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.An engineer models a technical drawing on a coordinate grid. A component is reflected in the line x = 2, and the image is then reflected in the line x = 6. Which single transformation has the same overall effect as these two reflections, for every point on the drawing?
- 2.OABC is a parallelogram, with OA = a and OC = c. X is the midpoint of the diagonal AC. Express the vector OX in terms of a and c, and use it to show that X also lies on the diagonal OB.
- 3.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 4.A rotation of 200° clockwise about the origin is applied, and the image is then rotated 250° anticlockwise about the origin. Work out the single angle of rotation, measured clockwise and between 0° and 360°, that has the same overall effect as the two rotations combined.
- 5.A pinhole camera is set up at a pitch-side stand, with its aperture at the point O = (2, 1) on a grid. Light from the corner F of a triangular flag, at the point (4, 2), passes through the aperture and forms an inverted image on the film behind it. The projection is an enlargement of scale factor −1.5 centred at O. Work out the coordinates of the image of corner F.
- 6.u is the column vector with top number 5 and bottom number 1. v is the column vector with top number 2 and bottom number 4. Work out u − v, giving your answer as a column vector in the form (top, bottom).
- 7.Triangle ABC is translated by the column vector with top number 3 and bottom number −5 to form triangle A′B′C′. Triangle A′B′C′ is then translated by the column vector with top number −7 and bottom number 2 to form triangle A″B″C″. Work out the single column vector that translates triangle ABC directly to triangle A″B″C″.
- 8.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 9.From an external point P, two tangents PA and PB touch a circle with centre O at points A and B. Angle APB = 44°. C is a point on the major arc AB. Using the fact that PA = PB, and the alternate segment theorem, work out the size of angle ACB.
- 10.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′?
- 11.Two cylindrical storage tins are geometrically similar. The smaller tin has height 4 cm and the larger tin has height 6 cm. Work out the ratio of the surface area of the smaller tin to the surface area of the larger tin, giving your answer in the form a : b in its simplest form.
- 12.A, B and C are points on a circle with centre O. B is on the minor arc AC. Angle ABC = 100°. Work out the size of the non-reflex angle AOC.
- 13.Triangle A has vertices (2, 1), (5, 1) and (2, 4). It is translated by the vector (−3, 2) and the image is then reflected in the x-axis. Work out the coordinates of the image of (5, 1).
- 14.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 15.Triangle ABC is similar to triangle PQR. Side AB is 8 cm and corresponds to side PQ, which is 12 cm. Side AC is 6 cm and corresponds to side PR. Work out the length of PR.
Answer key
- (d) A translation by the vector (8, 0) — Method: two reflections in PARALLEL lines combine into a single translation, perpendicular to the lines, of size twice the distance between them; two reflections in lines that CROSS combine into a rotation instead, never a translation. Working: the lines x = 2 and x = 6 are parallel, a distance of 6 − 2 = 4 apart. Doubling this distance gives 2 × 4 = 8, and the translation runs in the direction from the first line towards the second, so the vector is (8, 0). Answer: a translation by the vector (8, 0). Double the distance between the lines rather than using it directly, keep the direction running from the FIRST line reflected to the SECOND, and remember that two reflections in lines that never meet can only give a translation, never a rotation.
- (d) (1/2)a + (1/2)c — Method: X is the midpoint of AC, so OX = OA + (1/2)AC, with AC = c − a. Working: OX = a + 1/2(c − a) = a − (1/2)a + (1/2)c = (1/2)a + (1/2)c. Answer: OX = (1/2)a + (1/2)c. Since OB = a + c, this is exactly half of OB, so OX = (1/2)OB, meaning X lies on OB at its midpoint too — the two diagonals bisect each other. Forgetting to halve AC at all gives a + c, which is OB itself, not its midpoint; halving only the c-term gives (1/2)a + c; and a sign error on the c-term gives (1/2)a − (1/2)c. Halve the whole of AC, both terms together, and add it to OA rather than to a alone.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (a) 310° — Method: give the two rotations opposite signs since they turn in opposite senses — clockwise positive, anticlockwise negative — combine them into a single signed turn, then convert that turn into an angle measured clockwise between 0° and 360°. Working: the first rotation is 200° clockwise, so +200. The second is 250° anticlockwise, so −250. Combined: 200 − 250 = −50, meaning the net effect is a 50° turn anticlockwise. Measured clockwise instead, that same turn is 360 − 50 = 310°. Answer: 310°. Give the two rotations opposite signs before combining them, and convert a negative (anticlockwise) result into a clockwise angle by subtracting it from 360°, not from 180°: taking 50° away from a half turn gives 130°, which is a different rotation altogether; adding the two sizes as if both were clockwise gives 450°, which reduces to 90°; and reporting the anticlockwise size without converting it gives 50°.
- (a) (−1, −0.5) — Method: for an enlargement centred at a point O that is not the origin, first find the point's position relative to O, multiply that by the scale factor, then add O's coordinates back on — image = O + k × (F − O). Working: F relative to O is (4, 2) − (2, 1) = (2, 1). Multiplying by the scale factor, −1.5 × (2, 1) = (−3, −1.5). Adding O back on: (2, 1) + (−3, −1.5) = (−1, −0.5). Answer: (−1, −0.5). Measure F's position from the aperture O, not from the origin, apply the full signed scale factor −1.5 to that displacement, and add O's own coordinates back on afterwards: measuring from the origin instead of O, treating the factor as +1.5, or adding O onto the wrong displacement all place the image mark somewhere else.
- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (d) (−4, −3) — The combined translation is the sum of the two column vectors, added component by component: top numbers 3 + (−7) = −4, bottom numbers −5 + 2 = −3, giving (−4, −3). (10, −7) subtracts the second vector from the first instead of adding them. (−4, 3) gets the top number right but makes a sign error on the bottom, treating −5 + 2 as +3. (4, −3) gets the bottom number right but makes a sign error on the top, treating 3 + (−7) as +4.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (c) 68° — Since PA and PB are both tangents from the external point P, PA = PB, so triangle PAB is isosceles with equal base angles at A and B. The angles of triangle PAB sum to 180°, so angle PAB + angle PBA = 180° − 44° = 136°, and since the two base angles are equal, angle PAB = 136° ÷ 2 = 68°. Angle PAB is the angle between the tangent at A and the chord AB, so by the alternate segment theorem it equals the angle in the alternate segment, angle ACB = 68°. Using angle APB directly as angle ACB, without using the isosceles triangle to find angle PAB first, gives 44°. Halving angle APB directly, rather than subtracting it from 180° before halving, gives 22°. Finding 180° − 44° = 136° correctly but forgetting to divide by 2 for one base angle leaves 136°.
- (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.
- (b) 4 : 9 — The heights are in the ratio 4 : 6, which simplifies to 2 : 3. For similar solids, area scales with the square of the length ratio, so the surface area ratio is 2² : 3² = 4 : 9. 2 : 3 is just the simplified length ratio, before squaring has been done. 8 : 27 comes from cubing the ratio (2³ : 3³) instead of squaring it — that's the rule for volumes, not areas. 9 : 4 has the correct squared values but in the wrong order, giving the larger tin's area first instead of the smaller.
- (b) 160° — B is on the minor arc AC, so the angle at the centre theorem applies to the reflex angle AOC: reflex angle AOC = 2 × 100° = 200°. The non-reflex angle AOC is the rest of the full turn: 360° − 200° = 160°. Reporting the reflex angle itself, without subtracting it from 360°, gives 200°. Subtracting angle ABC from 180° instead, as if this were a cyclic quadrilateral, gives 80°. Halving angle ABC instead of doubling it gives 50°. Double first, then take the angle away from a full turn, and 160° is what's left.
- (c) (2, −3) — Method: apply the transformations in the order given, translation first, then reflection, to the point itself. Working: translating (5, 1) by the vector (−3, 2) moves the x-coordinate by −3 and the y-coordinate by 2, giving 5 − 3 = 2 and 1 + 2 = 3, so the image after the translation is (2, 3). Reflecting this point in the x-axis leaves the x-coordinate unchanged and reverses the sign of the y-coordinate, giving (2, −3). Answer: (2, −3). Do the two maps in the stated order and carry the point all the way through both of them: swapping the order, or stopping after the first map, lands you on a different point altogether.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
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