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 ferris wheel's circular frame has centre O. A support strut runs from O to a point A on the rim, and a second strut runs from O to point B on the rim, with angle AOB = 76°. A cabin is mounted at point C on the major arc AB, and D is a separate point on the minor arc AB. Work out the difference between angle ACB and angle ADB.
- 2.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.
- 3.Two traffic cones are geometrically similar. The smaller cone has height 6 m and the larger cone has height 9 m. Work out the ratio of the volume of the smaller cone to the volume of the larger cone, giving your answer in the form a : b in its simplest form.
- 4.Two similar triangular tiles have lengths in the ratio 3 : 5. The longest side of the smaller tile is 9 cm. Work out the length of the longest side of the larger tile.
- 5.Line 1 passes through (0, 1) and (2, 5). Line 2 passes through (3, 2) and (5, 6). Work out the gradient of each line. Then write down whether the two lines are parallel.
- 6.Triangle ABC is drawn. The interior angle bisectors from vertices A and B are constructed and meet at point I inside the triangle. Which statement about point I must be true?
- 7.A sector of a circle has radius 8 cm and takes up three-quarters of the full circle. Work out the perimeter of the sector, in terms of π.
- 8.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.
- 9.Quadrilateral ABCD has vertices A(1, 1), B(5, 1), C(5, 4) and D(1, 4). The quadrilateral is translated by the vector . Which of these points is NOT a vertex of the image?
- 10.In quadrilateral WXYZ, which of the following correctly names the angle at vertex Y using standard notation?
- 11.A dog is tied by a lead 4 m long to a fixed ring on a straight garden wall. The wall extends much further than the lead can reach in both directions, and the dog cannot cross through it. Describe the region the dog can reach.
- 12.A picture frame is a rhombus with a diagonal of 16 cm and a diagonal of 12 cm. The two diagonals meet at right angles at their midpoints. Work out the length of one side of the rhombus.
- 13.A rectangular photograph has an area of 96 cm². One of its sides is 8 cm long. Work out the length of the other side.
- 14.A gardener wants to plant a tree so that it is the same distance from two straight garden walls that meet at a corner, and also the same distance from two ornamental posts standing 4 m apart. Which construction locates this point?
- 15.In triangle ABC, AB = x cm, AC = (x + 3) cm, BC = 7 cm and angle BAC = 60°. Work out the value of x.
Answer key
- (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
- (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) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
- (c) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (a) 12π + 16 cm — A three-quarter sector's perimeter is the curved arc plus the two straight radii that close the shape. The full circumference is 2 × π × 8 = 16π cm, and three-quarters of that is 12π cm. Adding the two straight radii, 8 cm each, gives 12π + 16 cm. Leaving out the straight edges gives just 12π cm. Using one-quarter of the circumference, the piece left over rather than the piece asked for, gives 4π + 16 cm. Adding only one radius instead of two gives 12π + 8 cm.
- (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) (−1, 1) — Translating by $\binom{−2}{3}$ subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. This gives image vertices (−1, 4), (3, 4), (3, 7) and (−1, 7). The point (−1, 1) is not one of these: it has the correct new x-coordinate (1 − 2 = −1) but keeps the original y-coordinate (1) instead of adding 3, as if only the horizontal part of the vector had been applied.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
- (b) A semicircle of radius 4 m, away from the wall. — Every point the dog can reach is at most 4 m from the fixed ring, so without any wall the region would be a full circle of radius 4 m. The wall runs straight through the ring and blocks the dog from crossing it, and since the wall extends further than the lead in both directions, exactly half of that circle is cut off — leaving a semicircle of radius 4 m on the side of the wall the dog is tied on. (A full circle of radius 4 m ignores that the wall blocks half of the region; a quarter circle of radius 4 m would only be correct if the ring were fixed at a corner where two walls met, not along a single straight wall; a rectangle 4 m wide along the wall ignores that the lead lets the dog swing round in a curve, not stay a fixed distance out from the wall.)
- (c) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (b) 12 cm — Method: the area of a rectangle is one side multiplied by the other, so when the area and one side are known the other side is found by reversing that multiplication — divide the area by the side that is known. Working: 96 ÷ 8 = 12. Answer: 12 cm. The distractors: 88 cm comes from 96 − 8, subtracting the known side as though the area had been made by adding the two sides together; 768 cm comes from 96 × 8, running the area rule forwards on the two numbers given instead of reversing it; 40 cm comes from reading the 96 as a perimeter — halving it to 48 and taking the 8 cm side away — which reverses the perimeter rule rather than the area rule.
- (a) where the angle bisector meets the posts' perpendicular bisector — Being equidistant from the two walls means lying on the angle bisector of the corner; being equidistant from the two posts means lying on the perpendicular bisector of the 4 m segment joining them. A single point satisfying both conditions is wherever those two loci cross. "where the angle bisector meets the line joining the posts" uses the straight line between the posts instead of its perpendicular bisector — a point on that line is not generally equidistant from both posts. "the perpendicular bisector of the posts, alone" satisfies only the posts condition, ignoring the walls entirely. "the angle bisector of the corner, alone" satisfies only the walls condition, ignoring the posts entirely.
- (a) 5 — Method: BC faces the 60° angle, so put the two algebraic sides into the cosine rule and solve the equation that results. By hand, cos 60° = 0.5. Working: 7² = x² + (x + 3)² − 2 × x × (x + 3) × 0.5, so 49 = x² + x² + 6x + 9 − x² − 3x = x² + 3x + 9. That rearranges to x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0, giving x = −8 or x = 5. A length cannot be negative, so the negative root is rejected. Answer: x = 5. The distractors: 8 comes from expanding (x + 3)² as x² + 9, the error of assuming that squaring a bracket squares each term, which turns the equation into x² − 3x − 40 = 0; −8 is the negative root, kept by a candidate who solves the quadratic but never checks that x has to be a length; 37 comes from applying the cosine rule as though the 60° angle were at C facing AB, which gives x² = (x + 3)² + 49 − 7(x + 3) and a single linear solution.
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