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.
Non-calculator
Geometry and measures worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.Triangle T has vertices (1, 1), (3, 1) and (1, 4). It is mapped onto triangle T′ with vertices (5, −1), (3, −1) and (5, −4). Which single composition of two transformations maps T onto T′?
- 2.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 3.Two similar triangular flower beds have lengths in the ratio 3 : 7. The area of the smaller bed is 45 m². Work out the area of the larger bed.
- 4.OABC is a parallelogram, with OA = a and OC = c. M is the midpoint of AB. Express the vector OM in terms of a and c.
- 5.A designer plots a logo point at (3, 4) on a grid. She reflects it in the line y = 1 and then rotates the image 90° clockwise about the point (1, 1) to create a repeating tile pattern. Work out the coordinates of the point after both transformations.
- 6.In triangle ABC, AB = 5 cm, BC = 3 cm and AC = 7 cm. Work out the size of angle ABC.
- 7.A drone flies from its base in three stages, each stage measured in metres east and north as a column vector. Stage 1 is the column vector with top number 30 and bottom number 40. Stage 2 is the column vector with top number −10 and bottom number 20. Stage 3 is the column vector with top number 15 and bottom number −5. Work out the column vector that would take the drone in a single straight flight back to its base from where it ends up.
- 8.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
- 9.The perimeter of a square is 20 cm. Work out the area of the square.
- 10.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 11.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 12.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 13.In pentagon PQRST, which of the following correctly names the interior angle at vertex R, using standard three-letter angle notation?
- 14.A fish tank is a cuboid measuring 40 cm long, 25 cm wide and 30 cm high. Water is poured in to a depth of 20 cm. Work out the volume of water in the tank, in litres.
- 15.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector to pick up a beam, then along the vector to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
Answer key
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (a) 245 m² — Method: for similar figures the ratio of the areas is the square of the ratio of the lengths, so multiply the smaller area by the square of the length scale factor. Working: the length scale factor is 7 ÷ 3, so the area scale factor is 49 ÷ 9, and the larger area is 45 × 49 ÷ 9 = 5 × 49 = 245. Answer: 245 m². The distractors: 105 m² comes from multiplying by the length scale factor 7 ÷ 3 instead of by its square, the commonest slip on this topic; 315 m² comes from multiplying by 7 and forgetting to divide by 3; 405 m² comes from multiplying by 3² = 9, squaring the wrong part of the ratio.
- (c) a + (1/2)c — Method: OABC is a parallelogram, so OB = OA + AB, and since AB is equal and parallel to OC, AB = c; this gives OB = a + c. M is the midpoint of AB, so AM = (1/2)AB = (1/2)c. Working: OM = OA + AM = a + (1/2)c. Answer: OM = a + (1/2)c. Adding the whole of AB instead of half of it gives a + c, which is OB, not OM; halving the whole diagonal OB instead of just AB gives (1/2)a + (1/2)c, the midpoint of OB rather than of AB; and flipping the sign on the c-term gives a − (1/2)c, which points back the wrong way along AB. Halve only the side you are told to halve, and check the sign before you commit to an answer.
- (b) (−2, −1) — Method: apply the reflection to the point first, then rotate the image about the given centre, in the order the question states them. Working: reflecting (3, 4) in the line y = 1 keeps x = 3 and puts the image as far below the line as the point is above it: 4 is 3 units above y = 1, so the image is 3 units below, at 1 − 3 = −2 (the same as 2 × 1 − 4 = −2). The reflected point is (3, −2). Rotating (3, −2) by 90° clockwise about (1, 1): subtracting the centre gives 3 − 1 = 2 and −2 − 1 = −3, the clockwise rule swaps and negates these to give −3 and −2, and adding the centre back gives 1 + (−3) = −2 and 1 + (−2) = −1. The final image is (−2, −1). Answer: (−2, −1). Reflect before you rotate, exactly as the design process is described, and rotate about the CENTRE (1, 1) given in the question rather than the origin: either mistake, or reversing the two steps, sends the tile to a different point.
- (a) 120° — Method: three sides are known, so use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc, with a the side facing the angle wanted. Working: angle ABC lies between AB = 5 cm and BC = 3 cm and faces AC = 7 cm, so cos ABC = (5² + 3² − 7²) ÷ (2 × 5 × 3) = (25 + 9 − 49) ÷ 30 = −15 ÷ 30 = −0.5. The angle between 0° and 180° whose cosine is −0.5 is 180° − 60°. Answer: angle ABC = 120°. The distractors: 60° comes from taking the subtraction the other way round, (49 − 25 − 9) ÷ 30 = 0.5, which loses the minus sign that makes the angle obtuse; 90° comes from the instinct that three known sides always mean Pythagoras, and 5² + 3² = 34 is not 49, so the triangle is not right-angled; 150° comes from knowing the cosine is −0.5 but subtracting 30° from 180°, using the angle whose sine is 0.5 rather than the angle whose cosine is 0.5.
- (c) (−35, −55) — Add the three stages component by component to find the drone's position relative to base: (30+(−10)+15, 40+20+(−5)) = (35, 55). The flight back to base is the negative of this vector, reversing both numbers: (−35, −55). (35, 55) is the vector from base to the drone's position — it forgets to reverse direction for the return flight. (−35, 55) only reverses the top number. (35, −55) only reverses the bottom number.
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (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.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (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) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (a) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
Build your own mix at the worksheet builder.