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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 1–5) and Higher-only items (grades 4–9, 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).
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" where the mark scheme expects "36 cm2" 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.
Two angles on a straight line are x and 47∘. Work out x, giving a reason.
Solution: Angles on a straight line sum to 180∘ (the reason mark), so x=180∘−47∘=133∘ (the accuracy mark). A common slip is to subtract from 360∘ by confusing "on a line" with "at a point".
Work out the size of each interior angle of a regular decagon (10 sides).
Solution: The exterior angle of a regular polygon is n360∘=10360∘=36∘ (method). Each interior angle is 180∘−36∘=144∘ (accuracy). The alternative route via (10−2)×180∘÷10 gives the same 144∘ and would also earn full marks.
A shape is made of a rectangle 10 cm by 4 cm with a triangle of base 10 cm and height 3 cm sitting on top. Work out its total area.
Solution: Rectangle area =10×4=40 cm2 (method) and triangle area =21×10×3=15 cm2 (method). Total =40+15=55 cm2 (accuracy). A frequent error is forgetting the 21 in the triangle area.
A circle has diameter 20 cm. Work out its area, giving your answer in terms of π.
Solution: First halve the diameter: r=10 cm (method — this is the most-tested step). Then area =πr2=π×102=100π cm2 (accuracy). Leaving the answer as 100π respects "in terms of π"; evaluating it would lose the form mark.
A triangular prism has a cross-section of area 18 cm2 and length 12 cm. Work out its volume, and state the units.
Solution: Volume of a prism = area of cross-section × length =18×12=216 (method and accuracy), and the units are cm3 (unit mark). Quoting the wrong unit — cm2 instead of cm3 — is a needless lost mark here.
A right-angled triangle has the two shorter sides 6 cm and 8 cm. Work out the length of the hypotenuse.
Solution: By Pythagoras, hypotenuse =62+82=36+64=100=10 cm. The method mark is for squaring and adding; the accuracy mark is for 10 cm. This is the well-known 6–8–10 triple, a scaled 3–4–5.
Describe fully the single transformation that maps a shape with a vertex at (2,3) onto an image with the matching vertex at (2,−3), where every y-coordinate has changed sign.
Solution: Every x-coordinate is unchanged and every y-coordinate has its sign reversed, which is a reflection in the x-axis (the line 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.
A right-angled triangle has the side opposite an angle equal to 7 cm and the hypotenuse equal to 12 cm. Work out the angle, to 1 decimal place.
Solution: Opposite and hypotenuse means sine: sinθ=127 (method), so θ=sin−1(127)≈35.7∘ (accuracy). Forgetting the inverse function — typing sin(7/12) — gives a meaningless number and loses the accuracy mark.
[Higher] A, B and C are points on a circle with centre O. ∠AOC=130∘ and B lies on the major arc. Work out ∠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 ∠ABC=2130∘=65∘ (method and accuracy). The reason must be stated by name to earn the reasoning mark; a bare 65∘ earns only the answer mark.
[Higher] A triangle has sides a=8 cm, b=5 cm and the included angle C=60∘. Work out side c, to 1 decimal place.
Solution: Using the cosine rule c2=a2+b2−2abcosC=64+25−2×8×5×cos60∘ (method for correct substitution). Since cos60∘=0.5, this is 89−80×0.5=89−40=49 (method for the discriminant value). So c=49=7.0 cm (accuracy). Mis-bracketing — multiplying everything by the cosine — is the classic error; the cosine multiplies only the 2ab term.
[Higher] A sphere has radius 3 cm. Work out (a) its volume and (b) its surface area, each in terms of π.
Solution to (a): Volume =34πr3=34π×33=34π×27=36π cm3. The method mark is for correct substitution; the accuracy mark for 36π.
Solution to (b): Surface area =4πr2=4π×32=36π cm2. Note the powers: volume uses r3, surface area uses r2. Confusing the two is the most common error in sphere questions.
Answer: volume 36π cm3; surface area 36π cm2.
[Higher] In triangle OAB, OA=a and OB=b. M is the midpoint of AB. Show that OM=21(a+b).
Solution: First, AB=b−a (go from A back to O, then O to B). Since M is the midpoint, AM=21(b−a). Then OM=OA+AM=a+21(b−a)=21a+21b=21(a+b), as required. Because this is a "show that", every vector step must be visible; the marks reward AB, the midpoint step, and the final simplification.
Two similar triangles have corresponding sides in the ratio 2:3. The smaller triangle has a side of 8 cm. Work out the matching side of the larger triangle.
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