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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- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (d) Equilateral — Method: triangles are named by their sides — three equal sides, exactly two equal sides, or no equal sides — or by their angles. Working: the triangle described has three equal sides, which is the equilateral case, and each of its angles is 60°. Answer: equilateral. The distractors: isosceles is the name for a triangle with exactly two equal sides, and is chosen by candidates who remember only that it is the 'equal sides' word; scalene is the name for a triangle whose sides are all different, so it is the opposite of what is described; right-angled classifies a triangle by an angle of 90° rather than by its sides, and is chosen by candidates who recall that the angles of this triangle are all equal and take 'equal angles' to mean 'right angles'.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (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) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
- (a) 2 — Method: recall that the diagonals of a rhombus are always lines of symmetry, whatever its angles are. Working: a rhombus (all sides equal) always has its two diagonals as lines of symmetry, giving 2 lines of symmetry, whether or not the angles are 90°. Options: 0 wrongly assumes a non-square rhombus has no symmetry at all; 4 comes from the number of lines of symmetry a square has, mistaking this rhombus for a square; 1 comes from treating the rhombus like a kite, which has only one diagonal as a line of symmetry. Answer: 2.
- (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
- (a) £4.20 — 350 g is 3.5 lots of 100 g, since 350 ÷ 100 = 3.5, so the cost is 3.5 × £1.20 = £4.20. Multiplying the mass in grams directly by the price, without dividing by 100 first, gives £420.00. Working out 100 ÷ 350 instead of 350 ÷ 100 inverts the ratio and gives about £0.34. Rounding 350 g down to 300 g gives 3 × £1.20 = £3.60.
- (a) (3, 9) — Two column vectors are parallel only when one is the other multiplied by a single scalar — both numbers scaled by the same factor with the same sign. r has top:bottom ratio 2:6, which simplifies to 1:3. (3, 9) has ratio 3:9, also 1:3, so (3, 9) = 1.5 × r: it is parallel to r. (6, 2) swaps the top and bottom numbers of r, giving ratio 6:2 (3:1), which is not the same ratio. (2, 3) halves only the bottom number of r, giving ratio 2:3, not 1:3. (−2, 6) changes the sign of only the top number of r, giving ratio −2:6, which no longer matches 2:6 throughout.
- (d) 1250 g — There are 1000 g in a kilogram, so 1.25 kg = 1.25 × 1000 = 1250 g. Multiplying by 100 instead of 1000 gives 125 g. Converting only the whole 1 kg and forgetting the extra 0.25 kg gives 1000 g. Treating the 0.25 kg as 25 g instead of 250 g, a quarter of 1000, gives 1025 g.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (c) (4, 3) — Method: the midpoint of a segment is the mean of its two end points, so its x-coordinate is the mean of the two x-coordinates and its y-coordinate is the mean of the two y-coordinates. Working: for x, (1 + 7) ÷ 2 = 8 ÷ 2 = 4. For y, (3 + 3) ÷ 2 = 6 ÷ 2 = 3. The midpoint is therefore (4, 3). Answer: (4, 3). The distractors: (3, 3) comes from halving the difference of the x-coordinates, (7 − 1) ÷ 2 = 3, which measures half the distance instead of locating the point; (3.5, 3) comes from halving only the larger x-coordinate and leaving the smaller one out of the working; (4, 0) comes from averaging the x-coordinates correctly but then subtracting the y-coordinates, 3 − 3, rather than averaging them.
- (a) 9 — AB is a vertical segment, since A and B share the x-coordinate 1, and its length is the difference in y-coordinates: 5 − 1 = 4. BC is a horizontal segment, since B and C share the y-coordinate 5, and its length is the difference in x-coordinates: 6 − 1 = 5. The total path length is 4 + 5 = 9. 20 comes from multiplying the two lengths, 4 × 5, instead of adding them. 5 is only the length of BC, forgetting to include AB. 4 is only the length of AB, forgetting to include BC.
- (c) 20 cm² — The area of a triangle is half of base × height. First, base × height = 8 × 5 = 40. Half of 40 is 20 cm². 40 cm² forgets to halve and just gives base × height. 13 cm² adds the base and height together instead of multiplying them. 80 cm² doubles base × height instead of halving it.
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