Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation 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 Foundation
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- 1.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 2.In a right-angled triangle, θ is one of the acute angles. Write down the trigonometric ratio, in terms of the opposite and adjacent sides, that is used to find θ.
- 3.In a right-angled triangle one of the other two angles is 45°, and the side opposite that 45° angle is 7 cm. Work out the length of the hypotenuse. Give your answer to 1 decimal place.
- 4.A drone starts at the point (3.5, −2) on a grid measured in metres. It flies by the vector to check a first sensor. The drone then needs to fly in a straight line to reach the point (−1.7, 9) to check a second sensor. Work out the column vector of this second flight.
- 5.A circular coin has a diameter of 3 cm. Work out the circumference of the coin. Give your answer in terms of π.
- 6.A sector of a circle has an angle of 90° and an arc length of 15.7 cm. Using π = 3.14, work out the radius of the circle.
- 7.A rectangular garden bed measures 5 m by 3 m. A semicircular flower border is attached along one of the 3 m-wide ends, on the outside of the rectangle, so that the 3 m side is the diameter of the semicircle. Work out the total perimeter of the combined shape (the two long sides, the one remaining short side, and the curved edge). Use π = 3.14.
- 8.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 9.A rectangle has vertices A(0, 0), B(6, 0), C(6, 4) and D(0, 4). Work out the length of the diagonal AC. Give your answer correct to 2 decimal places.
- 10.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector to pick up a beam, then along the vector to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
- 11.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.
- 12.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 13.A garden flower bed is in the shape of a trapezium. Its two parallel edges are 3.5 m and 6.5 m long, and the perpendicular distance between them is 4 m. Bedding plants cost £2.50 per square metre to cover the bed. Work out the total cost of the bedding plants.
- 14.A cuboid-shaped shipping crate has a length of 12 m, a width of 5 m, and a volume of 360 m³. Work out the height of the crate, in metres.
- 15.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
Answer key
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (c) tan θ = opposite/adjacent — The tangent ratio is defined as tan θ = opposite/adjacent, so this is the ratio that connects the opposite and adjacent sides. "sin θ = opposite/adjacent" wrongly labels this ratio as sine, when sine actually connects the opposite side and the hypotenuse. "cos θ = opposite/adjacent" wrongly labels it as cosine, when cosine connects the adjacent side and the hypotenuse. "tan θ = adjacent/opposite" uses the correct ratio name but has the opposite and adjacent sides the wrong way round.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
- (c) 3π cm — Circumference = πd. With d = 3 cm, circumference = π × 3 = 3π cm. A student who uses the radius (1.5 cm) instead of the diameter in the formula gets 1.5π cm.
- (b) 10 cm — Arc length = (angle ÷ 360) × 2 × π × r, so 15.7 = (90 ÷ 360) × 2 × 3.14 × r = 1.57r, giving r = 15.7 ÷ 1.57 = 10 cm. (2.5 cm comes from forgetting the angle fraction and dividing by 2π alone; 20 cm comes from leaving out the factor of 2 in the arc length formula before dividing; 5 cm comes from dividing the arc length by π alone, ignoring both the angle fraction and the factor of 2.)
- (c) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (d) 7.21 units — Method: the diagonal AC is the hypotenuse of the right-angled triangle ABC, whose shorter sides are AB and BC, so Pythagoras' theorem gives its length. Working: AB runs from (0, 0) to (6, 0), so AB = 6; BC runs from (6, 0) to (6, 4), so BC = 4. Then AC² = 6² + 4² = 36 + 16 = 52, so AC = √52 = 7.2111…, which is 7.21 correct to 2 decimal places. Answer: 7.21 units. The distractors: 10.00 units comes from adding the two sides, 6 + 4, instead of adding their squares and taking the root; 4.47 units comes from subtracting the squares, √(36 − 16), which is the form of Pythagoras used to find a shorter side rather than the hypotenuse; 26.00 units comes from halving 52 in place of taking its square root.
- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
- (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.
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (b) 6 m — Volume of a cuboid = length × width × height, so height = volume ÷ (length × width) = 360 ÷ (12 × 5) = 360 ÷ 60 = 6 m. A pupil who divides the volume by the length only gets 360 ÷ 12 = 30 m. A pupil who divides the volume by the width only gets 360 ÷ 5 = 72 m. A pupil who subtracts length × width from the volume instead of dividing gets 360 − 60 = 300 m. The correct height is 6 m.
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
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