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.
Geometry and measures worksheet — GCSE Higher
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- 1.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 2.Point P has coordinates (6, 4). P is rotated 90° clockwise about the point (1, 2), and the image is then translated by the vector (−3, 5). Work out the coordinates of the final image of P.
- 3.A sign-maker is told a metal bracket outline must have two pairs of adjacent sides equal in length (not opposite pairs), exactly one pair of opposite angles equal, and exactly one line of symmetry. Which quadrilateral matches this description?
- 4.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 5.A map has a scale of 1 : 20 000. The distance between two villages is 3.4 km in real life. How long is this distance on the map, in centimetres?
- 6.A quadrilateral has two pairs of parallel sides. All four of its interior angles are right angles, but its sides are not all the same length. Which quadrilateral is this?
- 7.A solid is built from centimetre cubes: a base layer of 2 rows of 3 cubes each (a 3 by 2 rectangle of cubes), with one extra cube placed on top of one corner cube of that base. Looking down from directly above (the plan view), how many squares are visible?
- 8.A circle has centre O and radius 13 cm. AB is a chord of the circle, and M is the midpoint of AB. OM is drawn, and OM = 5 cm. Work out the length of AB.
- 9.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
- 10.The vector describes a translation. A student writes it as by mistake. Write down what error the student has made.
- 11.In triangle ABC, AB = x cm, AC = (x + 2) cm, BC = 15 cm and angle BAC = 100°. Work out the value of x. Give your answer to 1 decimal place.
- 12.A line joins the points A(0, 0, 0) and B(9, 12, 8) in a three-dimensional coordinate system in which the z-axis is vertical. Work out the angle this line makes with the horizontal (the xy-plane). Give your answer correct to 1 decimal place.
- 13.A cuboid has length 3 cm, width 4 cm and height 12 cm. Work out the length of the diagonal that runs from one corner of the cuboid to the opposite corner.
- 14.Point P is translated by the vector (5, −2), and the image is then rotated 90° clockwise about the origin to give the point (3, 9). Work out the coordinates of P.
- 15.Triangle T has a vertex at (8, 4). It is enlarged by a scale factor of 1/2, centre (2, 4). Work out the coordinates of the image of this vertex.
Answer key
- (c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (a) 17 cm — Convert the real distance to centimetres first: 3.4 km = 340 000 cm. Then divide by the scale factor, 20 000, since the map is 20 000 times smaller than real life: 340 000 ÷ 20 000 = 17, so the map distance is 17 cm. Choosing 1700 cm divides by 200 instead of 20 000, losing two zeros from the scale factor (340 000 ÷ 200 = 1700). Choosing 170 cm divides by 2 000 instead of 20 000, losing one zero from the scale factor (340 000 ÷ 2 000 = 170). Choosing 0.17 cm divides by 2 000 000 instead of 20 000, adding two extra zeros to the scale factor (340 000 ÷ 2 000 000 = 0.17).
- (c) Rectangle — A rectangle has two pairs of parallel sides and four right angles, but does not require all sides to be equal — this matches exactly, so Rectangle is correct. A square also has four right angles and parallel sides, but additionally requires all four sides to be equal, which contradicts 'not all the same length', so it is wrong. A rhombus has two pairs of parallel sides and all four sides equal, but its angles are not generally 90° unless it is also a square, so it does not match the right-angle condition here. A kite has no pairs of parallel sides at all, so it does not match the first condition given.
- (d) 6 — The plan view shows every square of the base footprint, whether or not there is a taller stack above it — the base layer alone already covers a 3 by 2 rectangle of cubes, which is 6 squares. The extra cube on top of a corner cube sits directly above a square that is already counted, so it adds no NEW square to the plan — height does not show up in a plan view, only footprint does. "7" comes from wrongly counting the extra cube as an additional square. "5" comes from missing one square of the base rectangle, perhaps forgetting a corner. "3" comes from counting only one row of the base rectangle and forgetting that the base is two rows deep.
- (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
- (d) 8.8 — Method: substitute into the cosine rule BC² = AB² + AC² − 2 × AB × AC × cos(BAC) and solve the resulting quadratic in x, keeping only the positive root. Working: 15² = x² + (x + 2)² − 2x(x + 2) cos 100°, which expands to a quadratic with two roots, x = 8.8 and x = −10.8; since x is a length, x = 8.8. Expanding (x + 2)² as x² + 4 instead of x² + 4x + 4, missing the middle term, gives x = 9.6; using +2x(x + 2) cos 100° instead of −2x(x + 2) cos 100° (a sign error on the cosine term) gives x = 10.6; and reporting the size of the rejected negative root instead of discarding it gives x = 10.8. Always expand a squared bracket fully before collecting terms.
- (a) 28.1° — The angle a line makes with the horizontal plane lies in the right-angled triangle formed by the vertical rise, the horizontal distance travelled, and the line itself. The horizontal distance from A to B is √(9² + 12²) = √(81 + 144) = √225 = 15, using only the x- and y-coordinates. The vertical rise is the z-coordinate, 8, so tan(angle) = 8 ÷ 15, giving angle = 28.1° (1 d.p.). Inverting the ratio, tan(angle) = 15 ÷ 8, gives 61.9° instead — the complement of the angle, not the angle with the horizontal. Using only the x-coordinate as if it were the whole horizontal distance, tan(angle) = 8 ÷ 9, gives 41.6°. Using the y-coordinate alone in the same way, tan(angle) = 8 ÷ 12, gives 33.7°.
- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
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