Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher 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 Higher
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- (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.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (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°.
- (a) 30° — Method: the exterior angles of any convex polygon add up to 360°, and in a regular polygon they are all equal, so divide 360° by the number of sides. Working: 360 ÷ 12 = 30. Answer: 30°. The distractors: 150° is the interior angle, 180 − 30, which answers for the wrong angle at the vertex; 15° comes from dividing 180 by 12, using the angles on a straight line instead of the full turn; 36° comes from dividing 360 by 12 − 2 = 10, carrying the subtraction of 2 out of the interior angle sum formula into a calculation that does not need it.
- (a) 10.4 cm — The scale factor from the smaller triangle to the larger triangle is 8 ÷ 5 = 1.6, so the larger side is 6.5 × 1.6 = 10.4 cm. The distractor 4.0625 cm comes from using the ratio the wrong way round, 6.5 × 5 ÷ 8 = 4.0625. The distractor 9.5 cm comes from adding the difference between the ratio numbers (8 − 5 = 3) to the given length, 6.5 + 3 = 9.5. The distractor 13 cm comes from doubling the given length, treating the scale factor as 2 instead of 1.6.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (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) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (c) No — 50° is angle P, not between PQ and QR. — Method: 1/2ab sin C only gives the area when C is the angle between the two named sides. Working: PQ and QR meet at vertex Q, so the angle between them is angle PQR, not angle QPR (which is the angle at vertex P, between sides PQ and PR). Since angle QPR is not included between PQ and QR, the student's calculation 1/2 × 8 × 11 × sin 50° does not give the area of the triangle — the verdict is No. Claiming that any angle works with any two sides ignores that the formula is only true for the INCLUDED angle. Claiming the formula needs all three sides is a different, incorrect statement about the formula itself — Area = 1/2ab sin C is valid with exactly two sides and their included angle, when that angle is actually known.
- (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.)
- (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.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (b) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
- (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
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