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.
Geometry and measures worksheet — GCSE Foundation
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- 1.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).
- 2.Write down the exact value of cos 45°.
- 3.A trapezium-shaped allotment plot has two parallel sides that face each other, and two sloping sides that are equal in length (an isosceles trapezium). One of the angles next to the shorter parallel side is 118°. Work out the angle next to the other end of the same shorter parallel side.
- 4.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 5.The column vector describes a translation. Write down the movement it represents.
- 6.Lines AB and CD are parallel. A straight line EF crosses AB at point P and crosses CD at point Q. At P, angle APE = 65°. Angle APE and angle BPQ are vertically opposite. Angle BPQ and angle PQD are co-interior (allied) angles. Work out angle PQD.
- 7.A circular decoration has a radius of 10 cm. A ribbon costs £0.50 per centimetre and is fixed along the curved edge (the arc) of a sector-shaped section with an angle of 90°. Using π = 3.14, work out the cost of the ribbon for this section.
- 8.A rectangular bookmark measures 2 cm by 25 cm. Work out the area of the bookmark.
- 9.A solid has a plan view that is a circle, and a front elevation that is a triangle. Which solid is this?
- 10.A triangle has vertices A(0, 0), B(10, 0) and C(5, 12). Work out the area of the triangle.
- 11.A square tile has sides of length 4.5 cm. Work out the perimeter of the tile.
- 12.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
- 13.In triangle ABC the base BC is 10 cm long and the sloping side AB is 13 cm long. The perpendicular height from A down to BC is 12 cm. Work out the area of triangle ABC.
- 14.A straight line touches the edge of a circle at exactly one point and does not cross into the circle at all. Write down the term for this line.
- 15.A ship sails 12 km on a bearing of 070° from a harbour to a buoy. It then sails a further 9 km, still on a bearing of 070°, from the buoy to a lighthouse. How far is the lighthouse from the harbour, in kilometres?
Answer key
- (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) √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).
- (d) 118° — In an isosceles trapezium, the two angles next to the same parallel side are equal, because the sloping sides are equal in length. So the angle at the other end of the shorter parallel side also equals 118°.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (a) 6 left, 2 up — In a column vector $\binom{x}{y}$, the top number gives the horizontal movement (negative means left) and the bottom number gives the vertical movement (positive means up), so $\binom{-6}{2}$ means 6 units left and 2 units up. '6 right, 2 up' ignores the negative sign on the top number. '2 left, 6 up' swaps the top and bottom numbers. '6 left, 2 down' ignores the positive sign on the bottom number.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (c) £7.85 — Arc length = (90 ÷ 360) × 2 × 3.14 × 10 = 0.25 × 62.8 = 15.7 cm. Cost = 15.7 × £0.50 = £7.85. (£31.40 comes from finding the full circumference and forgetting the angle fraction; £15.70 comes from using the diameter, 20 cm, in place of the radius; £157.00 comes from multiplying the arc length by the radius instead of by the cost per centimetre.)
- (b) 50 cm² — Method: the area of a rectangle is length × width. Working: 2 × 25 = 50. Answer: 50 cm². The distractors: 54 cm comes from working out the perimeter, 2 × (2 + 25), which is a length and not an area; 27 cm comes from adding the two sides, 2 + 25, instead of multiplying them; 25 cm² comes from halving the product, 50 ÷ 2, which is the rule for the area of a triangle rather than of a rectangle.
- (a) a cone — A cone, viewed from directly above, shows its circular base as a circle; viewed from the front, its curved surface narrows to a point at the apex, giving a triangle outline — so a circle plan with a triangle elevation identifies a cone. "a cylinder" has a circular plan too, but its front elevation is a rectangle, not a triangle, since its sides run straight up rather than narrowing to a point. "a sphere" gives a circle from every direction, plan and every elevation alike, not a triangle from the front. "a square-based pyramid" gives a triangle from the front, but its plan view is a square, not a circle.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (c) 18 cm — Method: a square has four equal sides, so the perimeter is 4 × the side length. Working: 4 × 4.5 = 4 × 4 + 4 × 0.5 = 16 + 2 = 18. Answer: 18 cm. The distractors: 9 cm comes from 2 × 4.5, adding only one pair of sides; 16 cm comes from rounding the side length down to 4 cm before multiplying, so the 0.5 cm on each side is lost; 20.25 cm² comes from working out 4.5 × 4.5, which is the area of the tile and carries a squared unit.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (b) 60 cm² — Method: the area of a triangle is half the base multiplied by the perpendicular height, and here the perpendicular height is the 12 cm, not the sloping side. Working: 10 × 12 = 120, then 120 ÷ 2 = 60. Answer: 60 cm². The distractors: 65 cm² comes from using the sloping side of 13 cm as the height, (10 × 13) ÷ 2; 120 cm² comes from using the right two lengths but forgetting to halve, 10 × 12; 78 cm² comes from taking 13 cm and 12 cm as the base and the height and ignoring BC altogether, (13 × 12) ÷ 2.
- (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
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