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.
Answer key: Geometry and measures worksheet — GCSE Higher
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- (c) VW — Method: within one circle a chord's distance from the centre is fixed by its length, because a chord passing nearer the centre cuts further across the circle; so the chord that lies closest to the centre is simply the longest one listed. Working: the four lengths are 6 cm, 10 cm, 14 cm and 15 cm. Placing them in order, the greatest is 15 cm, and that length belongs to VW, so VW lies closest to the centre. Answer: VW. The distractors: PQ comes from reversing the rule and taking the shortest chord to be the one tucked nearest the centre; TU comes from knowing that the very longest chord is a diameter, deciding that such a chord passes through the centre rather than lying close to it, ruling the 15 cm chord out on that ground and taking the next longest; RS comes from reading 'closest to the centre' as 'nearest the middle of the list of lengths' and picking a middling value.
- (d) SAS, vertically opposite angle included — AE equals CE and BE equals DE give two pairs of equal sides, and angle AEB equals angle CED because they are vertically opposite angles formed where AC and BD cross; vertically opposite angles are always equal without needing to be measured. This included angle sits between the two known sides in each triangle, giving SAS, so 'SAS, vertically opposite angle included' is correct. 'ASA, vertically opposite angle at E' is wrong because ASA needs two pairs of equal angles with the side between them, but only one angle is known in each triangle here, and the two other known facts are sides, not angles. 'SSS, three equal side pairs' is wrong because only two pairs of sides are given; there is no third pair of equal sides. 'Cannot prove — no angle measured' is wrong because vertically opposite angles are always equal automatically when two straight lines cross, so no separate measurement is needed.
- (b) 4 — The scale factor is the distance from the centre to the image, divided by the distance from the centre to the object: (5 − 1) ÷ (2 − 1) = 4 ÷ 1 = 4. (2.5 comes from dividing the raw y-coordinates, 5 ÷ 2, without first subtracting the centre's coordinate; 3 comes from subtracting the two distances instead of dividing them; 0.25 comes from dividing the distances the wrong way round.)
- (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) 8.75 cm — The scale factor of the enlargement is 11.2 ÷ 3.2 = 3.5. Applying this to the second line, 2.5 × 3.5 = 8.75 cm. The distractor 10.50 cm comes from adding the difference between the first line's two lengths (11.2 − 3.2 = 8) to the second line's original length, 2.5 + 8 = 10.5. The distractor 0.71 cm comes from using the scale factor the wrong way round, 2.5 × (3.2 ÷ 11.2) = 0.71 (to 2 d.p.). The distractor 8.70 cm comes from directly subtracting 11.2 − 2.5 = 8.7, muddling the two different lines instead of scaling the second one.
- (a) 32.2° — Method: all three sides are known, so rearrange the cosine rule as cos A = (b² + c² − a²) ÷ 2bc, where a is the side facing the angle you want. Working: angle ABC sits between AB = 8 cm and BC = 11 cm and faces AC = 6 cm, so cos ABC = (8² + 11² − 6²) ÷ (2 × 8 × 11) = (64 + 121 − 36) ÷ 176 = 149 ÷ 176 = 0.84659, and the inverse cosine of 0.84659 is 32.157°. Answer: angle ABC = 32.2° to 1 decimal place. The distractors: 102.6° comes from subtracting the square of the longest side, 11, rather than the square of the side the angle actually faces, giving (64 + 36 − 121) ÷ 96; 57.8° comes from taking the inverse sine of 0.84659 instead of the inverse cosine; 147.8° comes from rearranging with the subtraction the wrong way round, (36 − 64 − 121) ÷ 176 = −0.84659, which turns an acute angle into its supplement.
- (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) Minor segment — Method: compare the sizes of the two regions cut off by the chord, and recall the term used for the smaller one. Working: the chord creates two segments; the smaller region is called the minor segment and the larger one the major segment. A student who answers major segment has picked the larger region by mistake instead of the smaller one. A student who answers minor arc has named the curved boundary rather than the two-dimensional region it encloses. A student who answers semicircle has wrongly assumed the chord must pass through the centre. Answer: minor segment.
- (d) 60° — Method: the 4 cm side is next to the angle wanted and the 8 cm side is the hypotenuse, so the ratio built from them is cos θ = adjacent ÷ hypotenuse, and the angle comes from the inverse cosine. Working: cos θ = 4 ÷ 8 = 0.5, so θ = cos⁻¹(0.5). Answer: 60°. The distractors: 30° comes from using sin⁻¹(0.5), which treats the 4 cm side as the side opposite the angle when it is the side next to it; 45° comes from assuming the two acute angles of the triangle must be equal; 90° comes from writing down the right angle the question already gives instead of the angle it asks for.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (a) $\binom{5}{−6}$ — The vector is (image − original) in each coordinate: (2 − (−3), −1 − 5) = (5, −6). $\binom{−5}{6}$ comes from working out original − image instead of image − original. $\binom{5}{6}$ gets the x-component right but makes a sign error on the y-component. $\binom{−5}{−6}$ gets the y-component right but makes a sign error on the x-component, working out −3 − 2 = −5 instead of 2 − (−3) = 5.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (d) 1535 cm² — The wiper sweeps out a sector of radius 40 cm, the blade length, through an angle of 110°. Sector area is angle ÷ 360 × π × radius²: 110 ÷ 360 × 3.14 × 1600 = 1535.1 cm², which rounds to 1535 cm². Forgetting to square the radius, using radius instead of radius², gives 38 cm². Using 110 ÷ 180 instead of 110 ÷ 360 for the fraction gives 3070 cm². Treating the 40 cm blade length as a diameter, so using a radius of 20 cm, gives 384 cm².
- (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.
- (a) 28.6 m — The perimeter of a sector is the two straight radii plus the curved arc. This sector's 90° angle is one quarter of a full turn, so its arc length is one quarter of the full circle's circumference. The full circumference is 2 × 3.14 × 8 = 50.24 m, and one quarter of that is 50.24 ÷ 4 = 12.56 m. Add the two 8 m radii: 12.56 + 8 + 8 = 28.56 m, which rounds to 28.6 m. Choosing 12.6 m gives the arc length alone (rounded), forgetting the two straight edges of the sector. Choosing 20.6 m adds only one radius to the arc length instead of two, missing one of the two straight sides. Choosing 41.1 m comes from using the diameter, 16 m, as if it were the radius when working out the arc length (2 × 3.14 × 16 = 100.48, one quarter of which is 25.12), then adding the two correct 8 m radii (25.12 + 8 + 8 = 41.12).
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