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.
Geometry and measures worksheet — GCSE Higher
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- 1.An enlargement with scale factor 5 maps the point (4, 1) to the point (12, 9). Which of these points is invariant (does not move) under this enlargement?
- 2.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 3.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 4.A cone is cut by a flat plane parallel to its circular base, partway up between the base and the apex. What shape is the cross-section?
- 5.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 6.A circle has diameter 10 cm. Using π = 3.14, work out the circumference of the circle.
- 7.In triangle ABC, AB = 6.4 cm, BC = 9.1 cm and AC = 12.8 cm. Work out the size of the largest angle in the triangle. Give your answer to 1 decimal place.
- 8.A zip wire is fixed at an angle of 50° to the horizontal, running from a high platform down to the ground. The zip wire itself, from the platform to the ground, is 12 m long. Work out the horizontal distance covered by the zip wire. Give your answer correct to 1 decimal place.
- 9.Triangle ABC is similar to triangle PQR. AB corresponds to PQ, and BC corresponds to QR. AB = 5 cm, PQ = 15 cm and BC = 7 cm. Work out the length of QR.
- 10.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 11.A right-angled triangle has two sides of length 5 cm and 12 cm, and the angle between those two sides is 90°. Work out the length of the hypotenuse.
- 12.Two similar triangles have lengths in the ratio 5 : 8. A side of the smaller triangle is 6.5 cm. Work out the length of the corresponding side of the larger triangle.
- 13.A drone starts at the point (2, −3) on a coordinate grid. It flies by the vector and then by the vector . Write down the column vector that would take the drone straight back to its starting point.
- 14.In triangle OAB, OA = a and OB = b. P is the point on OA such that OP = (1/3)a, and Q is the point on OB such that OQ = (1/3)b. Express the vector PQ in terms of a and b.
- 15.A shape is translated by the vector , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector . Work out the coordinates of the image of that same vertex under this second translation.
Answer key
- (d) (2, −1) — Method: for an enlargement, image = centre + k × (point − centre), so the centre satisfies centre = (image − k × point) ÷ (1 − k). Working: with k = 5, point (4, 1) and image (12, 9): 5 × (4, 1) = (20, 5); (12, 9) − (20, 5) = (−8, 4); dividing by 1 − 5 = −4 gives (2, −1). Answer: (2, −1), the centre of the enlargement, is the only invariant point since the scale factor is not 1. Subtracting the point itself instead of k times the point, (12, 9) − (4, 1) = (8, 8), then dividing by −4 gives (−2, −2); dividing by k − 1 = 4 instead of 1 − k = −4 gives (−2, 1); and simply taking the midpoint of the point and its image ignores the scale factor altogether and gives (8, 5). The centre of an enlargement is never just the midpoint between a point and its image unless the scale factor happens to be −1 — always use the full centre formula and keep the scale factor k in it.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (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) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
- (d) 110.1° — Method: the largest angle in any triangle faces the longest side, so identify that angle first and then use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc. Working: the longest side is AC = 12.8 cm, which is faced by angle ABC, and the two sides meeting at B are 6.4 cm and 9.1 cm, so cos ABC = (6.4² + 9.1² − 12.8²) ÷ (2 × 6.4 × 9.1) = (40.96 + 82.81 − 163.84) ÷ 116.48 = −40.07 ÷ 116.48 = −0.34401. A negative cosine means an obtuse angle, and the inverse cosine of −0.34401 is 110.12°. Answer: the largest angle is 110.1° to 1 decimal place. The distractors: 28.0° is the angle facing the shortest side, chosen by a candidate who thinks the largest angle sits opposite the smallest side; 69.9° comes from dropping the minus sign and using cos = 0.34401, which turns the obtuse angle into its supplement; 81.4° comes from assuming the angles share out 180° in the same ratio as the sides, 180 × 12.8 ÷ 28.3, which is not how a triangle behaves.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (d) 21 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 15 ÷ 5 = 3. QR = BC × scale factor = 7 × 3 = 21 cm. A student who divides BC by the scale factor instead of multiplying gets 7 ÷ 3 = 2.33 cm. A student who multiplies BC by AB instead of by the scale factor gets 7 × 5 = 35 cm.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
- (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) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (c) (1/3)b − (1/3)a — Method: PQ runs from P to Q, so PQ = OQ − OP. Working: PQ = (1/3)b − (1/3)a. Answer: PQ = (1/3)b − (1/3)a. Subtracting the other way round gives (1/3)a − (1/3)b, the same vector pointing back from Q to P instead of P to Q; using 2/3 instead of the 1/3 that OP and OQ were actually given as gives (2/3)b − (2/3)a; and using the full vectors a and b with no scaling at all gives b − a, which is AB, not PQ. Always subtract START from END, OQ − OP, and carry the fraction given in the question through to your final vector.
- (d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
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