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.A circular plate has a diameter of 20 cm. Work out the area of the plate. Use π = 3.14.
- 3.A sector of a circle has radius 15 cm and angle 216°. Using π = 3.14, work out the arc length of the sector.
- 4.In triangle ABC, AB = 6 cm, angle ABC = 40° and AC = 9 cm. Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC, which gives two possible values for angle ACB: 25.4° or 154.6° (each to 1 decimal place). By considering the angle sum of the triangle, decide which of these two values angle ACB can actually take, and hence work out angle BAC. Give your answer to 1 decimal place.
- 5.In kite WXYZ, WX = WZ and XY = ZY (so at each of X and Z, one side from each of the two unequal pairs meets). Angle X = 100°. Work out angle Z.
- 6.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder.
- 7.A dog is tied by a lead 4 m long to a fixed ring on a straight garden wall. The wall extends much further than the lead can reach in both directions, and the dog cannot cross through it. Describe the region the dog can reach.
- 8.A shape is reflected in the line y = −x. Which of these points is invariant under this reflection?y = −x
- 9.A roof truss has two horizontal parallel rafters, one above the other. A straight strut crosses both rafters. Where the strut crosses the lower rafter, the angle above the rafter and to the left of the strut is 65°. Work out the size of the angle above the upper rafter and to the right of the strut, where the strut crosses it.
- 10.In triangle ABC, BC = 8 cm, AC = 10 cm and angle BAC = 40°. Given that angle ABC is acute, work out its size. Give your answer to 1 decimal place.
- 11.In triangle ABC, AB = 8 cm, AC = 7 cm and the area of the triangle is 24 cm². Given that angle BAC is acute, work out the size of angle BAC. Give your answer to 1 decimal place.
- 12.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 13.m is the column vector with top number 3 and bottom number −4. Which of these column vectors is a scalar multiple of m?
- 14.Triangle A has vertices (2, 1), (5, 1) and (2, 4). It is translated by the vector (−3, 2) and the image is then reflected in the x-axis. Work out the coordinates of the image of (5, 1).
- 15.Triangle S has vertices (2, 2), (5, 2) and (2, 5). It is mapped onto triangle S′ with vertices (2, 5), (5, 5) and (2, 2). Which single composition of two transformations maps S onto S′?
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.
- (a) 314 cm² — Method: the area of a circle is πr², and the radius is half the diameter, so halve the 20 cm before squaring. Working: r = 20 ÷ 2 = 10 cm, so the area is 3.14 × 10² = 3.14 × 100 = 314. Answer: 314 cm². The distractors: 1256 cm² comes from putting the diameter straight into πr² without halving it, 3.14 × 20²; 628 cm² comes from halving correctly but then using 2πr², a mixture of the circumference and area formulae; 62.8 cm² comes from working out πd = 3.14 × 20, which is the circumference of the plate rather than its area.
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (a) 114.6° — Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC = 6 × sin(40°) ÷ 9. Since sin(40°) ≈ 0.64279, this gives sin(ACB) ≈ 3.8567 ÷ 9 ≈ 0.42852, so angle ACB ≈ 25.4° or its supplement, 154.6°. Testing the obtuse candidate: 40° + 154.6° = 194.6°, which already exceeds 180°, so angle BAC would have to be negative — impossible, so 154.6° is rejected. With angle ACB ≈ 25.4°, angle BAC = 180° − 40° − 25.4° = 114.6°. 25.4° is angle ACB, not angle BAC that the question asks for. 14.6° comes from using the invalid 154.6° candidate anyway and then wrongly turning the resulting negative angle sum (−14.6°) positive instead of rejecting it. 65.4° comes from inverting the sine rule ratio — dividing AC × sin(ABC) by AB instead of AB × sin(ABC) by AC — which gives a different, incorrect candidate for angle ACB entirely.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
- (c) 282.6 cm³ — Volume of a cylinder = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. (94.2 cm³ comes from using πrh and forgetting to square the radius; 90 cm³ comes from using r²h and leaving π out altogether; 1130.4 cm³ comes from using the diameter, 6 cm, in place of the radius.)
- (b) A semicircle of radius 4 m, away from the wall. — Every point the dog can reach is at most 4 m from the fixed ring, so without any wall the region would be a full circle of radius 4 m. The wall runs straight through the ring and blocks the dog from crossing it, and since the wall extends further than the lead in both directions, exactly half of that circle is cut off — leaving a semicircle of radius 4 m on the side of the wall the dog is tied on. (A full circle of radius 4 m ignores that the wall blocks half of the region; a quarter circle of radius 4 m would only be correct if the ring were fixed at a corner where two walls met, not along a single straight wall; a rectangle 4 m wide along the wall ignores that the lead lets the dog swing round in a curve, not stay a fixed distance out from the wall.)
- (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) 115 — Method: use corresponding angles to carry the 65° angle from the lower rafter up to the upper rafter, then use angles on a straight line to move to the other side of the strut. Working: the angle above the upper rafter and to the left of the strut corresponds to the given angle, so it is 65°; the angle above the upper rafter and to the right of the strut lies on a straight line with it, so it is 180 − 65 = 115. Answer: 115°. A candidate who assumes the angle stays 65° without allowing for the move from the left of the strut to the right of it gives 65. A candidate who uses 90° instead of 180°, working out 90 − 65, gets 25. A candidate who adds instead of subtracting, working out 180 + 65, gets 245.
- (a) 53.5° — Method: use the sine rule BC/sin A = AC/sin B, since BC is opposite angle A and AC is opposite angle B. Working: sin(ABC) = AC × sin(BAC) / BC = 10 × sin 40° / 8, which gives angle ABC = 53.5° or its supplement 180 − 53.5 = 126.5°; since angle ABC is acute, the answer is 53.5°. Taking the obtuse supplement instead gives 126.5°; putting the sides the wrong way round in the ratio (sin B = BC × sin A / AC instead of AC × sin A / BC) gives 30.9°; and using sin 50° in place of sin 40° gives 73.2°. Every sine-rule ratio has two possible angle solutions that add to 180° — the word 'acute' tells you which one to keep.
- (a) 59.0° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject: sin C = 2 × Area ÷ (a × b). Working: sin C = 2 × 24 ÷ (8 × 7) = 0.857, so C = sin⁻¹(0.857) = 59.0° (1 d.p.), which is acute as the question requires. Answer: 59.0°. Taking the obtuse angle instead of the acute one asked for, 180 − 59.0 = 121.0°; forgetting to double the area before dividing gives sin C = 24 ÷ (8 × 7) = 0.4286, whose acute angle is 25.4°; and combining that same forgotten-doubling error with the obtuse branch gives 180 − 25.4 = 154.6°. Always double the area first, and then pick the acute branch, since that is what this question asks for.
- (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.
- (b) (6, −8) — Method: a scalar multiple of m has the same ratio between its top and bottom numbers as m does. Working: m = (3, −4); multiplying both parts by 2 gives 2 × 3 = 6 and 2 × (−4) = −8, so (6, −8) is a scalar multiple of m. Answer: (6, −8). The vector (6, −4) needs a multiplier of 2 for the top number but only 1 for the bottom number, so it is not a multiple. The vector (−6, −8) needs a multiplier of −2 for the top number but 2 for the bottom number, so it is not a multiple. The vector (9, −8) needs a multiplier of 3 for the top number but 2 for the bottom number, so it is not a multiple.
- (c) (2, −3) — Method: apply the transformations in the order given, translation first, then reflection, to the point itself. Working: translating (5, 1) by the vector (−3, 2) moves the x-coordinate by −3 and the y-coordinate by 2, giving 5 − 3 = 2 and 1 + 2 = 3, so the image after the translation is (2, 3). Reflecting this point in the x-axis leaves the x-coordinate unchanged and reverses the sign of the y-coordinate, giving (2, −3). Answer: (2, −3). Do the two maps in the stated order and carry the point all the way through both of them: swapping the order, or stopping after the first map, lands you on a different point altogether.
- (a) Reflect in the x-axis, then translate by (0, 7). — Reflecting in the x-axis sends (x, y) to (x, −y); applied to S's vertices (2, 2), (5, 2) and (2, 5) this gives (2, −2), (5, −2) and (2, −5). Translating this image by the vector (0, 7) adds 7 to every y-coordinate, giving (2, 5), (5, 5) and (2, 2), which matches S′ exactly. Using the correct reflection but translating by (7, 0) instead moves the image sideways rather than upwards, giving (9, −2), (12, −2) and (9, −5) — nowhere near S′. Reflecting in the y-axis instead of the x-axis changes the sign of the x-coordinate rather than the y-coordinate, so translating that image by (0, 7) gives (−2, 9), (−5, 9) and (−2, 12), the wrong shape entirely. Rotating 180° about the origin instead of reflecting sends every coordinate to its negative, so translating by (0, 7) gives (−2, 5), (−5, 5) and (−2, 2) — the y-coordinates match S′ but the x-coordinates do not.
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