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) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
- (c) 1 — Method: the instruction states that a curved surface counts as a face, so count the surfaces of the sphere on that basis. Working: a sphere has exactly one continuous curved surface and no flat surfaces at all. A student who answers 0 has ignored the instruction and refused to count the curved surface. A student who answers 2 has confused the sphere with a cylinder, which has two flat circular faces. A student who answers 3 has imagined extra hidden surfaces that do not exist. Answer: 1 face.
- (d) £100.48 — Area of the rug = (angle ÷ 360) × π × r² = (180 ÷ 360) × 3.14 × 16 = 0.5 × 50.24 = 25.12 m². Cost = 25.12 × £4 = £100.48. (£200.96 comes from finding the area of a full circle, 3.14 × 16 = 50.24 m², and forgetting the angle fraction before costing it; £25.12 comes from forgetting to square the radius, using 0.5 × 3.14 × 4 = 6.28 m², and then costing that; £50.24 comes from using the arc length formula, 0.5 × 2 × 3.14 × 4 = 12.56, in place of the area, and costing that.)
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
- (d) (−4, −3) — Reflecting in the y-axis keeps the y-coordinate the same and changes the sign of the x-coordinate, so (4, −3) maps to (−4, −3). "(4, 3)" changes the sign of the y-coordinate instead, which is what happens when reflecting in the x-axis. "(−4, 3)" changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the y-axis. "(3, −4)" swaps the two coordinates around instead of reflecting either of them.
- (a) a circle of radius 4 cm, centred at O — Every point exactly 4 cm from a fixed point O sweeps out a full circle of radius 4 cm centred at O — that is the definition of this locus. "a straight line 4 cm long, starting at O" only covers points in one direction, not every direction. "a square with sides of 4 cm, centred at O" would only touch the true locus at a few points on its perimeter, since most of a square's edge is not 4 cm from its centre. "two points, each 4 cm from O" comes from picturing only the points directly left and right of O, forgetting every other direction round it.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (a) 6 — A cuboid has six faces in total: a top, a bottom, a front, a back, and two ends — each one is a rectangle in the net, giving six rectangles altogether. Choosing 3 counts only the three PAIRS of congruent rectangles (top/bottom, front/back, two ends) rather than all six individual faces. Choosing 5 forgets one face, as if the net were missing its lid. Choosing 12 is the number of edges of a cuboid, not the number of rectangles in its net.
- (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.
- (a) (2, 6) — To translate by the vector (−3, 4), add −3 to the x-coordinate and add 4 to the y-coordinate: (5 − 3, 2 + 4) = (2, 6). A pupil who adds 3 instead of subtracting for the x-coordinate gets (8, 6). A pupil who swaps the x- and y-components of the vector gets (5 + 4, 2 − 3) = (9, −1). A pupil who subtracts both components instead of adding the y-component gets (5 − 3, 2 − 4) = (2, −2). The correct image is (2, 6).
- (b) 117 — Method: two angles meeting at a point on a straight line add up to 180°. Working: 180 − 63 = 117. Answer: 117°. A candidate who thinks the two angles on a straight line must be equal gives 63. A candidate who uses 90° instead of 180°, working out 90 − 63, gets 27. A candidate who uses 360° instead of 180°, working out 360 − 63, gets 297.
- (c) The midpoint of AB — M is a single point lying exactly halfway along the straight line AB, so 'the midpoint of AB' correctly describes it. 'The vertex of AB' is wrong because a vertex is a corner point where two edges or lines meet, which does not apply to a point on a single straight line. 'A perpendicular of AB' is wrong because perpendicular describes two lines meeting at right angles, not a single point. 'A plane of AB' is wrong because a plane is a flat two-dimensional surface, not a point.
- (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
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