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 birthday cake is a cylinder of radius 10 cm and height 8 cm, with a thin ribbon fixed exactly once around the curved side, at half the height, and joined with no overlap. Work out the length of ribbon needed. Use π = 3.14.
- 2.A furniture catalogue drawing uses a scale of 1 cm to 1.5 m. A sofa in the drawing is measured as 7 cm long. What is the real length of the sofa, in metres?
- 3.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 4.A composite solid is made from a cylinder of radius 3 cm and height 10 cm, with a cone of the same radius and vertical height 4 cm fixed on top, apex upward. Work out the total volume of the solid. Use π = 3.14 and give your answer correct to 1 decimal place.
- 5.A sector of a circle has radius 12 cm and angle 45°. Using π = 3.14, work out the area of the sector.
- 6.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
- 7.A solid cone has a base radius of 6 cm and a vertical height of 10 cm. Work out the volume of the cone. Use π = 3.14 and give your answer correct to 1 decimal place.
- 8.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.
- 9.A student draws a net using 5 identical squares arranged in a row of four with one extra square attached to the side of one of them. Can this net be folded to make a closed cube?
- 10.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 11.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.
- 12.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 13.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.
- 14.Write down the exact value of sin 60°.
- 15.A sector of a circle has angle 120° and an arc length of 31.4 cm. Using π = 3.14, work out the radius of the circle.
Answer key
- (a) 62.8 cm — The ribbon goes once around the circular cross-section, so its length equals the circumference: 2πr = 2 × 3.14 × 10 = 62.8 cm.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (b) 320.3 cm³ — Cylinder volume = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. Cone volume = (1/3)πr²h = (1/3) × 3.14 × 9 × 4 = (1/3) × 113.04 = 37.68 cm³. Total = 282.6 + 37.68 = 320.28 cm³, which rounds to 320.3 cm³.
- (c) 56.52 cm² — Sector area = (45 ÷ 360) × 3.14 × 12² = 0.125 × 3.14 × 144 = 56.52 cm². (4.71 cm² comes from forgetting to square the radius; 452.16 cm² comes from finding the area of the whole circle and forgetting the angle fraction; 9.42 cm² comes from using the arc length formula instead of the sector area formula.)
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (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.
- (b) No, it needs one more square — A closed cube has exactly 6 faces, so its net must be made of exactly 6 identical squares, arranged so each one unfolds to a separate face with none overlapping. This net has only 5 squares, so it is one square short and cannot be folded into a closed cube. Choosing 'Yes, it folds into a cube' ignores that a cube needs 6 faces, not 5. Choosing 'No, it has one square too many' miscounts in the wrong direction — 5 is one too FEW, not one too many. Choosing 'Yes, but only if two squares overlap' is not a valid net: a net's faces must not overlap when folded.
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (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.
- (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) 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) √3/2 — Method: sin 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2 and base 1, and Pythagoras gives its height as √3 because 2² − 1² = 3. Working: sine is the opposite side over the hypotenuse, the side opposite the 60° angle is the height √3, and the hypotenuse is 2, so sin 60° = √3 ÷ 2. Answer: sin 60° = √3/2. Swapping sine for cosine at this angle gives cos 60° = 1/2; taking the value from the other special triangle, the right-angled isosceles one, gives sin 45° = √2/2; and reading the tangent row instead of the sine row gives tan 60° = √3.
- (d) 15 cm — Arc length = (angle ÷ 360) × 2 × π × r. Here 120 ÷ 360 = 1/3, and 2 × 3.14 = 6.28, so 31.4 = (1/3) × 6.28 × r. Multiplying both sides by 3 gives 6.28 × r = 94.2, so r = 94.2 ÷ 6.28 = 15 cm. (5 cm comes from forgetting the angle fraction altogether and dividing the arc length by 2 × π alone: 31.4 ÷ 6.28 = 5; 30 cm comes from leaving out the factor of 2, dividing by (1/3) × 3.14 = 1.0467 instead of (1/3) × 6.28: 31.4 ÷ 1.0467 = 30; 7.5 cm comes from correctly finding a radius of 15 cm but then treating that 15 cm as a diameter and halving it.)
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