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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Answer key: Geometry and measures worksheet — GCSE Higher
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- (a) $\binom{-5}{-4}$ — Method: one translation followed by another is a single translation, and the two vectors are added: top to top, bottom to bottom. Working: across, 4 − 9 = −5; up, −7 + 3 = −4. Answer: $\binom{-5}{-4}$. Subtracting the second vector instead of adding it gives 13 on top and −7 − 3 = −10 underneath. Adding the top numbers correctly but subtracting the bottom ones gives −10 underneath with −5 on top. Adding 4 and 9 as though both were positive and then keeping the minus sign of the larger gives −13 on top.
- (c) The scale factor is 2, not 1, so the sides are not equal — Congruent shapes must be exactly the same size as well as the same shape, which means a scale factor of 1. Here the scale factor between the triangles is 2, so the sides are different lengths and the triangles cannot be congruent, even though they are similar. 'Similar triangles are never congruent' is too strong — a scale factor of exactly 1 would make them both similar and congruent. 'The angles are not necessarily equal' is wrong, since similar triangles always have equal matching angles. 'Congruent triangles must have a right angle' is an unrelated, false fact about congruence.
- (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.
- (a) a rounded rectangle: 5 m by 4 m with semicircular ends — Points within 2 m of the straight part of the fence form a rectangle running the 5 m length of the fence and 4 m wide (2 m on each side); points within 2 m of each END of the fence, beyond that rectangle, form a semicircle of radius 2 m there, since the nearest point of the fence to them is just that one end. Together this gives a rounded, stadium-shaped region. "a rectangle, 9 m by 4 m" extends the rectangle by 2 m at each end instead of rounding it, wrongly including corner points that are actually more than 2 m from every part of the fence. "a circle of radius 2 m" treats the whole 5 m fence as a single point. "a rectangle, 5 m by 2 m" uses 2 m as the full width instead of the distance on EACH side, so it only covers one side of the fence.
- (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
- (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) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (b) (6, 3) — For an enlargement centred on the origin, multiply both coordinates by the scale factor: (2 × 3, 1 × 3) = (6, 3). A pupil who adds the scale factor to each coordinate instead of multiplying gets (2 + 3, 1 + 3) = (5, 4). A pupil who multiplies only the x-coordinate gets (6, 1). A pupil who multiplies only the y-coordinate gets (2, 3). The correct image is (6, 3).
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (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) = $\binom{4}{−7}$.
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (b) Square — A square has all four sides equal, all four angles equal to 90°, and diagonals that are equal in length and bisect each other at right angles — every part of the description matches, so Square is correct. A rhombus has all four sides equal and diagonals bisecting at right angles, but its interior angles are not generally 90° (only a square, a special rhombus, has that), so it does not fully match. A rectangle has four 90° angles and equal diagonals, but its sides are not all equal in general, so it fails the equal-sides condition. A kite has two pairs of adjacent equal sides rather than all four sides equal, and its diagonals are not generally equal in length, so it fails both conditions.
- (c) (1, 0) — To enlarge about a centre other than the origin, find the vector from the centre to the point, scale that vector, then add it back to the centre. The vector from (2, 2) to A(4, 6) is (2, 4). Scaling by −1/2 gives (−1, −2). Adding this to the centre (2, 2) gives the image point (1, 0). (3, 4) comes from using +1/2 instead of −1/2, so the image lands on the same side as A instead of the opposite side. (−2, −3) comes from scaling A's coordinates directly about the origin, ignoring that the centre is (2, 2). (−2, −6) comes from using a scale factor of −2 instead of −1/2.
- (c) 2 — A point 4 cm from A lies on a circle of radius 4 cm centred at A; a point 3 cm from B lies on a circle of radius 3 cm centred at B. Since AB = 5 cm, and 4 + 3 = 7 is greater than 5 while 4 − 3 = 1 is less than 5, the two circles genuinely cross each other, at two separate points, one on each side of line AB. "1" comes from wrongly assuming the circles only touch rather than cross, which would need 4 + 3 to equal exactly 5. "0" comes from wrongly assuming the circles miss each other completely. "4" comes from counting where each circle crosses the line AB itself (two points each) instead of counting where the two circles cross each other.
- (a) 7.7 m — AB is parallel to DC, so those two sides are each parallel to another side. BC and AD are stated to be not parallel to each other, so neither one is parallel to any other side — these are the two sides that need edging. Adding these: 3.2 + 4.5 = 7.7 m, so 7.7 m is correct. 7.6 m comes from an arithmetic slip when adding 3.2 and 4.5. 15.4 m comes from doubling the correct total, mistakenly assuming edging strip is needed along both faces of each side. 4.5 m comes from using only the longer of the two non-parallel sides and forgetting to add the shorter one.
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