $r = 10$r=10 cm (method — this is the most-tested step). Then area $= \pi r^{2} = \pi \times 10^{2} = 100\pi$=πr2=π×102=100π cm$^{2}$2 (accuracy). Leaving the answer as $100\pi$100π respects "in terms of $\pi$π"; evaluating it would lose the form mark.

Question 5 — Volume of a prism (3 marks, Foundation/Higher, grades 4–5)

A triangular prism has a cross-section of area $18$18 cm$^{2}$2 and length $12$12 cm. Work out its volume, and state the units.

Solution: Volume of a prism $=$= area of cross-section $\times$× length $= 18 \times 12 = 216$=18×12=216 (method and accuracy), and the units are cm$^{3}$3 (unit mark). Quoting the wrong unit — cm$^{2}$2 instead of cm$^{3}$3 — is a needless lost mark here.

Question 6 — Pythagoras (3 marks, Foundation/Higher, grades 4–5)

A right-angled triangle has the two shorter sides $6$6 cm and $8$8 cm. Work out the length of the hypotenuse.

Solution: By Pythagoras, hypotenuse $= \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$=62+82​=36+64​=100​=10 cm. The method mark is for squaring and adding; the accuracy mark is for $10$10 cm. This is the well-known $6$6–$8$8–$10$10 triple, a scaled $3$3–$4$4–$5$5.

Question 7 — Transformations (3 marks, Foundation/Higher, grades 4–5)

Describe fully the single transformation that maps a shape with a vertex at $(2, 3)$(2,3) onto an image with the matching vertex at $(2, -3)$(2,−3), where every $y$y-coordinate has changed sign.

Solution: Every $x$x-coordinate is unchanged and every $y$y-coordinate has its sign reversed, which is a reflection in the $x$x-axis (the line $y = 0$y=0). The marks are for naming the transformation type and for giving the mirror line; a reflection without its mirror line scores no marks.

Question 8 — Trigonometry (3 marks, Foundation/Higher, grades 5–6)

A right-angled triangle has the side opposite an angle equal to $7$7 cm and the hypotenuse equal to $12$12 cm. Work out the angle, to $1$1 decimal place.

Solution: Opposite and hypotenuse means sine: $\sin\theta = \dfrac{7}{12}$sinθ=127​ (method), so $\theta = \sin^{-1}\left(\dfrac{7}{12}\right) \approx 35.7^{\circ}$θ=sin−1(127​)≈35.7∘ (accuracy). Forgetting the inverse function — typing $\sin(7/12)$sin(7/12) — gives a meaningless number and loses the accuracy mark.

Question 9 — Circle theorem [Higher] (3 marks, Higher, grades 5–6)

[Higher] $A$A, $B$B and $C$C are points on a circle with centre $O$O. $\angle AOC = 130^{\circ}$∠AOC=130∘ and $B$B lies on the major arc. Work out $\angle ABC$∠ABC, giving a reason.

Solution: The angle at the centre is twice the angle at the circumference on the same arc (the reason mark), so $\angle ABC = \dfrac{130^{\circ}}{2} = 65^{\circ}$∠ABC=2130∘​=65∘ (method and accuracy). The reason must be stated by name to earn the reasoning mark; a bare $65^{\circ}$65∘ earns only the answer mark.

Question 10 — Cosine rule [Higher] (4 marks, Higher, grades 6–7)

[Higher] A triangle has sides $a = 8$a=8 cm, $b = 5$b=5 cm and the included angle $C = 60^{\circ}$C=60∘. Work out side $c$c, to $1$1 decimal place.

Solution: Using the cosine rule $c^{2} = a^{2} + b^{2} - 2ab\cos C = 64 + 25 - 2 \times 8 \times 5 \times \cos 60^{\circ}$c2=a2+b2−2abcosC=64+25−2×8×5×cos60∘ (method for correct substitution). Since $\cos 60^{\circ} = 0.5$cos60∘=0.5, this is $89 - 80 \times 0.5 = 89 - 40 = 49$89−80×0.5=89−40=49 (method for the discriminant value). So $c = \sqrt{49} = 7.0$c=49​=7.0 cm (accuracy). Mis-bracketing — multiplying everything by the cosine — is the classic error; the cosine multiplies only the $2ab$2ab term.

Question 11 — Volume and surface area [Higher] (4 marks, Higher, grades 6–7)

[Higher] A sphere has radius $3$3 cm. Work out (a) its volume and (b) its surface area, each in terms of $\pi$π.

Solution to (a): Volume $= \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3}\pi \times 3^{3} = \dfrac{4}{3}\pi \times 27 = 36\pi$=34​πr3=34​π×33=34​π×27=36π cm$^{3}$3. The method mark is for correct substitution; the accuracy mark for $36\pi$36π.

Solution to (b): Surface area $= 4\pi r^{2} = 4\pi \times 3^{2} = 36\pi$=4πr2=4π×32=36π cm$^{2}$2. Note the powers: volume uses $r^{3}$r3, surface area uses $r^{2}$r2. Confusing the two is the most common error in sphere questions.

Answer: volume $36\pi$36π cm$^{3}$3; surface area $36\pi$36π cm$^{2}$2.

Question 12 — Vectors [Higher] (4 marks, Higher, grades 6–7)

[Higher] In triangle $OAB$OAB, $\vec{OA} = \mathbf{a}$OA=a and $\vec{OB} = \mathbf{b}$OB=b. $M$M is the midpoint of $AB$AB. Show that $\vec{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$OM=21​(a+b).

Solution: First, $\vec{AB} = \mathbf{b} - \mathbf{a}$AB=b−a (go from $A$A back to $O$O, then $O$O to $B$B). Since $M$M is the midpoint, $\vec{AM} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})$AM=21​(b−a). Then $\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$OM=OA+AM=a+21​(b−a)=21​a+21​b=21​(a+b), as required. Because this is a "show that", every vector step must be visible; the marks reward $\vec{AB}$AB, the midpoint step, and the final simplification.

Question 13 — Similar shapes (3 marks, Foundation/Higher, grades 4–5)

Two similar triangles have corresponding sides in the ratio $2 : 3$2:3. The smaller triangle has a side of $8$8 cm. Work out the matching side of the larger triangle.