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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (c) 18 — Rearranging F + V − E = 2 gives E = F + V − 2. Substitute F = 8 and V = 12: 8 + 12 − 2 = 18 edges. Choosing 20 comes from adding the faces and vertices but forgetting to subtract the 2 (8 + 12 = 20). Choosing 22 comes from adding the 2 instead of subtracting it (8 + 12 + 2 = 22). Choosing 16 comes from subtracting 2 twice by mistake (8 + 12 − 2 − 2 = 16).
- (a) Two sides equal, with the angle between them equal — Method: recall what each letter in SAS stands for, and the order they must appear in. Working: SAS stands for Side, Angle, Side — two pairs of equal sides, with the equal angle positioned between them. Options: 'two sides and any one angle' describes SSA, which is not valid, since the angle must specifically be the included one; 'two angles with the side between them' describes ASA, a different condition; 'all three pairs of sides' describes SSS, also different. Answer: two sides equal, with the angle between them equal.
- (b) 18.84 m² — The pond has radius 2.5 m (half of the 5 m diameter), so the outer edge of the path has radius 2.5 + 1 = 3.5 m. Path area = area of outer circle − area of pond = 3.14 × 3.5² − 3.14 × 2.5² = 38.465 − 19.625 = 18.84 m².
- (c) 2 — A cylinder has 3 faces in total: 1 curved surface and 2 flat circular faces, the base and the lid. The label covers only the curved surface, leaving the base and the lid uncovered — that is 2 faces. A candidate who forgets one of the two flat faces answers 1. A candidate who mistakenly splits the curved surface into two separate uncovered faces and then adds the 2 flat faces answers 4. A candidate who assumes the label covers the whole tin answers 0. The number of faces not covered by the label is 2.
- (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) = $\binom{4}{−7}$.
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (b) second quadrant — The point (−3, 5) has a negative x-coordinate and a positive y-coordinate, and this combination lies in the second quadrant. The first quadrant needs both coordinates positive. The third quadrant needs both coordinates negative. The fourth quadrant needs a positive x-coordinate and a negative y-coordinate.
- (a) (5, 2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, −2) maps to (5, 2). (−5, −2) changes the sign of the x-coordinate instead, which is what happens when reflecting in the y-axis. (−5, 2) changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the x-axis. (2, 5) comes from dropping the minus sign and then swapping the two numbers round, so neither coordinate has actually been reflected.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
- (b) 045° — North is 000° and east is 090°; halfway between them, measured clockwise, is 045°. Choosing 090° gives east itself, not the midpoint between north and east. Choosing 135° gives the midpoint between east (090°) and south (180°) instead of between north and east. Choosing 225° is the back bearing of 045° (045° + 180°), not the bearing itself.
- (d) (−4, −3) — The combined translation is the sum of the two column vectors, added component by component: top numbers 3 + (−7) = −4, bottom numbers −5 + 2 = −3, giving (−4, −3). (10, −7) subtracts the second vector from the first instead of adding them. (−4, 3) gets the top number right but makes a sign error on the bottom, treating −5 + 2 as +3. (4, −3) gets the bottom number right but makes a sign error on the top, treating 3 + (−7) as +4.
- (b) 3 — The real width is 0.6 × 500 = 300 cm, which converts to 3 m by dividing by 100. A candidate who uses the wrong side of the rectangle, 1.2 cm, instead of the 0.6 cm width, gets 1.2 × 500 = 600 cm = 6 m. A candidate who multiplies correctly but converts the 300 cm to metres by dividing by 1000 instead of 100 gets 0.3 m. A candidate who converts by dividing by 10 instead of 100 gets 30 m. The real width of the bay is 3 m.
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