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 circular pizza base has a radius of 12 cm. Work out the area of the pizza base. Use π = 3.14.
- 2.A trapezium has parallel sides of length 6 cm and 10 cm, and a perpendicular height of 4 cm. Work out its area.
- 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 ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 5.A gardener wants to buy edging trim to go all the way around a circular flower bed with a diameter of 8 m. Work out the length of edging trim needed, using π = 3.14. Give your answer to the nearest metre.
- 6.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.
- 7.PT is a tangent to a circle with centre O, touching the circle at T. OT is a radius. Angle OPT = 27°, where P is a point outside the circle. Work out the size of angle POT.
- 8.A triangular sail for a small boat has two sides of 4.2 m and 3.6 m, with an angle of 115° between them. One litre of waterproofing paint covers 3 m² of sail. Work out the least number of whole tins of paint (1 litre each) needed to cover the sail.
- 9.A scale drawing uses a scale of 1 : 60. A path on the drawing is measured as 9.8 cm long. What is the real length of the path, in metres, to 1 decimal place?
- 10.Which of these statements about solving a triangle is correct?
- 11.A sector of a circle has radius 9 cm and angle 120°. Using π = 3.14, work out the arc length of the sector, to 1 decimal place.
- 12.A zip wire is fixed at an angle of 50° to the horizontal, running from a high platform down to the ground. The zip wire itself, from the platform to the ground, is 12 m long. Work out the horizontal distance covered by the zip wire. Give your answer correct to 1 decimal place.
- 13.A bead starts at position (2, −1) on a grid, in centimetres. It is moved by the column vector u, with top number 3 and bottom number 5, and then moved by the column vector v, with top number −7 and bottom number 2. Work out the coordinates of the bead's final position.
- 14.A rhombus-shaped floor tile has sides of length 12 cm and an interior angle of 55°. Work out the area of the tile. Give your answer to 1 decimal place.
- 15.A student is asked to rearrange the cosine rule a² = b² + c² − 2bc cos A to make cos A the subject. They write: cos A = (a² − b² − c²) / (2bc). Is the student's rearrangement correct?
Answer key
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (b) 32 cm² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the height. Add the parallel sides: 6 + 10 = 16. Multiply by the height: 16 × 4 = 64. Half of 64 is 32 cm². 64 cm² forgets to halve and just gives (6+10)×4. 8 cm² averages the two parallel sides, (6+10)÷2 = 8, but forgets to multiply by the height. 20 cm² treats it as a triangle using only the longer parallel side as the base: half of 10 × 4.
- (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) = $\binom{4}{−7}$.
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (c) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (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.
- (b) 63° — Method: a tangent meets the radius drawn to the point of contact at a right angle, so triangle OPT has a 90° angle at T; the three angles of the triangle then sum to 180°. Working: angle OTP = 90°, angle OPT = 27°, so angle POT = 180 − 90 − 27 = 63 degrees. Answer: 63°. The tangent-radius angle is a fixed 90°, not something to assume equal to another angle in the triangle, and the three angles of ANY triangle sum to 180°, never 360°: that total belongs to a quadrilateral, not a triangle.
- (a) 3 — Method: find the sail's area with Area = (1/2)ab sin C, then divide by how much one litre covers, and round up to a whole number of tins since paint can only be bought by the tin. Working: Area = 1/2 × 4.2 × 3.6 × sin 115° = 6.85 m² (2 d.p.); 6.85 ÷ 3 = 2.28 tins, which rounds up to 3 tins. Answer: 3. Rounding 2.28 to the nearest whole number instead of rounding up gives 2, which is not enough paint to finish the sail; forgetting the 1/2 in the area formula gives an area of 13.7 m², needing 5 tins; and assuming one litre covers exactly 1 m² of sail, ignoring the coverage rate given, rounds the area itself up to 7 tins. Whenever a quantity can only be bought in whole units, round up even when the decimal part is small — 2.28 tins means you need a third tin.
- (a) 5.9 — The real length is 9.8 × 60 = 588 cm. Converting to metres, by dividing by 100, gives 5.88 m, which rounds to 5.9 m to 1 decimal place. A candidate who rounds 5.88 down instead of up gets 5.8 m. A candidate who forgets to convert from centimetres to metres gets 58.8. A candidate who divides by 60 instead of multiplying gets 0.16, to 2 decimal places. The real length, to 1 decimal place, is 5.9 m.
- (d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
- (d) 18.8 cm — Arc length is a fraction of the full circumference. Here 120 ÷ 360 = 1/3, and the full circumference is 2 × 3.14 × 9 = 56.52 cm, so the arc is 56.52 ÷ 3 = 18.84 cm, which rounds to 18.8 cm. Forgetting the fraction and giving the full circumference rounds to 56.5 cm. Using 9 cm as a diameter instead of a radius gives an arc of 9.4 cm. Using 120 ÷ 180 instead of 120 ÷ 360 for the fraction gives an arc of 37.7 cm.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (d) 118.0 cm² — Method: a rhombus is made of two congruent triangles either side of a diagonal, each with area 1/2 × 12 × 12 × sin 55°, so the whole rhombus has area 12 × 12 × sin 55° (side² × sin of the interior angle). Working: area = 12² × sin 55° = 118.0 cm². Stopping at one triangle's area, 1/2 × 12² × sin 55°, and forgetting to double it gives 59.0 cm²; using cos 55° instead of sin 55° gives 82.6 cm²; and multiplying the two sides together with no trig term at all gives 144.0 cm². Splitting the rhombus into its two triangles is the safest way to see why the 1/2 disappears from the whole-shape formula.
- (b) No — should divide by −2bc, not +2bc; sign is wrong. — Method: rearrange a² = b² + c² − 2bc cos A step by step and compare with the student's version. Working: subtracting b² + c² from both sides gives a² − b² − c² = −2bc cos A, then dividing both sides by −2bc gives cos A = (a² − b² − c²) / (−2bc), which is the same as cos A = (b² + c² − a²) / (2bc). The student divided by +2bc instead of −2bc, so their expression is the negative of the correct one — the verdict is No. Saying the algebra is fine because 'either sign order gives a valid result' ignores that only one of the two signed expressions matches the original equation. Saying the denominator should be bc rather than 2bc is a different, unrelated error — the coefficient 2bc in the original formula is correct and must stay.
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