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 ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 2.A survey team measures the angle between two fences with a protractor and gets a reading of 254°. What type of angle is this?
- 3.Angle ABC is 130°. The line BF divides angle ABC into two equal parts. Work out the size of angle FBC.
- 4.A rhombus has all four sides equal in length, but it is not a square — its angles are not all 90°. How many lines of symmetry does this rhombus have?
- 5.A locus is described as the set of points exactly 4 cm from a fixed point O. What shape is this locus?
- 6.The diagram shows the plan, front elevation and side elevation of a solid. Which solid could this be?
- 7.A circle has a radius of 3 cm. Work out the diameter of the circle.
- 8.A triangular prism has two triangular end faces and three rectangular faces joining them. How many faces does it have in total?
- 9.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?
- 10.Triangle ABC has AB = 10 cm, BC = 7 cm and angle A = 40°. Triangle DEF has DE = 10 cm, EF = 7 cm and angle D = 40°. Sam says the two triangles are congruent by SAS. Which statement about Sam's claim is correct?
- 11.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 12.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.
- 13.p is the column vector with top number 5 and bottom number 2. q is the column vector with top number 1 and bottom number 6. Work out p + q, giving your answer as a column vector in the form (top, bottom).
- 14.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.
- 15.A robot on a grid moves by the vector , then by the vector , then by the vector . Write down the single column vector that has the same overall effect as these three moves.
Answer key
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (c) Reflex — Method: compare the measured angle to the key boundaries of 90 degrees, 180 degrees and 360 degrees. Working: 254 degrees is greater than 180 degrees but less than 360 degrees, so it lies in the reflex range. A student who answers obtuse knows the angle is bigger than 90 degrees but has not realised it is also bigger than 180 degrees; they may have worked with the smaller angle at the same point, 360 - 254 = 106 degrees, instead of the reading given. A student who answers acute has misread the scale and taken the reading as 54 degrees rather than 254 degrees. A student who answers right has guessed a commonly recognised angle type without comparing the size properly. Answer: reflex.
- (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.
- (a) 2 — Method: recall that the diagonals of a rhombus are always lines of symmetry, whatever its angles are. Working: a rhombus (all sides equal) always has its two diagonals as lines of symmetry, giving 2 lines of symmetry, whether or not the angles are 90°. Options: 0 wrongly assumes a non-square rhombus has no symmetry at all; 4 comes from the number of lines of symmetry a square has, mistaking this rhombus for a square; 1 comes from treating the rhombus like a kite, which has only one diagonal as a line of symmetry. Answer: 2.
- (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) A cuboid — Method: work out which solid has flat faces only, no curved surfaces and no point where edges meet, since only that gives rectangles for all three views. Working: a solid whose plan, front elevation and side elevation are all rectangles has three pairs of flat rectangular faces meeting at right angles — that is a cuboid. Answer: a cuboid. The distractors: a cylinder is wrong because its plan view (from above) is a circle, not a rectangle. A cone is wrong because its plan view is a circle and its front and side elevations are triangles. A square-based pyramid is wrong because its front and side elevations come to a point at the apex, giving triangles rather than rectangles, even though its plan view could be a square.
- (a) 6 cm — Method: a diameter runs right across a circle through its centre, so it is made of two radii laid end to end, which gives diameter = 2 × radius. Working: the radius is 3 cm, so the diameter is 2 × 3 = 6 cm. Answer: 6 cm. The distractors: 1.5 cm comes from dividing by 2 instead of multiplying by it, which is the relationship applied in the wrong direction; 3 cm comes from copying the radius straight down, treating the two words as names for the same measurement; 5 cm comes from adding 2 to the radius instead of multiplying the radius by 2.
- (a) 5 — Method: add the two triangular end faces to the three rectangular side faces. Working: 2 triangular faces + 3 rectangular faces = 5 faces in total. A student who answers 6 has confused this with a cuboid's face count. A student who answers 9 has counted the edges instead of the faces. A student who answers 3 has only counted the rectangular faces and forgotten the two triangular ends. Answer: 5 faces.
- (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.
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (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).
- (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) (6, 8) — Method: add column vectors by adding the top numbers together and the bottom numbers together. Working: top numbers 5 + 1 = 6; bottom numbers 2 + 6 = 8. Answer: p + q = (6, 8). A candidate who subtracts instead of adds gets (4, −4). A candidate who swaps the top and bottom of the correct answer gets (8, 6). A candidate who only adds the top numbers and leaves the bottom number of p unchanged gets (6, 2).
- (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.
- (d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
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