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, B, C and D are points on a circle with centre O, placed in that order around the circle so that ABCD is a cyclic quadrilateral. B lies on the major arc AC and D lies on the minor arc AC. Angle AOC = 104°. Work out the size of angle ADC.
- 2.Triangles PQS and RQS share the common side QS. PQ equals RQ, and PS equals RS. Which condition proves that triangle PQS is congruent to triangle RQS?
- 3.Two points on a circle divide its circumference into two arcs of different lengths. Write down the name given to the shorter of the two arcs.
- 4.A solid pyramid has a square base of side 6 cm. Its apex is directly above the centre of the base, at a vertical height of 8 cm. Work out the volume of the pyramid.
- 5.Point Q has coordinates (3, 5). It is translated by the vector (1, −2), and the image is then enlarged by scale factor 3, centre the origin. Work out the coordinates of the final image of Q.
- 6.OAB is a triangle, with OA = a and OB = b. E lies on OA produced beyond A, such that A is the midpoint of OE. F lies on AB such that FB is twice AF. G is the midpoint of OB. By finding the vectors EF and EG, show that E, F and G are collinear, and give the scalar k such that EF = k × EG.
- 7.A skateboard ramp is designed so that a straight rail RT is tangent to a circular curve at the point T, and TC is a chord of the curve from T to another point C on the curve. A second rail continues straight through T on the other side, so that the two rails together form a straight line, and the angle between that second rail and the chord TC is 109°. A support point D is placed on the curve, on the major arc TC. Work out the size of angle TDC, the angle subtended by the chord TC at D.
- 8.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 9.Point A has coordinates (2, 5). Point A is translated to point B using the column vector with top number 6 and bottom number −3. Write down the coordinates of point B.
- 10.A robot on a grid moves by the vector , then by the vector , then by the vector . Write down the single column vector that has the same overall effect as these three moves.
- 11.A plumber charges a £45 call-out fee, plus £28 for each hour worked. She works on a job for 3 hours 30 minutes, and her hours are billed by rounding up to the next whole hour. Work out the total amount she charges.
- 12.A rectangular garden gate is 3 m wide and 4 m high. A straight brace runs from one corner of the gate to the opposite corner. Work out the length of the brace.
- 13.OABC is a parallelogram, with OA = a and OC = c. M is the midpoint of AB. Express the vector MC in terms of a and c.
- 14.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 15.Points A, B, C and D lie on a circle with centre O, in that order around the circle, and AC is a diameter. Which of the following circle facts does NOT apply to this diagram?
Answer key
- (b) 128° — Angle AOC and angle ABC stand on the same arc AC (the minor arc, which does not contain B), so by the angle at the centre theorem angle ABC = 104° ÷ 2 = 52°. ABCD is a cyclic quadrilateral, so its opposite angles ABC and ADC sum to 180°: angle ADC = 180° − 52° = 128°. A candidate who finds angle ABC = 52° correctly but then treats opposite angles as equal, as in a parallelogram, instead of supplementary, writes down 52° and stops there. Skipping the halving step and using 104° as angle ABC gives 180° − 104° = 76°. Reading off the given centre angle itself as the final answer, without applying either theorem, gives 104°. Work through both theorems in order and you land on 128°.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (d) Minor arc — Method: compare the lengths of the two arcs formed by the two points, and recall the term for the shorter one. Working: the two points split the circumference into two arcs; the shorter one is the minor arc and the longer one is the major arc. A student who answers major arc has picked the longer arc by mistake. A student who answers minor segment has confused the curved boundary with the enclosed two-dimensional region. A student who answers chord has named the straight line joining the two points instead of the curved arc. Answer: minor arc.
- (b) 96 cm³ — Method: the volume of a pyramid is one third of the base area multiplied by the vertical height. Work out the area of the square base, multiply by the height, then divide by 3. Working: the base area is 6 × 6 = 36 cm², then 36 × 8 = 288, and 288 ÷ 3 = 96. Answer: 96 cm³. The distractors: 288 cm³ comes from multiplying the base area by the height and forgetting the one third, which is the volume of a cuboid with the same base and height; 144 cm³ comes from halving that 288 instead of taking a third of it; 16 cm³ comes from using the base edge of 6 cm in place of the base area, (6 × 8) ÷ 3.
- (b) (12, 9) — Apply the transformations in the order given: first translate, then enlarge. Translating (3, 5) by the vector (1, −2) gives (3 + 1, 5 − 2) = (4, 3). Enlarging this by scale factor 3 about the origin multiplies both coordinates by 3: (4 × 3, 3 × 3) = (12, 9). Enlarging first and translating afterwards reverses the order and gives (3 × 3 + 1, 5 × 3 − 2) = (10, 13), a different point because the two transformations do not commute. Enlarging the original point by scale factor 3 while forgetting to translate it at all gives (3 × 3, 5 × 3) = (9, 15). Reversing the signs of the translation vector before applying it gives (3 − 1, 5 + 2) = (2, 7), which then enlarges to (2 × 3, 7 × 3) = (6, 21).
- (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.
- (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (a) (8, 2) — Method: translating a point by a column vector means adding the vector's top number to the x-coordinate and its bottom number to the y-coordinate. Working: x-coordinate 2 + 6 = 8; y-coordinate 5 + (−3) = 2. Answer: B = (8, 2). A candidate who ignores the negative sign and adds 3 instead of −3 gets (8, 8). A candidate who translates in the reverse direction, subtracting the vector from A instead of adding it, gets (−4, 8). A candidate who swaps the vector's top and bottom numbers before adding gets (−1, 11).
- (d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (b) 5 m — Method: the brace, the width and the height form a right-angled triangle in which the brace faces the right angle, so it is the hypotenuse and Pythagoras' theorem applies, a² + b² = c². Working: c² = 3² + 4² = 9 + 16 = 25, so c = √25 = 5. Answer: 5 m. The distractors: 7 m comes from adding the two sides, 3 + 4, instead of adding their squares; 25 m comes from stopping at c² = 25 and forgetting to take the square root; 12 m comes from multiplying 3 × 4, which gives the area of the gate in square metres and not a length across it.
- (c) (1/2)c − a — Method: in parallelogram OABC, AB is equal and parallel to OC, so AB = c; M is the midpoint of AB, so AM = (1/2)c and OM = OA + AM = a + (1/2)c. MC runs from M to C, so MC = OC − OM. Working: MC = c − (a + (1/2)c) = (1/2)c − a. Answer: MC = (1/2)c − a. Subtracting in the wrong order gives a − (1/2)c, the same vector pointing the opposite way, from C to M rather than M to C; forgetting to halve the c-term gives c − a, which is AC, not MC; and adding instead of subtracting gives (1/2)c + a, which is OM itself. Always subtract the vector for the START of the journey, OM, from the vector for its END point, OC — and keep the fraction from the halving step.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (c) Tangent-chord angle = angle in the alternate segment. — This diagram has a diameter AC, a centre O, and a cyclic quadrilateral ABCD, but no tangent anywhere in it. 'Angle in a semicircle = 90°' applies directly, because AC is a diameter. 'Opposite angles of a cyclic quadrilateral sum to 180°' applies directly, because ABCD is a cyclic quadrilateral. 'Angle at the centre = twice angle at circumference' applies directly, because O is given as the centre of the circle. The tangent-chord fact relates the angle between a tangent and a chord to an angle elsewhere in the circle, and since this diagram has no tangent, there is nothing in it for that fact to describe — so it is the one theorem that does not apply here.
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