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.Loose sweets are sold at £1.20 per 100 g. Work out the cost of 350 g of sweets.
- 2.To construct a perpendicular to a line at a given point P on the line, compasses are first opened to draw two arcs crossing the line, one on each side of P. What is the correct next step?
- 3.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?
- 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.Triangle ABC has AB = 5 cm, BC = 7 cm and CA = 9 cm. Triangle DEF has DE = 5 cm, EF = 7 cm and FD = 9 cm. Which condition proves the two triangles are congruent?
- 6.A plumber charges a £45 call-out fee, plus £28 for each hour worked. She works on a job for 3 hours 30 minutes, and her hours are billed by rounding up to the next whole hour. Work out the total amount she charges.
- 7.In parallelogram PQRS, angle P = 65°. Which reason correctly explains why angle R = 65°?
- 8.An isosceles triangle ABC has AB equal to AC. AM is a line from vertex A, perpendicular to BC, meeting BC at point M. Which condition proves that triangle ABM is congruent to triangle ACM?
- 9.The curved part of a circle's circumference, between two points on the circle, has its own name. Write down this name.
- 10.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 11.Four angles meet at a point. Three of them measure 82°, 105° and 96°. Work out the size of the fourth angle.
- 12.The point A(3, 2) is translated by the vector (−5, 4) to give point B. Work out the coordinates of B.
- 13.A shop stacks tins in a display block. The front elevation of the block is 4 tins wide and 3 tins high. The side elevation shows the block is 2 tins deep. Every position in the block is filled. How many tins are used in total?
- 14.A hiker starts by walking on a bearing of 245°. She turns clockwise through 50°, then turns clockwise through a further 15°, and continues walking in a straight line. What bearing is she now walking on?
- 15.A sector of a circle has radius 5 cm and angle 72°. Work out the arc length of the sector, in terms of π.
Answer key
- (a) £4.20 — 350 g is 3.5 lots of 100 g, since 350 ÷ 100 = 3.5, so the cost is 3.5 × £1.20 = £4.20. Multiplying the mass in grams directly by the price, without dividing by 100 first, gives £420.00. Working out 100 ÷ 350 instead of 350 ÷ 100 inverts the ratio and gives about £0.34. Rounding 350 g down to 300 g gives 3 × £1.20 = £3.60.
- (b) Draw wider arcs from each crossing point, meeting above. — Once the two arcs cross the line at points either side of P, compasses are opened to a radius greater than before and arcs are drawn from each of those two points so that they meet above (or below) the line; joining that meeting point to P gives the perpendicular. (Joining the two crossing points with a straight line only retraces part of the original line, since both points already lie on it; drawing a circle centred at P through both crossing points does not locate any new point needed for the perpendicular; drawing an arc centred at P through only one crossing point repeats the first step instead of moving on to the second pair of arcs.)
- (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.
- (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.
- (d) SSS - three sides equal — All three pairs of corresponding sides are equal in length (5 cm, 7 cm and 9 cm in both triangles), so the triangles are congruent by SSS. SAS needs an angle to be given as well as two sides, but no angle is given here. ASA needs two angles and the side between them, but no angles are given at all. RHS needs a right angle and the hypotenuse, but no angle is stated to be a right angle.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (a) opposite angles of a parallelogram are equal — P and R are opposite vertices of the parallelogram, and opposite angles of a parallelogram are always equal, which is why angle R equals angle P, 65°. Co-interior angles adding up to 180° is the correct fact for angle Q or angle S, the angles adjacent to P along a side, not for the opposite angle R. Alternate angles are equal is a fact about a transversal crossing two parallel lines, which explains other angle relationships in the parallelogram, not the one between opposite angles P and R directly. Angles on a straight line adding up to 180° applies to two angles that sit together on one straight line, which P and R do not.
- (d) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
- (a) arc — A curved part of a circle's circumference, between two points, is called an arc. A chord is the straight line joining those two points, not the curved part. A segment is the region enclosed between a chord and an arc, and a sector is the region enclosed between two radii and an arc — both segment and sector are regions, not curved lengths.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (a) (−2, 6) — Translating by the vector (−5, 4) means adding −5 to the x-coordinate and adding 4 to the y-coordinate: (3 + (−5), 2 + 4) = (−2, 6). (8, 6) comes from treating −5 as +5, adding instead of subtracting on the x-coordinate. (−2, −2) keeps the x-coordinate correct but subtracts 4 from the y-coordinate instead of adding it. (7, −3) comes from swapping the two components of the vector, applying 4 to the x-coordinate and −5 to the y-coordinate.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (d) 310 — Method: since both turns are clockwise, add both angles to the starting bearing. Working: 245° + 50° = 295°; 295° + 15° = 310°. A student who answers 295 has only added the first turn and forgotten the second one. A student who answers 180 has subtracted both turns instead of adding them. A student who answers 320 has added the two turns as 75° instead of 65° by misreading the second turn. Answer: 310°.
- (d) 2π cm — The arc is a fraction of the whole circumference. The fraction is 72 ÷ 360 = 1/5 of the circle, and the full circumference is 2 × π × 5 = 10π cm. So the arc length is 1/5 × 10π = 2π cm. Taking the whole circumference and forgetting the fraction gives 10π cm. Using 72 ÷ 180 instead of 72 ÷ 360 gives 4π cm. Treating the 5 cm as a diameter instead of a radius gives π cm.
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