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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- (b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
- (b) 104 — Method: angles on a straight line add up to 180°. Working: 180° − 76° = 104°. A student who answers 76 has mistaken this for the vertically opposite angle, which is equal, rather than the adjacent angle on a straight line. A student who answers 90 has wrongly assumed the two paths must be perpendicular. A student who answers 14 has subtracted 76° from 90° instead of from 180°. Answer: 104°.
- (c) 5 cm — Diameter = circumference ÷ π, so 31.4 ÷ 3.14 = 10 cm, and the radius is half the diameter, so 10 ÷ 2 = 5 cm. 10 cm is the diameter itself, given as the radius by forgetting the final halving step. 15.7 cm comes from halving the circumference, 31.4 ÷ 2 = 15.7, and stopping there, treating half the circumference as the radius without ever dividing by π. 2.5 cm comes from halving the correct radius again, effectively dividing by 2 twice instead of once.
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (d) a rectangle — The shed is a prism, so its two triangular ends are identical and parallel, a fixed distance apart along the shed's length; the floor they stand on is therefore bounded by the base of one triangle, the base of the other, and the two straight edges joining them — a rectangle. Looking straight down, the sloping roof projects onto that same rectangle rather than outside it, so the plan view is a plain rectangle, as long as the shed and as wide as its gable end. "a triangle" is the shape of the END wall, seen from the front or back, not from above. "a triangle with a rectangle attached" wrongly combines a side-elevation feature with the plan — the plan does not show the triangular end wall at all, since looking down hides it completely. "two triangles joined at their bases" describes neither the plan nor any single elevation of this shed.
- (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) 80 cm² — Method: a composite shape made of two rectangles that do not overlap has an area equal to the sum of the two rectangle areas, so work out each area and add them. Working: the first rectangle has area 12 × 4 = 48 cm² and the second has area 4 × 8 = 32 cm², so the total is 48 + 32 = 80. Answer: 80 cm². The distractors: 48 cm² comes from working out the larger rectangle only and treating it as the whole shape; 32 cm² comes from working out the smaller rectangle only; 28 cm comes from adding the four given lengths, 12 + 4 + 4 + 8, which is a distance rather than an area and so carries a plain centimetre unit.
- (a) 3/4 — 1 litre = 1000 ml, so 750 ml is 750/1000 of a litre. Dividing both the numerator and denominator by 250 simplifies this to 3/4. Writing the fraction upside down, as the litre out of the 750 ml, gives 4/3. Finding the fraction of the litre that is NOT filled, 250/1000, gives 1/4. Dividing the numerator by 250 but the denominator by only 100, an inconsistent simplification, gives 3/10.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
- (b) No — the angle given is not the included angle — Method: check whether the given angle sits between the two given sides, since SAS requires the included angle. Working: sides AB and BC meet at vertex B, so the angle between them is angle B — but the angle given is angle A, which is not between the two given sides, and the same mismatch happens in triangle DEF. Options: 'two sides and one angle match' restates SAS's ingredients without checking their positions, which is exactly Meera's mistake; 'SSS needs three equal sides' is a true fact about a different condition, but it is not the reason Meera is wrong here; 'SAS allows any equal angle' states a rule that is not how SAS works, since the angle must be the included one. Answer: no, the angle given is not the included angle.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (a) 31.4 — Method: first find the circumference of one turn using π × diameter, then multiply by the number of turns. Working: circumference = 3.14 × 0.5 = 1.57 m; total distance = 1.57 × 20 = 31.4 m. A student who answers 15.7 has mistakenly halved the diameter again before multiplying, using 0.25 m instead of 0.5 m. A student who answers 3.14 has simply written down π itself, without completing the circumference or multiplying by the number of turns. A student who answers 62.8 has mistakenly doubled the diameter to 1 m before multiplying, treating the given length as a radius. Answer: 31.4 m.
- (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).
- (a) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
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