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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- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (c) 36 cm² — The area of a parallelogram is base × height = 9 × 4 = 36 cm². 26 cm² is 2 × (9 + 4), the perimeter you would get by treating the perpendicular height as though it were a side length — and a perimeter is a length, not an area. 18 cm² uses the triangle formula, half of base × height, by mistake. 72 cm² doubles the correct area.
- (b) The arcs only meet exactly on AB, not above or below it. — With a radius of exactly half of AB, the arc centred at A and the arc centred at B each reach precisely to the midpoint of AB, so they meet only at that one point, on the line AB itself — there are no two intersection points above and below the line to join, so the perpendicular bisector cannot be drawn. (The claim that this radius is always too short to draw any arc is wrong, since a radius equal to half of AB is a perfectly valid, positive length for a compass arc; the claim that the radius must equal the full length of AB is wrong — a longer radius than half of AB would also work, it does not have to equal AB exactly; the claim that the arcs would not cross line AB at all is wrong, since they do cross it — that is exactly the problem, they meet only on it.)
- (a) 20 cm — The scale factor from the smaller flag to the larger flag is 5 ÷ 2 = 2.5, so the larger width is 8 × 2.5 = 20 cm. The distractor 3.2 cm comes from using the ratio the wrong way round, 8 × 2 ÷ 5 = 3.2. The distractor 11 cm comes from adding the difference between the ratio numbers (5 − 2 = 3) to the given width, 8 + 3 = 11. The distractor 40 cm comes from multiplying by 5 without dividing by 2 first, 8 × 5 = 40.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (c) 5√3 m — The cable, the pole and the ground form a right-angled triangle: the ground distance (5 m) is adjacent to the 60° angle, and the height of the pole is opposite it, so height = 5 × tan 60° = 5 × √3 = 5√3 m. 5√3/2 m comes from using sin 60° = √3/2 instead of tan 60°. 5/√3 m comes from using tan 30° = 1/√3, the reciprocal-angle value, instead of tan 60°. 10√3 m comes from doubling the correct height by mistake.
- (d) 0 — A vertex is a sharp corner point where edges meet. A cylinder has two curved, circular edges but no sharp corner points at all, so it has 0 vertices, making that the correct answer. '2' wrongly treats the two circular edges themselves as vertices, but an edge is not the same as a vertex. '4' overcounts by treating each circular edge as if it had two end-vertices, which does not apply to a continuous curved edge. '1' wrongly imagines the curved surface itself forming a single corner point, which it does not.
- (b) P, since OP = 5 and OQ = 6 — Using the distance formula, OP = √(3² + 4²) = √(9 + 16) = √25 = 5, and OQ = √(6² + 0²) = √36 = 6. Since 5 is less than 6, P is closer to the origin. Naming Q as closer, with OQ = 5 and OP = 6, has the two distances swapped around the wrong point. Naming P as closer but with OP = 6 and OQ = 5 also has the two values swapped, even though it names the right point. The distances are not equal, since 5 is not the same as 6, so P and Q are not equally distant from the origin.
- (b) 270° — Bearings are measured clockwise from north (000°). Due north is 000°, due east is 090°, due south is 180° and due west is 270°. Choosing 090° gives the bearing for due east, not west. Choosing 180° gives the bearing for due south. Choosing 000° gives the bearing for due north.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (a) 62° — Angle KPQ and angle MQP are co-interior (allied) angles between parallel lines, so they add up to 180°: angle MQP = 180° − 118° = 62°. 118° comes from treating the angles as equal, as if this were a corresponding or alternate angle pair, instead of using the co-interior rule. 90° comes from wrongly assuming the crossing line meets JK and LM at right angles. 59° comes from halving 118° instead of subtracting it from 180°.
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
- (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) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
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