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 chord divides a circle into two regions, one larger than the other. Write down the term used for the smaller of these two regions.
- 2.A right-angled triangle has a hypotenuse of 29 cm and one shorter side of 20 cm. Work out the length of the other shorter side.
- 3.Triangle ABC has AB = 8 cm, angle B = 50° and BC = 6 cm. Triangle DEF has DE = 8 cm, angle E = 50° and EF = 6 cm. Which condition proves the two triangles are congruent?
- 4.A sector of a circle is bounded by two straight edges and one curved edge. Write down the mathematical term for each of the two straight edges.
- 5.Two sides of a triangle are 20 cm and 8 cm long. Write down which one of these lengths is possible for the third side.
- 6.In quadrilateral WXYZ, which of the following correctly names the angle at vertex Y using standard notation?
- 7.A hexagonal prism has 8 faces and 12 vertices. Using F + V − E = 2, work out how many edges it has.
- 8.A trapezium-shaped garden bed has parallel sides of 2.5 m and 4.5 m, and a perpendicular width of 3 m. A gardener wants to know the area of the bed before ordering topsoil. Work out the area of the garden bed.
- 9.In triangle ABC and triangle DEF, AB = DE, AC = DF, and the angle at A equals the angle at D. Write down the congruence condition that proves the two triangles are congruent.
- 10.A hiker walks on a bearing of 065°. She then turns clockwise through 90° and continues walking in a straight line. What bearing is she now walking on?
- 11.Triangle ABC has angle A = 40°, angle B = 65° and AB = 10 cm. Triangle DEF has angle D = 40°, angle E = 65° and DE = 10 cm. Which condition proves the two triangles are congruent?
- 12.The point (2, 5) is translated 3 units right and 4 units down. Work out the coordinates of the image point.
- 13.A scale drawing of a park uses a scale of 1 : 2000. A path is drawn 8.5 cm long on the drawing. A cyclist rides the length of the path and then rides straight back again along the same path. How far does the cyclist travel in total, in metres?
- 14.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 15.A polygon has one more side than a heptagon. What is the polygon called?
Answer key
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (c) SAS - two sides and the included angle equal — Two sides (AB and BC) and the angle between them (angle B) are equal in both triangles, so this is SAS. SSS would need all three sides given, but only two sides are stated. ASA needs two angles and the side between them, but only one angle is given. AAS needs two angles and a side, but again only one angle is given here.
- (c) Radius — Method: recall that a sector is formed using two straight lines drawn from the centre out to the circle's edge. Working: each straight edge of a sector runs from the centre of the circle to a point on the circumference. A student who answers chord has confused a straight edge from the centre with one joining two points on the circumference. A student who answers diameter has wrongly assumed the two straight edges must form a single full diameter. A student who answers arc has named the curved edge instead of the straight edges. Answer: radius (radii).
- (d) 15 cm — Method: any two sides of a triangle must together be longer than the third, so the third side must be longer than the difference of the two given sides and shorter than their sum. Working: the difference is 20 − 8 = 12 cm and the sum is 20 + 8 = 28 cm, so the third side must be between 12 cm and 28 cm, and 15 cm lies inside that range. Answer: 15 cm. The distractors: 12 cm is exactly the difference, so the three lengths would lie flat along a straight line and never close into a triangle; 5 cm is shorter than the difference — 5 + 8 = 13 cm cannot reach across the 20 cm side — and is chosen by candidates who check no lower limit at all; 30 cm is longer than the sum of the other two, so those two sides could never meet, and it is chosen by candidates who check no upper limit.
- (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.
- (c) 18 — Rearranging F + V − E = 2 gives E = F + V − 2. Substitute F = 8 and V = 12: 8 + 12 − 2 = 18 edges. Choosing 20 comes from adding the faces and vertices but forgetting to subtract the 2 (8 + 12 = 20). Choosing 22 comes from adding the 2 instead of subtracting it (8 + 12 + 2 = 22). Choosing 16 comes from subtracting 2 twice by mistake (8 + 12 − 2 − 2 = 16).
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (d) SAS — Method: a congruence condition is named by the parts that are given equal and the order in which they sit round the triangle, so count the sides and the angles first. Working: AB = DE and AC = DF are two pairs of equal sides, and the equal angle at A and D lies between AB and AC, so the given parts read side, included angle, side. Answer: SAS. The distractors: SSS needs three pairs of equal sides, and the third pair, BC and EF, is not given — it follows from the proof rather than being part of it; ASA reads the two equal sides as two equal angles, swapping which facts are which; RHS applies only when the triangles contain a right angle and the equal pair includes the hypotenuse, and nothing here says the angle at A is 90°.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (d) ASA - two angles and the included side equal — Two angles (A and B) are given, and AB is the side between them, so this is ASA. SAS needs two sides and the angle between them, but only one side is given. AAS also uses two angles and a side, but the side must NOT be between the two angles — here AB is between angle A and angle B, so it is ASA, not AAS. RHS needs a right angle, and neither 40° nor 65° is 90°.
- (c) (5, 1) — Moving right increases the x-coordinate, and moving down decreases the y-coordinate, so (2, 5) becomes (2 + 3, 5 − 4) = (5, 1). (5, 9) comes from adding 4 to the y-coordinate instead of subtracting, as if the point moved up rather than down. (−1, 1) comes from subtracting 3 from the x-coordinate instead of adding, as if the point moved left rather than right. (2, 1) keeps the x-coordinate unchanged and only applies the vertical move, forgetting the horizontal move altogether.
- (b) 340 — The real length of the path is 8.5 × 2000 = 17000 cm, which converts to 170 m by dividing by 100. The cyclist rides the path there and back, so the total distance is 170 × 2 = 340 m. A candidate who forgets the return journey gives only the one-way distance, 170 m. A candidate who halves the one-way distance instead of doubling it, reading “there and back” as splitting the journey, gets 85 m. A candidate who doubles the scale factor to 4000 by mistake, getting a one-way length of 340 m, and then doubles that for the return journey, gets 680 m. The total distance the cyclist travels is 340 m.
- (a) 143° — Angles on a straight line add up to 180°. Set up 37° + x = 180°. Subtract: x = 180° − 37° = 143°. 53° comes from using 90° as the total, as if the two angles made a right angle, instead of the 180° of a straight line.
- (c) octagon — A heptagon has 7 sides. One more than 7 is 8, and the polygon with 8 sides is an octagon. hexagon (6 sides) comes from subtracting one instead of adding one. heptagon (7 sides) is just the number of sides already given in the question, not one more than that. nonagon (9 sides) comes from adding two instead of one.
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