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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Answer key: Geometry and measures worksheet — GCSE Higher
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- (c) 12.4 cm — Using the cosine rule, BC² = AB² + AC² − 2 × AB × AC × cos(A) = 9² + 6² − 2 × 9 × 6 × cos(110°) = 81 + 36 − 108 × cos(110°). Since cos(110°) ≈ −0.34202, 108 × cos(110°) ≈ −36.94, so BC² ≈ 117 + 36.94 = 153.94. Taking the square root, BC ≈ 12.4072, which rounds to 12.4 cm. 8.9 cm comes from treating cos(110°) as if it were positive (using +0.342 instead of −0.342), which wrongly subtracts instead of adds and gives BC² ≈ 80.06. 153.9 cm is BC² itself, rounded, with the square root never taken. 11.6 cm comes from leaving out the factor of 2 in the formula, computing BC² = 81 + 36 − 9 × 6 × cos(110°) ≈ 135.47 instead.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (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) = $\binom{4}{−7}$.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (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.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (a) 53.5° — Method: use the sine rule BC/sin A = AC/sin B, since BC is opposite angle A and AC is opposite angle B. Working: sin(ABC) = AC × sin(BAC) / BC = 10 × sin 40° / 8, which gives angle ABC = 53.5° or its supplement 180 − 53.5 = 126.5°; since angle ABC is acute, the answer is 53.5°. Taking the obtuse supplement instead gives 126.5°; putting the sides the wrong way round in the ratio (sin B = BC × sin A / AC instead of AC × sin A / BC) gives 30.9°; and using sin 50° in place of sin 40° gives 73.2°. Every sine-rule ratio has two possible angle solutions that add to 180° — the word 'acute' tells you which one to keep.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (b) 43.8 km — Method: find the angle of the triangle at Q from the two bearings, then use the cosine rule on the two known sides and that included angle. Working: the bearing of P from Q is the back bearing of 040°, which is 040° + 180° = 220°, and the bearing of R from Q is 115°, so angle PQR = 220° − 115° = 105°. Then PR² = 24² + 31² − 2 × 24 × 31 × cos 105° = 576 + 961 − 1488 × (−0.25882) = 1537 + 385.12 = 1922.12, and the square root of 1922.12 is 43.842. Answer: PR = 43.8 km to 1 decimal place. The distractors: 33.9 km comes from subtracting the bearings, 115° − 040° = 75°, and using 75° as the angle at Q, which is the angle the ship turns through, not the angle inside the triangle; 19.9 km comes from using the first bearing itself, 040°, as the angle at Q; 55.0 km comes from adding the two legs, 24 + 31, which would be the distance only if the ship had sailed in a straight line.
- (a) 114.6° — Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC = 6 × sin(40°) ÷ 9. Since sin(40°) ≈ 0.64279, this gives sin(ACB) ≈ 3.8567 ÷ 9 ≈ 0.42852, so angle ACB ≈ 25.4° or its supplement, 154.6°. Testing the obtuse candidate: 40° + 154.6° = 194.6°, which already exceeds 180°, so angle BAC would have to be negative — impossible, so 154.6° is rejected. With angle ACB ≈ 25.4°, angle BAC = 180° − 40° − 25.4° = 114.6°. 25.4° is angle ACB, not angle BAC that the question asks for. 14.6° comes from using the invalid 154.6° candidate anyway and then wrongly turning the resulting negative angle sum (−14.6°) positive instead of rejecting it. 65.4° comes from inverting the sine rule ratio — dividing AC × sin(ABC) by AB instead of AB × sin(ABC) by AC — which gives a different, incorrect candidate for angle ACB entirely.
- (c) It is a parallelogram but not a rectangle — Method: match the given properties to the definition of each named quadrilateral. Working: having both pairs of opposite sides parallel and equal in length is exactly the definition of a parallelogram; since none of the angles are right angles, it cannot also be a rectangle, which needs four right angles. Options: 'must be a rectangle' wrongly assumes every parallelogram has right angles; 'must be a rhombus' wrongly assumes equal opposite sides means all four sides are equal, but only the opposite pairs are stated as equal here; 'trapezium' is wrong because a trapezium has exactly one pair of parallel sides, while this shape has two pairs, so it is a parallelogram and not a trapezium. Answer: it is a parallelogram but not a rectangle.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
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