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.
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Geometry and measures worksheet — GCSE Higher
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- 1.A tile is described as having two pairs of parallel sides, four equal sides, and diagonals that cross at right angles but are not equal in length. Give a reason why this tile cannot be a square.
- 2.A circular table mat has diameter 20 cm. Yasmin has a 70 cm length of ribbon and sews it around the edge of the mat with no overlap. Using π = 3.14, work out how much ribbon is left over after going around the mat once.
- 3.A furniture catalogue drawing uses a scale of 1 cm to 1.5 m. A sofa in the drawing is measured as 7 cm long. What is the real length of the sofa, in metres?
- 4.Point P has coordinates (6, 4). P is rotated 90° clockwise about the point (1, 2), and the image is then translated by the vector (−3, 5). Work out the coordinates of the final image of P.
- 5.A chocolate bar is a prism. Its cross-section is a triangle with a base of 6 cm and a perpendicular height of 5 cm, and the bar is 12 cm long. Work out the volume of the bar.
- 6.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
- 7.Point P(2, 5) is enlarged with centre (2, 1) to give image point P′(2, −7). Work out the scale factor of the enlargement.
- 8.A parallelogram-shaped tile has two sides of 6 cm and 9 cm, with an angle of 60° between them. Work out the area of the tile. Give your answer to 1 decimal place.
- 9.A straight fence is 5 m long. Describe the shape of the region made up of every point on the ground within 2 m of any part of the fence.
- 10.A cube has edges of length 5 cm. Work out the length of the diagonal that runs from one corner of the cube to the opposite corner. Give your answer as a surd in its simplest form.
- 11.Square B has vertices (1, 1), (4, 1), (4, 4) and (1, 4). It is rotated 90° clockwise about the origin, and the image is then translated by the vector (2, −1). Work out the coordinates of the image of (4, 1).
- 12.Work out the length of the straight line segment joining the points (2, −4) and (−3, −4).
- 13.Point P is translated by the vector (5, −2), and the image is then rotated 90° clockwise about the origin to give the point (3, 9). Work out the coordinates of P.
- 14.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 15.A zip-wire is fixed at an angle of 60° to the horizontal ground. The zip-wire is 12 m long. Using the exact value of cos 60°, work out the exact horizontal distance it covers.
Answer key
- (a) A square's diagonals are equal in length — A square has two pairs of parallel sides, four equal sides, and diagonals that are equal in length and cross at right angles. This tile's diagonals are not equal, so it is a rhombus that is not a square.
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (a) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (d) −2 — Method: the scale factor is the ratio of the image vector to the object vector, both measured FROM THE CENTRE of enlargement, keeping every sign. Working: the vector from the centre (2, 1) to P(2, 5) is (0, 4); the vector from the centre to P′(2, −7) is (0, −8). The scale factor is −8 ÷ 4 = −2. Answer: −2. Measure both vectors from the CENTRE, not from the origin, divide the IMAGE vector by the OBJECT vector and not the other way round, and keep the negative sign: a negative scale factor is not the same size as its positive counterpart with the sign dropped.
- (b) 46.8 cm² — Method: for a parallelogram, not a triangle, the two sides and the angle between them give Area = ab sin C — there is no 1/2. Working: Area = 6 × 9 × sin 60° = 46.8 cm² (1 d.p.). Answer: 46.8 cm². Using the triangle formula, (1/2)ab sin C, on a parallelogram by mistake gives half the true area, 23.4 cm²; using cos 60° instead of sin 60° gives 27.0 cm²; and adding the two sides before multiplying by sin 60° gives 13.0 cm². A parallelogram is exactly two of the triangles this formula was built for, so never carry the 1/2 across from the triangle version.
- (a) a rounded rectangle: 5 m by 4 m with semicircular ends — Points within 2 m of the straight part of the fence form a rectangle running the 5 m length of the fence and 4 m wide (2 m on each side); points within 2 m of each END of the fence, beyond that rectangle, form a semicircle of radius 2 m there, since the nearest point of the fence to them is just that one end. Together this gives a rounded, stadium-shaped region. "a rectangle, 9 m by 4 m" extends the rectangle by 2 m at each end instead of rounding it, wrongly including corner points that are actually more than 2 m from every part of the fence. "a circle of radius 2 m" treats the whole 5 m fence as a single point. "a rectangle, 5 m by 2 m" uses 2 m as the full width instead of the distance on EACH side, so it only covers one side of the fence.
- (c) 5√3 cm — The space diagonal of a cube with edge a satisfies d² = a² + a² + a² = 3a², using Pythagoras' theorem in three dimensions. With a = 5, d² = 3 × 5² = 3 × 25 = 75, so d = √75 = √(25 × 3) = 5√3 cm. Finding the diagonal of one face instead, using only two of the three edges, gives d = √(5² + 5²) = √50 = 5√2 cm, which leaves out the third dimension. Adding the three edges directly, 5 + 5 + 5 = 15 cm, ignores that Pythagoras' theorem works with squares of lengths, not the lengths themselves. Squaring the edge and multiplying by 3 correctly, 3 × 5² = 75, but then forgetting to take the square root, leaves 75 cm — the squared length, not the diagonal itself.
- (b) (3, −5) — Method: apply the rotation to the point first, then translate the image, in the order the question gives them. Working: rotating (4, 1) by 90° clockwise about the origin sends (x, y) to (y, −x), so (4, 1) becomes (1, −4). Translating (1, −4) by the vector (2, −1) gives 1 + 2 = 3 and −4 − 1 = −5, so the final image is (3, −5). Answer: (3, −5). Use the CLOCKWISE rule, (x, y) → (y, −x), not the anticlockwise one, and apply the rotation before the translation, exactly as the question states them: reversing the order or the direction of turn both land on a different point.
- (b) 5 — Method: the two points have the same y-coordinate, so the segment joining them runs horizontally and its length is the gap between the two x-coordinates; a length is a distance, so it is never negative. Working: the x-coordinates are 2 and −3, so the gap is 2 − (−3) = 2 + 3 = 5. Both points have y = −4, so there is no vertical part to add on. Answer: 5. The distractors: 1 comes from dropping the minus sign on −3 and working out 3 − 2 instead; 0 comes from subtracting the y-coordinates, which are equal, in place of the x-coordinates; 6 comes from counting the grid lines from −3 across to 2 inclusive, which counts one more than the number of gaps between them.
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
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