Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation 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 Foundation
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- 1.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 2.A parallelogram has an area of 96 cm² and a base of 8 cm. Work out the perpendicular height of the parallelogram.
- 3.A train leaves Leeds at 08:47 and arrives in London at 11:05. Work out how long the journey takes.
- 4.A quadrilateral has angles of 100°, 85° and 95° at three of its vertices. Work out the fourth angle.
- 5.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.
- 6.A right-angled triangle has a hypotenuse of 29 cm and one shorter side of 20 cm. Work out the length of the other shorter side.
- 7.A chord divides a circle into two segments of different sizes. Write down the name given to the smaller of the two segments.
- 8.A treasure map has a scale of 1 cm to 4 m. A path from the start to a rock is drawn as two straight sections, measuring 3 cm and 2.5 cm. Work out the total real distance from the start to the rock, in metres.
- 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.Write down which one of these quadrilaterals always has diagonals that are equal in length and that cross at right angles.
- 11.In triangle ABC, angle ABC = 90° and angle BAC = 30°. The hypotenuse AC = 12 cm. Work out the exact length of AB.
- 12.Vector A is and vector B is . Write down how vector B compares to vector A.
- 13.A stained-glass window panel is designed as a kite. Two adjacent sides are each 25 cm, and the other two adjacent sides are each 40 cm. A frame is fitted around the whole panel. The frame costs £2.50 per metre, sold only in whole metres. Work out the total cost of the frame.
- 14.Work out the distance between the points (−3, 4) and (5, −2).
- 15.A map has a scale of 1 : 25000. A distance between two points measures 2 cm on the map. What is the real distance, in kilometres?
Answer key
- (a) 143° — Angles on a straight line add up to 180°. Set up 37° + x = 180°. Subtract: x = 180° − 37° = 143°. 53° comes from using 90° as the total, as if the two angles made a right angle, instead of the 180° of a straight line.
- (a) 12 cm — Area of a parallelogram = base × perpendicular height, so height = area ÷ base = 96 ÷ 8 = 12 cm. A student who adds the area and base instead of dividing gets 96 + 8 = 104 cm. A student who multiplies the area and base instead of dividing gets 96 × 8 = 768 cm. A student who uses the triangle area formula, area = 1/2 × base × height, instead of the parallelogram formula solves 96 = 1/2 × 8 × h and gets h = 24 cm.
- (d) 2 h 18 min — Method: count on from the departure time in whole steps, rather than subtracting the two clock readings as if they were ordinary decimals. Working: from 08:47 to 09:00 is 13 minutes; from 09:00 to 11:00 is 2 hours; from 11:00 to 11:05 is a further 5 minutes. 13 + 5 = 18, so the journey lasts 2 hours and 18 minutes. Answer: 2 h 18 min. Subtracting as decimals gives 11.05 − 8.47 = 2.58 and the false reading 2 h 58 min, because an hour holds 60 minutes and not 100. Taking the minutes the wrong way round, 47 take away 5, gives 2 h 42 min. Counting the hours as 11 − 8 = 3 and then attaching the 18 minutes gives 3 h 18 min.
- (d) 80° — The four angles of any quadrilateral add up to 360°. Three of the angles add up to 100 + 85 + 95 = 280°, so the fourth angle is 360 − 280 = 80°. "70°" comes from using 350° as the total instead of 360°, an easy slip on the quadrilateral angle sum. "280°" comes from stopping at the sum of the three given angles and writing that total down as the answer, instead of subtracting it from 360°. "260°" comes from subtracting only the one angle 100° from 360°, instead of subtracting all three given angles.
- (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) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (b) Minor segment — Method: compare the sizes of the two regions cut off by the chord, and recall the term used for the smaller one. Working: the chord creates two segments; the smaller region is called the minor segment and the larger one the major segment. A student who answers major segment has picked the larger region by mistake instead of the smaller one. A student who answers minor arc has named the curved boundary rather than the two-dimensional region it encloses. A student who answers semicircle has wrongly assumed the chord must pass through the centre. Answer: minor segment.
- (b) 22 — Method: add the two drawn lengths together first, then apply the scale to the total. Working: 3 cm + 2.5 cm = 5.5 cm; 5.5 cm × 4 = 22 m. A student who answers 10 has only converted one of the two sections (2.5 cm × 4) and forgotten the other. A student who answers 5.5 has added the two drawn lengths but forgotten to apply the scale at all. A student who answers 44 has doubled the correct answer, effectively applying the scale twice. Answer: 22 m.
- (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.
- (a) Square — Method: two conditions are being asked for at once, so test each shape against both — the two diagonals must always be the same length as each other, and they must always meet at 90°. Working: in a rectangle the diagonals are equal but they meet at 90° only in the special case where the rectangle is also a rhombus; in a rhombus the diagonals do meet at 90° but they are of different lengths unless the rhombus is also a rectangle; the shape that satisfies both conditions for every example of it is the one that is both, and its diagonals are equal and perpendicular. Answer: the square. The distractors: the rectangle is where a candidate stops who tests only the equal-length condition and never checks the angle at the crossing; the rhombus is where a candidate stops who tests only the right-angle condition and never checks the two lengths; the parallelogram is chosen by a candidate who remembers that the diagonals of a parallelogram bisect each other and treats bisecting each other as being equal to each other, which is a different property.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (d) B is the reverse of A — Every component of vector B is the negative of the matching component of vector A (−2 is the negative of 2, and 5 is the negative of −5), so B undoes the translation that A performs — B is the reverse of A. 'B is the same as A' ignores that both signs have flipped. 'B is twice A' confuses a sign change with a scale-factor change; the sizes of the components have not changed, only their signs. 'B is unrelated to A in direction' misses that the two vectors are directly linked, just in opposite directions.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (d) 10 — Method: the distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical gaps, so work out both gaps first, handling the negative coordinates carefully, and then apply Pythagoras' theorem. Working: the horizontal gap is 5 − (−3) = 5 + 3 = 8 and the vertical gap is 4 − (−2) = 4 + 2 = 6. Then d² = 8² + 6² = 64 + 36 = 100, so d = √100 = 10. Answer: 10. The distractors: 14 comes from adding the two gaps, 8 + 6, instead of adding their squares and taking the root; 100 comes from stopping at the sum of the squares and never taking the square root; 50 comes from reaching 100 correctly and then halving it instead of taking its square root, a candidate who has read the last step as “halve” rather than “root”.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
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