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.A trapezium has vertices (0, 2), (4, 2), (3, 5) and (1, 5). It is rotated 180° about the point (2, 2), and the image is then translated by the vector (1, −5). Work out the coordinates of the image of (0, 2).
- 2.A written description of a diagram states: 'AB is marked with a single arrow, CD is marked with a single arrow, and EF is marked with a double arrow.' Using the standard convention for marking parallel lines, what can be concluded?
- 3.Each of these has an exact value. Write down the one whose value is the greatest.
- 4.A cuboid has length 3 cm, width 4 cm and height 12 cm. Work out the length of the diagonal that runs from one corner of the cuboid to the opposite corner.
- 5.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.
- 6.A stained-glass window panel is designed as a kite. Two adjacent sides are each 25 cm, and the other two adjacent sides are each 40 cm. A frame is fitted around the whole panel. The frame costs £2.50 per metre, sold only in whole metres. Work out the total cost of the frame.
- 7.A cuboid measures 5 cm long (left to right), 3 cm deep (front to back) and 2 cm tall. Its front elevation is 5 cm wide by 2 cm high. Its side elevation is 3 cm wide by 2 cm high. What are the dimensions of its plan view, looking down from above?
- 8.In cyclic quadrilateral ABCD, with vertices in that order around the circle, the diagonal BD is drawn. In triangle ABD, angle ABD = 35° and angle ADB = 65°. Side DC is extended beyond C to a point E. Work out the size of angle BCE.
- 9.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.
- 10.Shape T has a vertex at (5, 3). T is enlarged by scale factor −1, centre (2, 1). Work out the coordinates of the image of the vertex (5, 3).
- 11.Triangle ABC has AB = 9 cm, BC = 6 cm and angle A = 35°. Triangle DEF has DE = 9 cm, EF = 6 cm and angle D = 35°. Which of the following correctly identifies the condition shown here?
- 12.A goat is tied by a rope 7 m long to a post at a corner of a rectangular field, where two fences meet at a right angle. The goat can reach anywhere inside the field that the rope allows. Using π = 3.14, work out the area the goat can graze, to the nearest square metre.
- 13.A circular decoration has a radius of 10 cm. A ribbon costs £0.50 per centimetre and is fixed along the curved edge (the arc) of a sector-shaped section with an angle of 90°. Using π = 3.14, work out the cost of the ribbon for this section.
- 14.Two triangular offcuts of wood, PQR and STU, are cut for a construction project. PQ = 8 cm, QR = 6 cm and angle PQR = 90°. ST = 8 cm, TU = 6 cm and angle STU = 90°. A carpenter wants to check the two pieces are identical in shape and size before using them as a matching pair. Using only the measurements given, and without working out any further lengths, which condition proves that triangle PQR is congruent to triangle STU?
- 15.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.
Answer key
- (b) (5, −3) — Method: rotate the point about the given centre first, then translate the image, in the stated order. Working: rotating (0, 2) by 180° about (2, 2) uses the rule (x, y) → (4 − x, 4 − y), since the centre doubles in each coordinate. This gives 4 − 0 = 4 and 4 − 2 = 2, so (0, 2) maps to (4, 2). Translating (4, 2) by the vector (1, −5) gives 4 + 1 = 5 and 2 − 5 = −3, so the final image is (5, −3). Answer: (5, −3). Rotate about the centre GIVEN in the question, (2, 2), not about the origin, and translate the rotated image afterwards, in that order: rotating about the wrong centre, swapping the order, or stopping after one step all give a different point.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (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 $\binom{-5}{-5}$.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (b) 5 cm by 3 cm — The plan view looks straight down on the cuboid's footprint, so it shows the length (5 cm, left to right) and the depth (3 cm, front to back) — the two dimensions that do not involve height. "5 cm by 2 cm" repeats the front elevation's dimensions, pairing the length with the height instead of the depth. "3 cm by 2 cm" repeats the side elevation's dimensions, again pairing the depth with the height rather than with the length. "5 cm by 5 cm" comes from mistakenly assuming the plan must be a square, pairing the length with itself instead of with the depth.
- (a) 80° — In triangle ABD, the three angles sum to 180°: angle BAD = 180° − 35° − 65° = 80°. ABCD is a cyclic quadrilateral, so the exterior angle at C, angle BCE, is equal to the interior angle at the opposite vertex, angle BAD: angle BCE = 80°. Adding the two given angles in triangle ABD instead of subtracting them from 180°, 35° + 65° = 100°, and then applying the exterior-angle rule correctly still gives the wrong angle BAD and so the wrong exterior angle, 100°. Halving the correct angle BAD, 80° ÷ 2 = 40°, confuses the exterior-angle rule with a different circle theorem in which one angle is half of another. Doubling it instead, 80° × 2 = 160°, makes the same kind of confusion in the other direction.
- (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.
- (b) (−1, −1) — An enlargement by scale factor −1, centre (2, 1), sends a point P to the point on the opposite side of the centre, the same distance away: the image is 2 × centre − P. For the vertex (5, 3), this gives (2 × 2 − 5, 2 × 1 − 3) = (4 − 5, 2 − 3) = (−1, −1). Treating the centre as though it were the origin, and simply negating the point's coordinates, gives (−5, −3) — this ignores that the true centre is (2, 1), not (0, 0). Using scale factor +1 instead of −1 leaves the point exactly where it started, at (5, 3). Adding the point's displacement from the centre instead of subtracting it gives (2 × 2 + 5, 2 × 1 + 3) = (9, 5). Double the centre and subtract the point, and the image is (−1, −1).
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (c) £7.85 — Arc length = (90 ÷ 360) × 2 × 3.14 × 10 = 0.25 × 62.8 = 15.7 cm. Cost = 15.7 × £0.50 = £7.85. (£31.40 comes from finding the full circumference and forgetting the angle fraction; £15.70 comes from using the diameter, 20 cm, in place of the radius; £157.00 comes from multiplying the arc length by the radius instead of by the cost per centimetre.)
- (b) SAS (two sides and the included angle) — PQ = ST and QR = TU are two pairs of matching sides, and the angle between them (angle Q and angle T) is 90° in both triangles — two sides and the angle between them are equal, so the triangles are congruent by SAS, using only the measurements given. RHS needs the hypotenuses to be known equal; here only the two legs are given, so you would first have to work out each hypotenuse by Pythagoras' theorem — extra working the question rules out. SSS has the same problem: the third side of each triangle is not given, only calculable. ASA needs two angles and the side between them, but here it is two sides and the angle between them that are given, not two angles.
- (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.
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