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.In quadrilateral ABCD the two sides AB and BC are each 6 cm long and meet each other at B. The two sides CD and DA are each 9 cm long and meet each other at D. No side of ABCD is parallel to any other side. Write down the mathematical name of this quadrilateral.
- 2.A rotation of 180° about the point (3, 2) can be written as a reflection in the line x = 3, followed by a reflection in a second line, for every point in the plane. Which line is the second line?
- 3.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = (4x + 20)°. Work out the size of angle ABC in terms of x, in its simplest form.
- 4.A shape is made by joining two rectangles edge to edge, with no overlap. One rectangle measures 12 cm by 4 cm and the other measures 4 cm by 8 cm. Work out the total area of the shape.
- 5.In triangle ABC, AB = 9 cm, BC = 6 cm and angle BAC = 35°. This description fits two different triangles. Work out the two possible sizes of angle ACB, each to 1 decimal place.
- 6.In triangle ABC, AB = 15 cm, angle BAC = 38° and the area of the triangle is 61 cm². Work out the length of BC. Give your answer to 1 decimal place.
- 7.A triangular sail for a small boat has a base of 2.4 m and a height of 1.75 m. Sailcloth costs £12.50 per m². Work out the total cost of the sailcloth needed for the sail.
- 8.A wooden cube has edges of length 4 cm. Work out the total surface area of the cube.
- 9.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 10.A boat sails from a harbour on a bearing of 090° for 24 km to a buoy, then changes course and sails on a bearing of 000° for 16 km to reach an island. Work out the direct distance from the harbour to the island. Give your answer correct to 1 decimal place.
- 11.A triangle has one unknown side or angle, marked x. In which one of these situations should the sine rule, rather than the cosine rule, be used to find x?
- 12.In triangle PQR, PQ = 8 cm and QR = 11 cm. Angle QPR = 50°. A student calculates the area of the triangle as Area = 1/2 × 8 × 11 × sin 50°. Is the student's method correct?
- 13.A carpenter measures two triangular braces. Brace 1 has a 55° angle, a 65° angle, and, opposite the 55° angle, a side of 8 cm. Brace 2 has matching 55° and 65° angles and, opposite its 55° angle, a side of 8 cm. The carpenter says that because the equal 8 cm side is not between the two equal angles in either brace, he cannot be sure the two braces are the same size. Is the carpenter correct?
- 14.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?
- 15.ABCD is an isosceles trapezium. The sides AB and DC are parallel, and the two sloping sides AD and BC are equal in length. Angle D is 112°. Work out the size of angle B.
Answer key
- (b) Kite — Method: name a quadrilateral by matching what is given — which sides are equal, whether those equal sides lie next to each other or opposite each other, and whether any sides are parallel — against the definitions of the special quadrilaterals. Working: the two 6 cm sides meet at B and the two 9 cm sides meet at D, so each pair of equal sides is a pair of neighbours rather than a pair of opposites, and the stem rules out any parallel sides. The quadrilateral with two pairs of equal adjacent sides and no parallel sides is a kite. Answer: kite. The distractors: a rhombus is chosen by candidates who see two pairs of equal sides and read that as all four sides being equal, which the two different lengths of 6 cm and 9 cm rule out; a parallelogram is chosen by candidates who remember that a parallelogram has two pairs of equal sides but not that in a parallelogram the equal sides are the opposite ones, and who pass over the statement that nothing is parallel; an isosceles trapezium is chosen by candidates who notice that the shape is symmetrical about the line BD and treat symmetry on its own as the mark of a trapezium, when a trapezium needs a pair of parallel sides.
- (d) y = 2 — Two reflections in perpendicular lines that cross at a point combine to a 180° rotation about that point. The line x = 3 is vertical, so the second line must be horizontal, and it must pass through the centre of rotation (3, 2) — that line is y = 2. Taking the y-coordinate of the centre but writing it against the wrong letter gives y = 3. Assuming the second line must also be vertical, like the first one, and just swapping in the other coordinate gives x = 2. Reaching for the standard mirror line y = x without checking that it actually passes through (3, 2) gives y = x — it does not pass through that point at all. The line that is both perpendicular to x = 3 and through (3, 2) is y = 2.
- (b) (2x + 10)° — By the angle at the centre theorem, angle ABC is half of angle AOC, because both stand on the same arc AC: angle ABC = (4x + 20)° ÷ 2 = (2x + 10)°. Writing down the centre angle itself, without halving at all, gives (4x + 20)°. Halving only the constant term and leaving the x-term unchanged gives (4x + 10)°. Doubling the centre angle instead of halving it gives (8x + 40)°. Halve every term in the expression, and (2x + 10)° is what you get.
- (c) 80 cm² — Method: a composite shape made of two rectangles that do not overlap has an area equal to the sum of the two rectangle areas, so work out each area and add them. Working: the first rectangle has area 12 × 4 = 48 cm² and the second has area 4 × 8 = 32 cm², so the total is 48 + 32 = 80. Answer: 80 cm². The distractors: 48 cm² comes from working out the larger rectangle only and treating it as the whole shape; 32 cm² comes from working out the smaller rectangle only; 28 cm comes from adding the four given lengths, 12 + 4 + 4 + 8, which is a distance rather than an area and so carries a plain centimetre unit.
- (d) 59.4° and 120.6° — Method: two sides and an angle that is not between them can describe two triangles, because an acute angle and its supplement have the same sine. Use the sine rule for the angle, then test whether the supplement also fits inside 180°. Working: AB = 9 cm faces angle ACB and BC = 6 cm faces the 35° angle, so sin ACB = 9 × sin 35° ÷ 6 = 5.1622 ÷ 6 = 0.86036. The inverse sine of 0.86036 is 59.358°, and its supplement is 180° − 59.358° = 120.642°. Both survive the angle-sum test, since 35° + 59.4° = 94.4° and 35° + 120.6° = 155.6°, each less than 180°. Answer: angle ACB is 59.4° or 120.6° to 1 decimal place. The distractors: 22.5° and 157.5° come from putting the sides the wrong way up, 6 × sin 35° ÷ 9 = 0.38238; 30.6° and 149.4° come from pressing the inverse cosine key on 0.86036 instead of the inverse sine key; 59.4° and 85.6° are the two unknown angles of the first triangle, 59.358° and 180° − 35° − 59.358°, given by a candidate who has found one triangle and reported its angles rather than the two possible sizes of the same angle.
- (b) 9.3 cm — Method: the area formula gives the second side that encloses the 38° angle, and once two sides and the angle between them are known the cosine rule gives the third side. Working: 61 = 1/2 × 15 × AC × sin 38°, so AC = 2 × 61 ÷ (15 × sin 38°) = 122 ÷ 9.2349 = 13.211 cm. Then BC² = 15² + 13.211² − 2 × 15 × 13.211 × cos 38° = 225 + 174.53 − 312.31 = 87.22, and the square root of 87.22 is 9.339. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 10.6 cm comes from forgetting to double the area when rearranging, so AC is taken as 6.606 cm before the cosine rule is applied; 26.7 cm comes from adding the last term of the cosine rule instead of subtracting it, 399.53 + 312.31; 13.2 cm is the length of AC, written down by a candidate who completes the first step and stops there.
- (b) £26.25 — First multiply the base and height: 2.4 × 1.75 = 4.2. The area of the triangular sail is half of that: half of 4.2 is 2.1 m². Then multiply by the cost per m²: 2.1 × £12.50 = £26.25. £52.50 forgets to halve in the area formula, giving an area of 4.2 m² and doubling the true cost. £30.00 multiplies the base length by the cost per m² (2.4 × £12.50) without ever finding the area. £25.00 rounds the area to 2 m² before multiplying by the cost, losing accuracy.
- (a) 96 cm² — Method: a cube has six identical square faces, so the total surface area is six times the area of one face. Working: one face has area 4 × 4 = 16 cm², and 6 × 16 = 96. Answer: 96 cm². The distractors: 16 cm² is the area of a single face, from stopping before multiplying by the six faces; 64 cm³ comes from working out the volume, 4 × 4 × 4, which is a different measure and carries a different unit; 24 cm² comes from multiplying the six faces by the edge length, 6 × 4, instead of by the area of a face.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (b) 28.8 km — A bearing of 090° is due east and a bearing of 000° is due north, so the two legs of the journey are at right angles to each other, meeting at the buoy. Pythagoras' theorem therefore applies directly, with the direct distance from the harbour to the island as the hypotenuse: distance² = 24² + 16² = 576 + 256 = 832. Taking the square root, distance = √832 = 28.8 km (1 d.p.). Adding the two legs of the journey directly, 24 + 16 = 40 km, treats the route as if it were a straight line, ignoring that the boat actually turns through a right angle partway. Using only the first leg of the journey, 24 km, ignores the second leg entirely. Subtracting the two legs instead of combining them with Pythagoras' theorem, √(24² − 16²) = √(576 − 256) = √320 = 17.9 km (1 d.p.), also gives the wrong distance.
- (c) Two angles and one side are known; x is another side — The sine rule needs a matching pair, a side and the angle opposite it, that you already know, so you can set up a ratio with the unknown. When two angles and one side are known, you can find the third angle from the angle sum, giving you an angle opposite the known side and an angle opposite x: the sine rule applies directly. When all three sides are known and x is an angle, there is no side-angle pair available at all, so the cosine rule, rearranged for an angle, is what's needed instead. When two sides and the included angle are known and x is the third side, again there is no matching side-angle pair yet, so the cosine rule finds the third side directly. When two sides and the included angle are known and x is one of the other angles, you still have no side-angle pair to start from — the cosine rule has to be used first, to find the third side, before any angle can be found. Only the two-angles-and-a-side case hands you a ready-made pair, which is exactly what the sine rule needs.
- (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) No — third angle is also fixed — Since both braces have angles of 55° and 65°, their third angles must both be 60°, because angles in a triangle sum to 180°. All three angles now match, so the braces have the same shape. Both 8 cm sides lie in the same position relative to those angles — opposite the 55° angle in each brace — so one matching pair of corresponding sides fixes the size as well, exactly as ASA or AAS would. The braces are therefore guaranteed to be congruent and the carpenter is incorrect: 'No — third angle is also fixed' is correct. 'Yes — side must be included' is wrong because the side does not have to lie physically between the two named angles; once the third angle is fixed, a corresponding equal side anywhere is enough. 'No — any two angles enough alone' is wrong because two equal angles with no side length at all would only show the triangles are similar, not congruent. 'Yes — third angle may differ' is wrong because the third angle is fixed at 60° by the angle sum and cannot vary.
- (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) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
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