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 pentagon has centre O. Each vertex is 8 cm from O, and each of the 5 triangles formed by joining O to the vertices is isosceles with an angle of 72° at O. Work out the area of the pentagon. Give your answer to 1 decimal place.
- 2.A triangular plot of land has sides AB = 13 m and AC = 10 m, with angle BAC = 72° between them. Work out the perimeter of the plot. Give your answer to 1 decimal place.
- 3.A map has a scale of 1 : 25000. A distance between two points measures 2 cm on the map. What is the real distance, in kilometres?
- 4.Two traffic cones are geometrically similar. The smaller cone has height 6 m and the larger cone has height 9 m. Work out the ratio of the volume of the smaller cone to the volume of the larger cone, giving your answer in the form a : b in its simplest form.
- 5.A parallelogram has an area of 136 cm² and a base of 17 cm. Work out the perpendicular height of the parallelogram.
- 6.A window is made from a rectangle with a semicircle on top. The rectangle is 40 cm wide and 60 cm tall. The semicircle has the same width as the rectangle. Work out the total area of the window. Use π = 3.14. Give your answer to the nearest whole number.
- 7.A support strut for a shed roof is in the shape of a right-angled triangle. The two shorter sides of the strut are 5 m and 9 m. Work out the length of the sloping strut (the hypotenuse). Give your answer correct to 1 decimal place.
- 8.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
- 9.A plumber charges a £45 call-out fee, plus £28 for each hour worked. She works on a job for 3 hours 30 minutes, and her hours are billed by rounding up to the next whole hour. Work out the total amount she charges.
- 10.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.
- 11.A sprinkler waters a sector-shaped patch of a garden with radius 8 m and angle 90°. Work out the perimeter of the watered sector, to 1 decimal place. (Use π = 3.14.)
- 12.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
- 13.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 14.A drone flies from base station D on a bearing of 065° for 14 km to a checkpoint E. At E it changes course and flies on a bearing of 165° for 20 km to a delivery point F. Work out the direct distance from D to F. Give your answer to 1 decimal place.
- 15.In triangle ABC, AB = 9 cm, AC = 6 cm and angle BAC = 110°. Work out the length of BC. Give your answer to 1 decimal place.
Answer key
- (c) 152.2 cm² — Method: find the area of one of the 5 isosceles triangles with 1/2ab sin C, then multiply by 5 for the whole pentagon. Working: one triangle has area 1/2 × 8 × 8 × sin 72° = 30.4338 cm², so the pentagon's area = 5 × 30.4338 = 152.2 cm² — keep the unrounded triangle area, since 5 × 30.4 would give 152.0. Reporting the area of a single triangle and forgetting to multiply by 5 gives 30.4 cm²; multiplying by 6 instead of 5, as for a hexagon, gives 182.6 cm²; and leaving out the 1/2 from each triangle before multiplying by 5 gives 304.3 cm². A regular pentagon splits into exactly 5 triangles at its centre, each with a 72° angle, since 360° ÷ 5 = 72°.
- (c) 36.7 m — Method: use the cosine rule to find the missing side BC, then add all three sides for the perimeter. Working: BC² = 13² + 10² − 2 × 13 × 10 × cos 72°, so BC = 13.7 m, and perimeter = 13 + 10 + 13.7 = 36.7 m. Forgetting the negative sign in the cosine rule (using + instead of −) gives BC = 18.7 m and a perimeter of 41.7 m; leaving AC out of the total and only adding AB and BC gives 26.7 m; and adding AB twice instead of AB and AC gives 39.7 m. Always add all three named sides once the missing one has been found.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (b) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (c) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 cm².
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (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.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (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) 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).
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) 22.3 km — The bearing of E from D is 065°, so the bearing of D from E (the back bearing) is 065° + 180° = 245°. The bearing of F from E is 165°, so the angle at E between ED and EF is 245° − 165° = 80°. Using the cosine rule, DF² = DE² + EF² − 2 × DE × EF × cos(80°) = 14² + 20² − 2 × 14 × 20 × cos(80°) = 196 + 400 − 560 × cos(80°). Since cos(80°) ≈ 0.17365, 560 × cos(80°) ≈ 97.24, so DF² ≈ 596 − 97.24 = 498.76. Taking the square root, DF ≈ 22.333, which rounds to 22.3 km. 26.3 km comes from using 100° as the angle at E — the difference between the two given bearings taken directly (165° − 65°) without converting to the back bearing first. 498.8 km is DF² itself, rounded, with the square root never taken. 23.4 km comes from leaving out the factor of 2 in the cosine rule formula, computing DF² = 196 + 400 − 14 × 20 × cos(80°) ≈ 547.38 instead.
- (c) 12.4 cm — Using the cosine rule, BC² = AB² + AC² − 2 × AB × AC × cos(A) = 9² + 6² − 2 × 9 × 6 × cos(110°) = 81 + 36 − 108 × cos(110°). Since cos(110°) ≈ −0.34202, 108 × cos(110°) ≈ −36.94, so BC² ≈ 117 + 36.94 = 153.94. Taking the square root, BC ≈ 12.4072, which rounds to 12.4 cm. 8.9 cm comes from treating cos(110°) as if it were positive (using +0.342 instead of −0.342), which wrongly subtracts instead of adds and gives BC² ≈ 80.06. 153.9 cm is BC² itself, rounded, with the square root never taken. 11.6 cm comes from leaving out the factor of 2 in the formula, computing BC² = 81 + 36 − 9 × 6 × cos(110°) ≈ 135.47 instead.
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