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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- (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.
- (a) 6 left, 2 up — In a column vector $\binom{x}{y}$, the top number gives the horizontal movement (negative means left) and the bottom number gives the vertical movement (positive means up), so $\binom{-6}{2}$ means 6 units left and 2 units up. '6 right, 2 up' ignores the negative sign on the top number. '2 left, 6 up' swaps the top and bottom numbers. '6 left, 2 down' ignores the positive sign on the bottom number.
- (d) Two lines that cross at right angles — Perpendicular lines are defined as two lines that cross at right angles (90°), so 'two lines that cross at right angles' is correct. 'Two lines that never meet' describes parallel lines, not perpendicular lines, so it is wrong. 'Two lines equal in length' confuses perpendicularity with two lines being the same length, which has nothing to do with the angle between them, so it is wrong. 'Two lines crossing at any angle' is too general, since crossing lines are simply called intersecting unless that angle is specifically 90°, so it is wrong.
- (c) 11.0 cm — Arc length = (angle ÷ 360) × 2 × π × r = (90 ÷ 360) × 2 × 3.14 × 7 = 0.25 × 43.96 = 10.99 cm, which rounds to 11.0 cm. (44.0 cm comes from finding the full circumference, 2 × 3.14 × 7 = 43.96 cm, and forgetting the angle fraction; 5.5 cm comes from leaving out the factor of 2, using π × r instead of 2 × π × r, giving 0.25 × 3.14 × 7 = 5.495 cm; 22.0 cm comes from using the diameter, 14 cm, in place of the radius, giving 0.25 × 2 × 3.14 × 14 = 21.98 cm.)
- (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
- (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) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (c) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (c) 18 — Method: divide the real length by 5 to find how many 'units' of 5 m it contains, then multiply by 2 cm for each unit. Working: 45 ÷ 5 = 9, so the real bridge is 9 lots of 5 m; each lot is represented by 2 cm on the model, so the model length is 9 × 2 = 18 cm. Options: 9 comes from stopping after the division, without multiplying by the 2 cm per unit; 90 comes from multiplying the real length by 2 directly, without dividing by 5 first; 4.5 comes from dividing by 5 and then dividing by 2 again, instead of multiplying by 2. Answer: 18.
- (a) 113.04 cm³ — Method: the volume of a sphere is (4 ÷ 3) × π × r³. Cube the radius, multiply by π, then multiply by 4 and divide by 3. Working: r³ = 3³ = 27, then 3.14 × 27 = 84.78, then 84.78 × 4 = 339.12 and 339.12 ÷ 3 = 113.04. Answer: 113.04 cm³. The distractors: 84.78 cm³ comes from stopping at πr³ and leaving out the four thirds; 37.68 cm³ comes from squaring the radius instead of cubing it, (4 ÷ 3) × 3.14 × 9; 28.26 cm³ comes from using πr², the area of a circle, and labelling it as a volume.
- (c) 27 cm² — Area of a rectangle = length × width = 4.5 × 6 = 27 cm². A pupil who adds the two sides instead of multiplying gets 4.5 + 6 = 10.5 cm². A pupil who works out the perimeter instead of the area gets 2 × (4.5 + 6) = 21 cm². A pupil who rounds 4.5 up to 5 before multiplying gets 5 × 6 = 30 cm². The correct area is 27 cm².
- (c) 78° — The angles in any quadrilateral add up to 360°. Add the three given angles: 92° + 84° + 106° = 282°. Angle S = 360° − 282° = 78°. A pupil who only adds angle P and angle Q, forgetting angle R, gets 360° − (92° + 84°) = 184°. A pupil who only adds angle Q and angle R, forgetting angle P, gets 360° − (84° + 106°) = 170°. A pupil who makes a carrying slip adding the three angles, getting 292° instead of 282°, gets 360° − 292° = 68°. The correct answer is 78°.
- (b) 8.64 m — The real length is 240 times the model length: 3.6 × 240 = 864 cm. Converting to metres, 864 cm ÷ 100 = 8.64 m. The distractor 864 m comes from finding the length correctly in centimetres but forgetting to convert to metres. The distractor 0.0864 m comes from dividing by 100 a second time, converting 864 cm to metres twice over (864 ÷ 10 000) instead of once. The distractor 86.4 m comes from converting centimetres to metres by dividing by 10 instead of 100.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (a) opposite angles of a parallelogram are equal — P and R are opposite vertices of the parallelogram, and opposite angles of a parallelogram are always equal, which is why angle R equals angle P, 65°. Co-interior angles adding up to 180° is the correct fact for angle Q or angle S, the angles adjacent to P along a side, not for the opposite angle R. Alternate angles are equal is a fact about a transversal crossing two parallel lines, which explains other angle relationships in the parallelogram, not the one between opposite angles P and R directly. Angles on a straight line adding up to 180° applies to two angles that sit together on one straight line, which P and R do not.
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