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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Geometry and measures worksheet — GCSE Higher
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- 1.Quadrilateral PQRS has angle P = 92°, angle Q = 84° and angle R = 106°. Work out the size of angle S.
- 2.A cuboid has three different edge lengths: a length, a width and a height, all different from each other. How many rectangles make up its net in total?
- 3.A scale drawing of a bridge uses a scale of 1 : 25. A support beam on the drawing measures 4.4 cm. What is the real length of the beam, in metres?
- 4.m is the column vector with top number 4 and bottom number 6. n is the column vector with top number −6 and bottom number −9. Given that n = k × m for some number k, work out the value of k.
- 5.A straight fence is 5 m long. Describe the shape of the region made up of every point on the ground within 2 m of any part of the fence.
- 6.A hiker starts by walking on a bearing of 245°. She turns clockwise through 50°, then turns clockwise through a further 15°, and continues walking in a straight line. What bearing is she now walking on?
- 7.Point A(1, 2) is enlarged to give image point A′(7, −4). The centre of enlargement lies on the x-axis. Work out the scale factor of the enlargement.
- 8.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 9.Lines AB and CD are parallel. A straight line EF crosses AB at point P and crosses CD at point Q. At P, angle APE = 65°. Angle APE and angle BPQ are vertically opposite. Angle BPQ and angle PQD are co-interior (allied) angles. Work out angle PQD.
- 10.A drone starts at the point (2, −3) on a coordinate grid. It flies by the vector and then by the vector . Write down the column vector that would take the drone straight back to its starting point.
- 11.A shape is reflected in the line y = −x. Which of these points is invariant under this reflection?y = −x
- 12.A shape is reflected in the line x = 1, and the image is then reflected in the line x = 5. Which single transformation is equivalent to this combination, for every point?
- 13.A designer enlarges a drawing of a model car for a poster. She first enlarges the drawing by a scale factor of 1.5, and then enlarges that result by a further scale factor of 2. On the original drawing, the position of a wheel relative to the front bumper is given by the column vector with top number 4 and bottom number −3, in centimetres. What is the corresponding column vector on the poster, in centimetres?
- 14.A transversal crosses a pair of parallel lines. At one line, the angle is (5x + 4)°. The corresponding angle at the other line is (3x + 24)°. Work out the value of x.
- 15.A tent has the cross-section of a right-angled triangle. The sloping side of the cross-section is 10 m long, and it makes an angle of 60° with the horizontal ground. Using the exact value of cos 60°, work out the horizontal distance covered by the sloping side.
Answer key
- (c) 78° — The angles in any quadrilateral add up to 360°. Add the three given angles: 92° + 84° + 106° = 282°. Angle S = 360° − 282° = 78°. A pupil who only adds angle P and angle Q, forgetting angle R, gets 360° − (92° + 84°) = 184°. A pupil who only adds angle Q and angle R, forgetting angle P, gets 360° − (84° + 106°) = 170°. A pupil who makes a carrying slip adding the three angles, getting 292° instead of 282°, gets 360° − 292° = 68°. The correct answer is 78°.
- (a) 6 — A cuboid has six faces in total: a top, a bottom, a front, a back, and two ends — each one is a rectangle in the net, giving six rectangles altogether. Choosing 3 counts only the three PAIRS of congruent rectangles (top/bottom, front/back, two ends) rather than all six individual faces. Choosing 5 forgets one face, as if the net were missing its lid. Choosing 12 is the number of edges of a cuboid, not the number of rectangles in its net.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (a) a rounded rectangle: 5 m by 4 m with semicircular ends — Points within 2 m of the straight part of the fence form a rectangle running the 5 m length of the fence and 4 m wide (2 m on each side); points within 2 m of each END of the fence, beyond that rectangle, form a semicircle of radius 2 m there, since the nearest point of the fence to them is just that one end. Together this gives a rounded, stadium-shaped region. "a rectangle, 9 m by 4 m" extends the rectangle by 2 m at each end instead of rounding it, wrongly including corner points that are actually more than 2 m from every part of the fence. "a circle of radius 2 m" treats the whole 5 m fence as a single point. "a rectangle, 5 m by 2 m" uses 2 m as the full width instead of the distance on EACH side, so it only covers one side of the fence.
- (d) 310 — Method: since both turns are clockwise, add both angles to the starting bearing. Working: 245° + 50° = 295°; 295° + 15° = 310°. A student who answers 295 has only added the first turn and forgotten the second one. A student who answers 180 has subtracted both turns instead of adding them. A student who answers 320 has added the two turns as 75° instead of 65° by misreading the second turn. Answer: 310°.
- (a) −2 — Let the centre be C = (c, 0). The vector from the centre to the image equals the scale factor k times the vector from the centre to the object: (7 − c, −4) = k(1 − c, 2). The y-coordinate gives −4 = 2k, so k = −2 — this doesn't depend on knowing c. (Checking: substituting k = −2 into the x-equation gives c = 3, consistent with a centre on the x-axis.) '2' comes from taking the magnitude of the ratio without noticing the image is on the opposite side of the centre from the object, so the sign should be negative. '−1/2' comes from inverting the scale factor, dividing the object's coordinate by the image's instead of the other way round. '3' is the x-coordinate of the centre, mistaken for the scale factor.
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (a) (4, −4) — A point is invariant under a reflection only if it lies exactly on the mirror line. The line y = −x consists of every point where the y-coordinate is the negative of the x-coordinate: (4, −4) satisfies this, since −4 = −(4), so it is invariant. (5, 5) lies on the line y = x, a different line altogether, not y = −x. (4, 4) has equal coordinates, but that alone does not put it on y = −x; it would need y = −4, not 4. (−4, −4) also has equal coordinates and lies on y = x, not y = −x — its coordinates would need opposite signs to sit on the given mirror line. Only a point whose coordinates are negatives of each other stays fixed under this reflection.
- (c) Translation by the vector (8, 0) — Method: reflecting twice in two parallel vertical 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 5 − 1 = 4 units apart, so the translation is 2 × 4 = 8 units in the positive x-direction. Answer: translation by the vector (8, 0). Using just the gap itself, without doubling it, gives (4, 0); translating in the negative x-direction, from the second line back towards the first, gives (−8, 0); and describing the combination as a single reflection in the line halfway between them, x = 3, 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) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
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