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 sector of a circle has an angle of 90° at the centre. Write down what fraction of the whole circle this sector represents.
- 2.How many edges does a cube have?
- 3.A construction for the perpendicular bisector of AB begins by opening a pair of compasses to more than half the length of AB, then drawing an arc centred at A. What is the next step?
- 4.A chord is drawn across a circle, dividing the circle into two regions. Write down the name of one of these two regions.
- 5.A kite string makes an angle of 30° with the ground. The kite is flying at a height of 6 m directly above a point on the ground. Using the exact value of sin 30°, work out the exact length of the kite string.
- 6.A rectangular garden gate is 3 m wide and 4 m high. A straight brace runs from one corner of the gate to the opposite corner. Work out the length of the brace.
- 7.A tent's sloping side makes an angle of 45° with the ground. The sloping side is 3√2 m long. Using the exact value of sin 45°, work out the exact height of the tent.
- 8.Two similar triangles have lengths in the ratio 5 : 8. A side of the smaller triangle is 6.5 cm. Work out the length of the corresponding side of the larger triangle.
- 9.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 10.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 11.A cylindrical drum and a cuboid box are being wrapped for a display. A shop assistant wraps only the flat faces of each item, not any curved surface. How many flat faces in total does the assistant wrap across both items?
- 12.m is the column vector with top number 4 and bottom number 6. n is the column vector with top number −6 and bottom number −9. Given that n = k × m for some number k, work out the value of k.
- 13.A locus is described as the set of points exactly 4 cm from a fixed point O. What shape is this locus?
- 14.Angle ABC is 130°. The line BF divides angle ABC into two equal parts. Work out the size of angle FBC.
- 15.A solid pyramid has a square base of side 10 cm and four identical triangular faces. Its apex is directly above the centre of the base, at a vertical height of 12 cm, and the slant height of each triangular face — the distance from the apex to the midpoint of a base edge — is 13 cm. Work out the total surface area of the pyramid.
Answer key
- (c) 1/4 — A full turn at the centre of a circle is 360°, so a sector's fraction of the circle is its angle divided by 360°: 90 ÷ 360 = 1/4. 1/2 would be the fraction for a sector with an angle of 180°, not 90°. 3/4 is the fraction of the rest of the circle, the major sector left over from the 270° that is not part of this sector. 1/8 would be the fraction for a sector with an angle of 45°, half of 90°.
- (a) 12 — A cube has 6 square faces, 12 edges and 8 vertices. The edges are the straight lines where two faces meet: 4 edges around the top face, 4 edges around the bottom face, and 4 vertical edges joining them, giving 4 + 4 + 4 = 12. A candidate who counts the vertices instead of the edges answers 8. A candidate who counts the faces instead answers 6. A candidate who counts only the edges around the top face, forgetting the bottom face and the vertical edges, answers 4. The correct number of edges is 12.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (a) segment — A chord cuts a circle into two regions, and each of these regions is called a segment. A sector is a different region, bounded by two radii and an arc, not by a chord. An arc is a curved part of the circumference, a length, not a region. A radius is a straight line from the centre to the edge, also a length, not a region.
- (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
- (b) 5 m — Method: the brace, the width and the height form a right-angled triangle in which the brace faces the right angle, so it is the hypotenuse and Pythagoras' theorem applies, a² + b² = c². Working: c² = 3² + 4² = 9 + 16 = 25, so c = √25 = 5. Answer: 5 m. The distractors: 7 m comes from adding the two sides, 3 + 4, instead of adding their squares; 25 m comes from stopping at c² = 25 and forgetting to take the square root; 12 m comes from multiplying 3 × 4, which gives the area of the gate in square metres and not a length across it.
- (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (a) 10.4 cm — The scale factor from the smaller triangle to the larger triangle is 8 ÷ 5 = 1.6, so the larger side is 6.5 × 1.6 = 10.4 cm. The distractor 4.0625 cm comes from using the ratio the wrong way round, 6.5 × 5 ÷ 8 = 4.0625. The distractor 9.5 cm comes from adding the difference between the ratio numbers (8 − 5 = 3) to the given length, 6.5 + 3 = 9.5. The distractor 13 cm comes from doubling the given length, treating the scale factor as 2 instead of 1.6.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (a) a circle of radius 4 cm, centred at O — Every point exactly 4 cm from a fixed point O sweeps out a full circle of radius 4 cm centred at O — that is the definition of this locus. "a straight line 4 cm long, starting at O" only covers points in one direction, not every direction. "a square with sides of 4 cm, centred at O" would only touch the true locus at a few points on its perimeter, since most of a square's edge is not 4 cm from its centre. "two points, each 4 cm from O" comes from picturing only the points directly left and right of O, forgetting every other direction round it.
- (a) 65° — Method: a line that divides an angle into two equal parts gives each part half of the original angle, so halve 130°. Working: 130 ÷ 2 = 65. Answer: 65°. The distractors: 130° is the whole of angle ABC, written down without halving it; 50° comes from working out 180 − 130, using the angles on a straight line instead of dividing the angle in two; 32.5° comes from dividing by 4 instead of by 2, as though the line split the angle into four equal parts.
- (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.
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