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
MathsUKwww.geekhero.co.uk
- 1.Triangle T has a vertex A at (3, 1). It is enlarged by a scale factor of −2, centre (1, 1). Work out the coordinates of the image of point A.
- 2.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 3.In one circle, chord PQ is 6 cm long, chord RS is 10 cm long, chord TU is 14 cm long and chord VW is 15 cm long. Write down which of these chords lies closest to the centre of the circle.
- 4.In triangle XYZ, the side opposite vertex X is labelled using the standard lowercase-letter convention. Which label is correct?
- 5.A plan of a school hall uses a scale of 1 : 250. A wall is drawn 3.4 cm long on the plan. Sam wants to know the wall's real length in metres. What is it?
- 6.The perimeter of a square is 20 cm. Work out the area of the square.
- 7.Two similar model cars have a length scale factor of 2 from the small model to the large model. Priya works out that the area scale factor is 2² = 4, and then says the volume scale factor must also be 4, since that is the number she just calculated. Is Priya correct?
- 8.In triangle ABC, AB = 7.4 cm, AC = 5.9 cm and angle BAC = 68°. Work out the length of BC. Give your answer to 1 decimal place.
- 9.A model of a bridge is built to a scale where every length on the model is 1/240 of the matching length on the real bridge. A support beam on the model measures 3.6 cm. Work out the length of the real support beam, giving your answer in metres.
- 10.Shape S has an area of 5 cm². Shape S is enlarged by a scale factor of −3 to give shape T. Work out the area of shape T.
- 11.Work out the exact value of tan 30° + tan 30°.
- 12.A ship's radio can be heard up to 30 km from the ship. A lighthouse's light can be seen up to 20 km from the lighthouse. The ship and the lighthouse are 40 km apart along the coast. Describe the region where BOTH the radio can be heard AND the light can be seen.
- 13.A tank holds 3.2 m³ of water. Work out this volume in litres.
- 14.Triangle ABC has AB = 5 cm and BC = 7 cm. Triangle DEF has DE = 5 cm and EF = 7 cm. Which extra fact would prove the triangles are congruent by SAS?
- 15.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = 108°. Work out the size of angle BCD.
Answer key
- (c) (−3, 1) — Method: for an enlargement about a centre, first find the vector from the centre to the point, multiply it by the scale factor, INCLUDING its sign, then add the result back onto the centre. Working: the vector from the centre (1, 1) to A(3, 1) is (3 − 1, 1 − 1) = (2, 0). Multiplying by the scale factor −2 gives −2 × 2 = −4 and −2 × 0 = 0, so the scaled vector is (−4, 0). Adding this to the centre gives 1 + (−4) = −3 and 1 + 0 = 1, so the image is (−3, 1). Answer: (−3, 1). A NEGATIVE scale factor keeps its sign all the way through the calculation: do not treat −2 as +2, and do not treat it as a fraction like 1/2, which is the rule for a scale factor between 0 and 1, not a negative one. Always measure the vector from the CENTRE of enlargement, never from the origin, unless the two happen to coincide.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (c) VW — Method: within one circle a chord's distance from the centre is fixed by its length, because a chord passing nearer the centre cuts further across the circle; so the chord that lies closest to the centre is simply the longest one listed. Working: the four lengths are 6 cm, 10 cm, 14 cm and 15 cm. Placing them in order, the greatest is 15 cm, and that length belongs to VW, so VW lies closest to the centre. Answer: VW. The distractors: PQ comes from reversing the rule and taking the shortest chord to be the one tucked nearest the centre; TU comes from knowing that the very longest chord is a diameter, deciding that such a chord passes through the centre rather than lying close to it, ruling the 15 cm chord out on that ground and taking the next longest; RS comes from reading 'closest to the centre' as 'nearest the middle of the list of lengths' and picking a middling value.
- (c) x — By convention, the side opposite a vertex is labelled with the lowercase version of that vertex's letter, so the side opposite X is labelled x. y is the label for the side opposite Y, not X. z is the label for the side opposite Z, not X. X is the vertex's own uppercase letter — the convention specifically switches to lowercase for the side, so the uppercase letter on its own is not correct.
- (b) 8.5 m — First apply the scale to convert the plan length to a real length in centimetres: 3.4 × 250 = 850 cm. Then convert centimetres to metres by dividing by 100: 850 ÷ 100 = 8.5, so the wall is 8.5 m long. Choosing 850 m applies the scale correctly but forgets to convert the answer from centimetres into metres. Choosing 0.85 m divides by 1000 instead of 100, confusing the centimetre-to-metre conversion with a metre-to-kilometre one. Choosing 3.4 m ignores the scale factor completely and just restates the plan length as if it were already the real length.
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (d) No — the volume factor is 2³ = 8, not the area factor 4. — For similar solids, area scales with the square of the length scale factor and volume scales with its cube — different powers of the same number, so they are not usually equal. The length scale factor here is 2, so the area scale factor is 2² = 4 (Priya's figure) but the volume scale factor is 2³ = 8. Priya is wrong: the correct volume scale factor is 8, not 4. The option claiming area and volume always scale by the same factor treats the two as interchangeable, which is only true for the length factor itself. The option giving 6 comes from multiplying the length factor by the number of dimensions (2 × 3) rather than cubing it. The option repeating '4' for volume simply reuses Priya's own area calculation.
- (a) 7.5 cm — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A, with BC as the side facing the 68° angle. Working: BC² = 7.4² + 5.9² − 2 × 7.4 × 5.9 × cos 68° = 54.76 + 34.81 − 87.32 × 0.3746 = 89.57 − 32.71 = 56.86, and the square root of 56.86 is 7.5405. Answer: BC = 7.5 cm to 1 decimal place. The distractors: 11.1 cm comes from adding the last term instead of subtracting it, 89.57 + 32.71 = 122.28, the commonest sign slip on the cosine rule; 56.9 cm is the value of BC² written down as though it were BC, stopping one step before the square root; 2.9 cm comes from pressing sin instead of cos, using 87.32 × sin 68° = 80.96 as the term to subtract.
- (b) 8.64 m — The real length is 240 times the model length: 3.6 × 240 = 864 cm. Converting to metres, 864 cm ÷ 100 = 8.64 m. The distractor 864 m comes from finding the length correctly in centimetres but forgetting to convert to metres. The distractor 0.0864 m comes from dividing by 100 a second time, converting 864 cm to metres twice over (864 ÷ 10 000) instead of once. The distractor 86.4 m comes from converting centimetres to metres by dividing by 10 instead of 100.
- (b) 45 cm² — Area scales with the square of the linear scale factor, and squaring a negative number gives a positive result: (−3)² = 9. The area of T is 5 × 9 = 45 cm². 15 cm² comes from multiplying the original area by the scale factor directly (5 × 3), without squaring. 9 cm² is the area scale factor itself, (−3)², with the multiplication by the original area 5 cm² left out. −15 cm² comes from multiplying 5 × (−3) and carrying the negative sign through, without squaring at all.
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (c) 3200 — Method: to change cubic metres to litres, multiply by 1000. Working: 3.2 × 1000 = 3200. Answer: 3200 litres. A candidate who multiplies by 100 instead of 1000 gets 320. A candidate who multiplies by 10000 instead of 1000 gets 32000. A candidate who does not convert at all gives 3.2.
- (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) 72° — Method: opposite angles in a cyclic quadrilateral always sum to 180°. Working: angle DAB and angle BCD are opposite angles of the cyclic quadrilateral ABCD, so angle BCD = 180 − 108 = 72 degrees. Answer: 72°. Opposite angles in a cyclic quadrilateral are SUPPLEMENTARY, not equal, so do not simply copy the given angle; and do not subtract from 360°, which is the rule for angles round a point, or confuse this with the angle-in-a-semicircle theorem, which gives a fixed 90° that has nothing to do with this quadrilateral.
Build your own mix at the worksheet builder.