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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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (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).
- (d) 8.75 cm — The scale factor of the enlargement is 11.2 ÷ 3.2 = 3.5. Applying this to the second line, 2.5 × 3.5 = 8.75 cm. The distractor 10.50 cm comes from adding the difference between the first line's two lengths (11.2 − 3.2 = 8) to the second line's original length, 2.5 + 8 = 10.5. The distractor 0.71 cm comes from using the scale factor the wrong way round, 2.5 × (3.2 ÷ 11.2) = 0.71 (to 2 d.p.). The distractor 8.70 cm comes from directly subtracting 11.2 − 2.5 = 8.7, muddling the two different lines instead of scaling the second one.
- (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.
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
- (b) 8.5 m — First apply the scale to convert the plan length to a real length in centimetres: 3.4 × 250 = 850 cm. Then convert centimetres to metres by dividing by 100: 850 ÷ 100 = 8.5, so the wall is 8.5 m long. Choosing 850 m applies the scale correctly but forgets to convert the answer from centimetres into metres. Choosing 0.85 m divides by 1000 instead of 100, confusing the centimetre-to-metre conversion with a metre-to-kilometre one. Choosing 3.4 m ignores the scale factor completely and just restates the plan length as if it were already the real length.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (c) 100.5 m — The curved part of a semicircular bend is half of a full circle's circumference. The full circumference would be 2 × 3.14 × 32 = 200.96 m, and half of that is 200.96 ÷ 2 = 100.48 m, which rounds to 100.5 m. Choosing 201.0 m uses the FULL circumference, forgetting to halve it for a semicircle. Choosing 50.2 m halves the radius as well as taking a semicircle, using 3.14 × 16 = 50.24 m instead of the correct radius of 32 m. Choosing 64.0 m uses the diameter, 2 × 32 = 64, as if it were the curved length, ignoring π and the semicircle shape entirely.
- (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) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (d) £100.48 — Area of the rug = (angle ÷ 360) × π × r² = (180 ÷ 360) × 3.14 × 16 = 0.5 × 50.24 = 25.12 m². Cost = 25.12 × £4 = £100.48. (£200.96 comes from finding the area of a full circle, 3.14 × 16 = 50.24 m², and forgetting the angle fraction before costing it; £25.12 comes from forgetting to square the radius, using 0.5 × 3.14 × 4 = 6.28 m², and then costing that; £50.24 comes from using the arc length formula, 0.5 × 2 × 3.14 × 4 = 12.56, in place of the area, and costing that.)
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (d) 1535 cm² — The wiper sweeps out a sector of radius 40 cm, the blade length, through an angle of 110°. Sector area is angle ÷ 360 × π × radius²: 110 ÷ 360 × 3.14 × 1600 = 1535.1 cm², which rounds to 1535 cm². Forgetting to square the radius, using radius instead of radius², gives 38 cm². Using 110 ÷ 180 instead of 110 ÷ 360 for the fraction gives 3070 cm². Treating the 40 cm blade length as a diameter, so using a radius of 20 cm, gives 384 cm².
- (d) 2π cm — Arc length is the fraction θ/360 of the full circumference, 2πr. Substitute θ = 60 and r = 6: 60 out of 360 is one sixth, and the full circumference is 2π × 6 = 12π. One sixth of 12π is 2π, so the arc length is 2π cm. Choosing 12π cm uses the full circumference without scaling it down by the fraction θ/360 first. Choosing 6π cm applies the SECTOR AREA formula, (θ/360) × πr², instead of the arc length formula — that calculation actually gives 6π, which is the area in cm², not a length. Choosing π cm uses πr instead of 2πr, missing the factor of 2 in the circumference formula.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
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