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.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
- 2.A cone is cut by a flat plane parallel to its circular base, partway up between the base and the apex. What shape is the cross-section?
- 3.A solid has 5 faces: one square face and four triangular faces, which all meet at a single point above the square. What is the name of this solid?
- 4.Triangle ABC has a right angle at B and hypotenuse AC = 17 cm, with AB = 8 cm. Triangle DEF has a right angle at E and hypotenuse DF = 17 cm. Which extra fact about DEF would complete the RHS condition for proving that ABC ≅ DEF, with the vertices matching in that order?
- 5.A surveyor is marking out field ABCD for a new crop. The plan states that side AB is parallel to side DC, but AB and DC are not equal in length, and sides BC and AD are not parallel to each other. Which term correctly describes the shape of field ABCD?
- 6.Work out the y-coordinate of the midpoint of the line segment joining (2, 4) and (8, 10).
- 7.Write down the exact value of cos 45°.
- 8.Quadrilateral PQRS has angle P = 92°, angle Q = 84° and angle R = 106°. Work out the size of angle S.
- 9.p is the column vector with top number 5 and bottom number 1. q is the column vector with top number −2 and bottom number 3. Work out p − 2q, giving your answer as a column vector in the form (top, bottom).
- 10.A trapezium has parallel sides of 16 cm and 24 cm, and a perpendicular height of 7 cm. Work out the area of the trapezium.
- 11.A right-angled triangle has its two shorter sides equal to 8 cm and 15 cm. Work out the value of cos θ, where θ is the angle opposite the 8 cm side. Give your answer as a fraction.
- 12.Here are four trigonometric ratios of special angles. Write down the one that does not have a value.
- 13.A quadrilateral has two pairs of parallel sides. All four of its interior angles are right angles, but its sides are not all the same length. Which quadrilateral is this?
- 14.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 15.A recipe uses 1.25 kg of flour. Work out how many grams of flour this is.
Answer key
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (a) Square-based pyramid — A solid with one square base and four triangular faces meeting at a single apex above the base is a square-based pyramid. A triangular prism has two triangular faces and three rectangular faces, not a square base with four triangles, so that is a different solid. A cube has six square faces, and a cuboid has six rectangular faces — neither has any triangular faces at all. The solid described is a square-based pyramid.
- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
- (b) trapezium — A quadrilateral with exactly one pair of parallel sides is a trapezium; field ABCD has AB parallel to DC and no other pair of parallel sides, so trapezium is correct. Parallelogram requires BOTH pairs of opposite sides to be parallel, but the plan says only AB and DC are parallel. Rhombus requires all four sides to be equal in length, which is not stated here. A kite is defined by two pairs of adjacent equal sides, not by having a pair of parallel sides, so it does not match this description either.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (b) √2/2 — cos 45° comes from a right-angled isosceles triangle with both shorter sides 1 and hypotenuse √2, giving cos 45° = 1/√2 = √2/2. 1/2 is the value of cos 60° (mixing up the two angles). √2 is the hypotenuse length itself, not divided by it (forgetting to divide by the hypotenuse). √3/2 is the value of cos 30° (using the wrong special triangle).
- (c) 78° — The angles in any quadrilateral add up to 360°. Add the three given angles: 92° + 84° + 106° = 282°. Angle S = 360° − 282° = 78°. A pupil who only adds angle P and angle Q, forgetting angle R, gets 360° − (92° + 84°) = 184°. A pupil who only adds angle Q and angle R, forgetting angle P, gets 360° − (84° + 106°) = 170°. A pupil who makes a carrying slip adding the three angles, getting 292° instead of 282°, gets 360° − 292° = 68°. The correct answer is 78°.
- (a) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
- (b) 140 cm² — Method: the area of a trapezium is the mean of the two parallel sides multiplied by the perpendicular height. Working: (16 + 24) ÷ 2 = 20, then 20 × 7 = 140. Answer: 140 cm². The distractors: 280 cm² comes from multiplying the sum of the parallel sides by the height, (16 + 24) × 7, and forgetting to halve; 168 cm² comes from using only the longer parallel side, 24 × 7, as though the shape were a rectangle; 47 cm comes from adding all three given lengths, 16 + 24 + 7, which gives a length rather than an area.
- (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
- (b) tan 90° — Method: write each ratio as one side of a right-angled triangle divided by another, and look for the one whose bottom line can be zero. Working: sine and cosine are both a side divided by the hypotenuse, and a hypotenuse is never zero, so sin 90° = 1 and cos 90° = 0 are both perfectly good values — a value of zero is still a value. Tangent is the opposite side divided by the adjacent side. As the angle opens out towards 90° the adjacent side shrinks to nothing, and a division by zero has no result, while at the other end of the row the adjacent side is the long one and tan 0° = 0. Answer: tan 90°.
- (c) Rectangle — A rectangle has two pairs of parallel sides and four right angles, but does not require all sides to be equal — this matches exactly, so Rectangle is correct. A square also has four right angles and parallel sides, but additionally requires all four sides to be equal, which contradicts 'not all the same length', so it is wrong. A rhombus has two pairs of parallel sides and all four sides equal, but its angles are not generally 90° unless it is also a square, so it does not match the right-angle condition here. A kite has no pairs of parallel sides at all, so it does not match the first condition given.
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (d) 1250 g — There are 1000 g in a kilogram, so 1.25 kg = 1.25 × 1000 = 1250 g. Multiplying by 100 instead of 1000 gives 125 g. Converting only the whole 1 kg and forgetting the extra 0.25 kg gives 1000 g. Treating the 0.25 kg as 25 g instead of 250 g, a quarter of 1000, gives 1025 g.
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