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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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (a) 12 cm — Area of a parallelogram = base × perpendicular height, so height = area ÷ base = 96 ÷ 8 = 12 cm. A student who adds the area and base instead of dividing gets 96 + 8 = 104 cm. A student who multiplies the area and base instead of dividing gets 96 × 8 = 768 cm. A student who uses the triangle area formula, area = 1/2 × base × height, instead of the parallelogram formula solves 96 = 1/2 × 8 × h and gets h = 24 cm.
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (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) (2, 6) — To add column vectors, add the top numbers together and add the bottom numbers together, separately: top = 3 + (−1) = 2, bottom = 2 + 4 = 6, giving (2, 6). A candidate who writes the top and bottom numbers of the correct answer in the wrong order gives (6, 2). A candidate who subtracts instead of adds, working out 3 − (−1) = 4 and 2 − 4 = −2, gives (4, −2). A candidate who adds the top numbers correctly but subtracts the bottom numbers by mistake, working out 2 − 4 = −2, gives (2, −2). The correct answer is (2, 6).
- (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.
- (c) A line has no endpoints; a segment has two — A line extends without end in both directions, whereas a line segment is the part of a line between two specific fixed endpoints, so 'a line has no endpoints; a segment has two' is correct. 'A line has two endpoints; a segment has none' reverses these two definitions, so it is wrong. 'A line segment is always curved' is wrong because a line segment is straight, not curved, and does not extend infinitely. 'A line segment is a closed shape' is wrong because a line segment is a straight length between two points, not a polygon.
- (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) 13 — Area of the trapezium = 1/2 × (8 + 12) × 5 = 1/2 × 100 = 50 m². Number of bags = 50 ÷ 4 = 12.5, which rounds up to 13 bags since seed is sold only in whole bags. A student who mistakenly uses 2 m² of coverage per bag instead of 4 m² finds 50 ÷ 2 = 25 bags.
- (b) 0 — At 0°, the opposite side of the right-angled triangle has shrunk to zero length, so sin 0° = 0. 1 is the value of sin 90° (mixing up the two angles). 1/2 is the value of sin 30°. 'Undefined' is what happens for tan 90°, not sin 0° — sin 0° has a perfectly good exact value.
- (c) 090 — Method: bearings are measured clockwise from north; east is a quarter turn clockwise. Working: a quarter turn is 90°, written with a leading zero as three figures. A student who answers 009 has kept the correct digits but put the leading zero in the wrong place. A student who answers 180 has confused east with south. A student who answers 270 has confused east with west. Answer: 090.
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
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