Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Geometry and measures worksheet — GCSE Higher
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- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (d) (2/5)a + (3/5)b — Method: OP = OA + AP, and since AP : PB = 3 : 2 splits AB into 5 equal parts, AP is 3/5 of the whole of AB, with AB = b − a. Working: OP = a + 3/5(b − a) = a − (3/5)a + (3/5)b = (2/5)a + (3/5)b. Answer: OP = (2/5)a + (3/5)b. Using the ratio the wrong way round, as though it read AP : PB = 2 : 3, gives (3/5)a + (2/5)b; adding (3/5)b onto the whole of a without subtracting a inside the bracket first gives a + (3/5)b; and treating the ratio as 1 : 1 gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from A, matching the ORDER the ratio is stated in, and subtract before you scale.
- (b) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (c) 68° — Since PA and PB are both tangents from the external point P, PA = PB, so triangle PAB is isosceles with equal base angles at A and B. The angles of triangle PAB sum to 180°, so angle PAB + angle PBA = 180° − 44° = 136°, and since the two base angles are equal, angle PAB = 136° ÷ 2 = 68°. Angle PAB is the angle between the tangent at A and the chord AB, so by the alternate segment theorem it equals the angle in the alternate segment, angle ACB = 68°. Using angle APB directly as angle ACB, without using the isosceles triangle to find angle PAB first, gives 44°. Halving angle APB directly, rather than subtracting it from 180° before halving, gives 22°. Finding 180° − 44° = 136° correctly but forgetting to divide by 2 for one base angle leaves 136°.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (c) 5√3 cm — The space diagonal of a cube with edge a satisfies d² = a² + a² + a² = 3a², using Pythagoras' theorem in three dimensions. With a = 5, d² = 3 × 5² = 3 × 25 = 75, so d = √75 = √(25 × 3) = 5√3 cm. Finding the diagonal of one face instead, using only two of the three edges, gives d = √(5² + 5²) = √50 = 5√2 cm, which leaves out the third dimension. Adding the three edges directly, 5 + 5 + 5 = 15 cm, ignores that Pythagoras' theorem works with squares of lengths, not the lengths themselves. Squaring the edge and multiplying by 3 correctly, 3 × 5² = 75, but then forgetting to take the square root, leaves 75 cm — the squared length, not the diagonal itself.
- (a) Translation by the vector (0, 6) — Method: reflecting twice in two parallel horizontal lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 4 − 1 = 3 units apart, so the translation is 2 × 3 = 6 units in the positive y-direction. Answer: translation by the vector (0, 6). Using just the gap itself, without doubling it, gives (0, 3); translating in the negative y-direction, from the second line back towards the first, gives (0, −6); and describing the combination as a single reflection in the line halfway between them, y = 2.5, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
- (b) (3, −5) — Method: apply the rotation to the point first, then translate the image, in the order the question gives them. Working: rotating (4, 1) by 90° clockwise about the origin sends (x, y) to (y, −x), so (4, 1) becomes (1, −4). Translating (1, −4) by the vector (2, −1) gives 1 + 2 = 3 and −4 − 1 = −5, so the final image is (3, −5). Answer: (3, −5). Use the CLOCKWISE rule, (x, y) → (y, −x), not the anticlockwise one, and apply the rotation before the translation, exactly as the question states them: reversing the order or the direction of turn both land on a different point.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (b) −2 — The gradient of the original line is (6 − 2) ÷ (3 − 1) = 4 ÷ 2 = 2. Reflecting in the x-axis sends every y-coordinate to its negative, which flips the sign of the gradient: the image line has gradient −2. Translating by (2, 0) is a horizontal shift, which does not change the line's steepness or direction at all, so the gradient stays at −2. Assuming the gradient is unaffected by the reflection gives 2, the original gradient carried straight through. Thinking a reflection in the x-axis turns a gradient into its positive reciprocal gives 1/2. Combining that same wrong idea with the sign flip from the reflection gives −1/2. Only the sign flips, from the reflection, and translating never changes a gradient at all, so the answer is −2.
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (c) AB, AC and angle BAC: Area = 1/2 × AB × AC × sin(BAC) — Method: Area = 1/2ab sin C only works when the angle used is the one included between the two sides being multiplied. Working: AB and AC meet at A, and angle BAC is the angle at A between them, so the statement pairing AB, AC and angle BAC is the correct one. The statement that three sides with no angle can still go into 1/2 AB × AC × sin(BAC) is wrong: with no angle known, sin(BAC) cannot be evaluated, so a different method must find an angle first. The statement pairing AB and BC with sin(BAC) is wrong: AB and BC meet at B, so the angle between them is angle ABC, not angle BAC — it names the wrong angle for the sides it uses. The statement pairing AB and AC with sin(ABC) is wrong for the same reason: AB and AC meet at A, so their included angle is angle BAC, and angle ABC is not between them at all.
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