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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- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (b) angle B = angle E — AB and BC meet at vertex B, so the angle INCLUDED between them is angle B; making angle B = angle E completes SAS. Angle A sits between AB and AC, not between AB and BC, so it is not the included angle needed for SAS here. Angle C sits between BC and CA, not between AB and BC, so it is not included either. AC = DF would give a third pair of equal sides, which proves congruence by SSS instead of SAS.
- (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.
- (b) (12, −8) — Method: to multiply a column vector by a number, multiply every part of the vector by that number. Working: top number 4 × 3 = 12; bottom number 4 × (−2) = −8. Answer: 4a = (12, −8). A candidate who adds 4 to each part instead of multiplying gets (7, 2). A candidate who multiplies only the top number by 4 and leaves the bottom number unchanged gets (12, −2). A candidate who multiplies only the bottom number by 4 and leaves the top number unchanged gets (3, −8).
- (b) circumference — The distance all the way around the outside edge of a circle is called the circumference. The diameter is a straight line across the circle through the centre, so it is a length through the circle, not around it. The radius is a straight line from the centre to the edge, again a length across, not around. The area is the amount of surface inside the circle, a region, not a length at all.
- (b) 4 — A triangular prism has 5 faces in total: 2 triangular ends and 3 rectangular faces. One rectangular face is the groundsheet, lying on the ground, so the other 5 − 1 = 4 faces are above the ground. A candidate who forgets one of the triangular ends when counting the remaining faces answers 3. A candidate who forgets to subtract the groundsheet at all answers 5. A candidate who only counts the two sloped rectangular faces, forgetting the two triangular ends, answers 2. The number of faces above the ground is 4.
- (c) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (a) 59° — Method: the three angles of a triangle add up to 180°, and in an isosceles triangle the two angles opposite the equal sides are equal, so take the known angle away from 180° and share what is left equally between the other two. Working: 180° − 62° = 118°, and 118° ÷ 2 = 59°. Answer: 59°. The distractors: 118° comes from taking 62° from 180° and stopping there, which gives the two angles together rather than one of them; 31° comes from halving the 62° that is given instead of halving what is left; 62° comes from assuming that all three angles of the triangle are equal to the one that is given.
- (c) (6, −5) — Method: the translation has already happened, so it must be undone: reverse the vector and apply the reverse to the point that is given. Working: reversing $\binom{-2}{6}$ gives $\binom{2}{-6}$, so the x-coordinate is 4 + 2 = 6 and the y-coordinate is 1 − 6 = −5. Check: from (6, −5) the given vector gives 6 − 2 = 4 and −5 + 6 = 1, which is the point named in the question. Answer: (6, −5). Applying the vector forwards instead of backwards gives (2, 7). Reversing the horizontal movement but not the vertical one gives (6, 7), and reversing the vertical movement but not the horizontal one gives (2, −5).
- (b) 3 — A cuboid with all different edge lengths has three planes of symmetry: one parallel to each pair of opposite faces, cutting the solid exactly in half. Choosing 9 is the number of planes of symmetry a CUBE has (where all edges are equal) — this cuboid's edges are all different, so it has fewer. Choosing 1 counts only one of the three planes and forgets the other two, each parallel to a different pair of faces. Choosing 6 double-counts each of the three planes, as if counting each one from both sides.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (c) Radius — Method: recall that a sector is formed using two straight lines drawn from the centre out to the circle's edge. Working: each straight edge of a sector runs from the centre of the circle to a point on the circumference. A student who answers chord has confused a straight edge from the centre with one joining two points on the circumference. A student who answers diameter has wrongly assumed the two straight edges must form a single full diameter. A student who answers arc has named the curved edge instead of the straight edges. Answer: radius (radii).
- (c) $\binom{5}{-2}$ — Method: the top number of a column vector is the change in the x-coordinate and the bottom number is the change in the y-coordinate, each worked out as image minus object. Working: across, 7 − 2 = 5; up, 3 − 5 = −2. So the point moves 5 to the right and 2 down. Answer: $\binom{5}{-2}$. Subtracting the other way round, object minus image, gives $\binom{-5}{2}$, which is the journey from B back to A. Recording the vertical change as 2 because the gap between 3 and 5 is 2, without noting the direction, gives $\binom{5}{2}$, a movement 2 upwards. Adding the coordinates instead of subtracting them gives $\binom{9}{8}$.
- (c) RHS applies: the right angle, hypotenuse and one matching side are all equal — Both triangles are right-angled, and the hypotenuses (AC and DF) are equal, and one further pair of matching sides (AB and DE) are equal. This matches the Right angle-Hypotenuse-Side condition, which needs only these three facts, not a third side.
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