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.A regular hexagon is divided into six identical triangles by joining its centre to each of the six vertices. Work out the size of the angle of one of these triangles at the centre of the hexagon.
- 2.Triangle ABC has AB = 9 cm, BC = 6 cm and angle A = 35°. Triangle DEF has DE = 9 cm, EF = 6 cm and angle D = 35°. Which of the following correctly identifies the condition shown here?
- 3.Point P has coordinates (4, 2). P is rotated 90° clockwise about the point (1, 1), and the image is then reflected in the line x = 1. Work out the coordinates of the final image of P.
- 4.Triangle T has an area of 6 cm². It is enlarged by scale factor −2 about the origin, and the image is then rotated 90° about the origin. Work out the area of the final image, and state whether it is the same way round as T or a mirror image of it.
- 5.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = (2x + 10)° and angle DCB = (3x − 5)°. Work out the size of angle DAB.
- 6.A courier drone delivers a parcel. It starts at the point (2, 1) on a map grid measured in kilometres. It flies by the vector to a warehouse, then by the vector to the delivery address. Work out the coordinates of the delivery address.
- 7.A shop stacks tins in a display block. The front elevation of the block is 4 tins wide and 3 tins high. The side elevation shows the block is 2 tins deep. Every position in the block is filled. How many tins are used in total?
- 8.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 9.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.
- 10.OABC is a parallelogram, with OA = a and OC = c. X is the midpoint of the diagonal AC. Express the vector OX in terms of a and c, and use it to show that X also lies on the diagonal OB.
- 11.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = (4x + 20)°. Work out the size of angle ABC in terms of x, in its simplest form.
- 12.A tap fills a tank at a rate of 18 litres per minute. The tank holds 0.45 m³. Work out how long the tap takes to fill the tank.
- 13.A solid pyramid has a square base of side 10 cm and four identical triangular faces. Its apex is directly above the centre of the base, at a vertical height of 12 cm, and the slant height of each triangular face — the distance from the apex to the midpoint of a base edge — is 13 cm. Work out the total surface area of the pyramid.
- 14.Two similar triangles have lengths in the ratio 3 : 4. The perimeter of the smaller triangle is 24 cm. Work out the perimeter of the larger triangle.
- 15.ABCD is an isosceles trapezium. The sides AB and DC are parallel, and the two sloping sides AD and BC are equal in length. Angle D is 112°. Work out the size of angle B.
Answer key
- (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (d) (0, −2) — To rotate (4, 2) by 90° clockwise about (1, 1), first find its position relative to the centre: (4 − 1, 2 − 1) = (3, 1). A 90° clockwise rotation sends (a, b) to (b, −a), so (3, 1) becomes (1, −3); adding the centre back gives (1 + 1, 1 − 3) = (2, −2). Reflecting (2, −2) in the line x = 1 gives (2 × 1 − 2, −2) = (0, −2). Doing the two transformations in the opposite order, reflecting first and then rotating, gives a different result, (2, 4), which shows the order matters. Stopping after the rotation and forgetting the reflection gives (2, −2). Stopping after only reflecting P in x = 1 and forgetting the rotation entirely gives (−2, 2). Rotate first, then reflect, in that order, and the final image is (0, −2).
- (d) 24 cm²; same way round as T — An enlargement scales area by the square of the scale factor, whatever its sign: area factor = (−2)² = 4, so the image's area is 6 × 4 = 24 cm². A negative scale factor is equivalent to an enlargement by the positive scale factor together with a rotation of 180°, so it does not create a mirror image — it is still a direct transformation, and combining it with a further rotation cannot create one either, since rotations never change a shape's orientation. Using the scale factor itself rather than its square gives area 6 × 2 = 12 cm², which is too small. Believing that a negative scale factor reflects the shape gives 'a mirror image of T' alongside the correct area of 24 cm², and combining both mistakes gives 12 cm² together with 'a mirror image of T'.
- (b) 80° — Method: opposite angles of a cyclic quadrilateral sum to 180°, so form and solve an equation for x, then substitute back to find angle DAB. Working: (2x + 10) + (3x − 5) = 180, so 5x + 5 = 180, x = 35, and angle DAB = 2(35) + 10 = 80°. Using 360° instead of 180° as the total (as if the two angles were the whole circle rather than a pair of opposite angles) gives x = 71 and angle DAB = 152°; solving correctly for x but reporting x itself instead of substituting it back into 2x + 10 gives 35°; and mishandling the signs when combining the two expressions, using (2x + 10) + (3x + 5) = 180 instead of (2x + 10) + (3x − 5) = 180, gives x = 33 and angle DAB = 76°. Opposite angles of a cyclic quadrilateral — not adjacent ones — are the pair that sums to 180°.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (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).
- (d) (1/2)a + (1/2)c — Method: X is the midpoint of AC, so OX = OA + (1/2)AC, with AC = c − a. Working: OX = a + 1/2(c − a) = a − (1/2)a + (1/2)c = (1/2)a + (1/2)c. Answer: OX = (1/2)a + (1/2)c. Since OB = a + c, this is exactly half of OB, so OX = (1/2)OB, meaning X lies on OB at its midpoint too — the two diagonals bisect each other. Forgetting to halve AC at all gives a + c, which is OB itself, not its midpoint; halving only the c-term gives (1/2)a + c; and a sign error on the c-term gives (1/2)a − (1/2)c. Halve the whole of AC, both terms together, and add it to OA rather than to a alone.
- (b) (2x + 10)° — By the angle at the centre theorem, angle ABC is half of angle AOC, because both stand on the same arc AC: angle ABC = (4x + 20)° ÷ 2 = (2x + 10)°. Writing down the centre angle itself, without halving at all, gives (4x + 20)°. Halving only the constant term and leaving the x-term unchanged gives (4x + 10)°. Doubling the centre angle instead of halving it gives (8x + 40)°. Halve every term in the expression, and (2x + 10)° is what you get.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
- (c) 32 cm — Method: a perimeter is lengths added together, so it scales by the length scale factor itself, which is 4 ÷ 3 going from the smaller triangle to the larger one — not by its square. Working: 24 ÷ 3 = 8, and 8 × 4 = 32. Answer: 32 cm. The distractors: 18 cm comes from multiplying by 3 ÷ 4, scaling from the larger triangle down to the smaller one; 25 cm comes from adding the difference between the parts of the ratio, 4 − 3 = 1, to the perimeter; 8 cm comes from dividing by 3 and stopping there, before multiplying by 4.
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
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