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 plan of a car park is drawn to a scale of 1 : 500. A parking bay is drawn as a rectangle measuring 1.2 cm by 0.6 cm. The real width of the bay corresponds to the 0.6 cm side. What is the real width of the bay, in metres?
- 2.A map has a scale of 1 : 20 000. The distance between two villages is 3.4 km in real life. How long is this distance on the map, in centimetres?
- 3.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 4.A textbook contains the instruction 'Draw AB' and, on the next line, the statement 'AB = 5 cm'. Which statement about the notation AB is correct?
- 5.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 6.A trapezium has an area of 45 cm², a perpendicular height of 5 cm, and one parallel side of length 10 cm. Work out the length of the other parallel side.
- 7.A sector of a circle has radius 6 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector.
- 8.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?
- 9.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 10.A sector of a circle has an angle of 90° and an area of 78.5 cm². Using π = 3.14, work out the radius of the circle.
- 11.An ice cream is made from a cone of radius 3 cm and height 10 cm, topped with a hemisphere of the same radius sitting exactly on top of the cone. Work out the total volume of the ice cream. Use π = 3.14. Give your answer to the nearest whole number. Volume of a cone = 1/3 × πr²h. Volume of a sphere = 4/3 × πr³.
- 12.The bearing of a campsite B from a walker's position A is 070°. What is the bearing of A from B?
- 13.A solid cone has a base radius of 6 cm and a vertical height of 10 cm. Work out the volume of the cone. Use π = 3.14 and give your answer correct to 1 decimal place.
- 14.A sector of a circle has radius 9 cm and angle 120°. Using π = 3.14, work out the arc length of the sector, to 1 decimal place.
- 15.A plan of a school hall uses a scale of 1 : 250. A wall is drawn 3.4 cm long on the plan. Sam wants to know the wall's real length in metres. What is it?
Answer key
- (b) 3 — The real width is 0.6 × 500 = 300 cm, which converts to 3 m by dividing by 100. A candidate who uses the wrong side of the rectangle, 1.2 cm, instead of the 0.6 cm width, gets 1.2 × 500 = 600 cm = 6 m. A candidate who multiplies correctly but converts the 300 cm to metres by dividing by 1000 instead of 100 gets 0.3 m. A candidate who converts by dividing by 10 instead of 100 gets 30 m. The real width of the bay is 3 m.
- (a) 17 cm — Convert the real distance to centimetres first: 3.4 km = 340 000 cm. Then divide by the scale factor, 20 000, since the map is 20 000 times smaller than real life: 340 000 ÷ 20 000 = 17, so the map distance is 17 cm. Choosing 1700 cm divides by 200 instead of 20 000, losing two zeros from the scale factor (340 000 ÷ 200 = 1700). Choosing 170 cm divides by 2 000 instead of 20 000, losing one zero from the scale factor (340 000 ÷ 2 000 = 170). Choosing 0.17 cm divides by 2 000 000 instead of 20 000, adding two extra zeros to the scale factor (340 000 ÷ 2 000 000 = 0.17).
- (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) AB means the segment or its length, shown by context — Method: recall the standard convention for a two-letter label such as AB. Working: AB names the segment AND its length; the sentence around it shows which is meant — 'draw AB' means the segment, 'AB = 5 cm' means the length. Options: 'a different symbol is needed' and 'must be written as |AB|' both describe stricter rules than the convention GCSE actually uses; 'AB can only mean the segment' ignores that the same label is also used for the length. Answer: AB means the segment or its length, shown by context.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (c) 8 cm — Area of a trapezium = (sum of parallel sides) ÷ 2 × height, so 45 = (a + 10) ÷ 2 × 5. Dividing 45 by 5 gives 9, so (a + 10) ÷ 2 = 9, meaning a + 10 = 18, so a = 18 − 10 = 8 cm. A pupil who correctly finds that the two parallel sides add up to 18 but forgets to subtract the known side of 10 cm gives the sum of both parallel sides, 18 cm, as the answer. A pupil who then adds 10 again by mistake instead of subtracting gets 18 + 10 = 28 cm. A pupil who halves the correct answer by mistake gets 8 ÷ 2 = 4 cm. The correct length of the other parallel side is 8 cm.
- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (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°.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (a) 10 cm — Sector area is (angle ÷ 360) × π × radius². Here 90 ÷ 360 = 1/4, and 1/4 × 3.14 = 0.785, so radius² is 78.5 ÷ 0.785 = 100, and the radius is √100 = 10 cm. Stopping after finding radius² and not taking the square root gives 100 cm. Treating 78.5 as the area of the WHOLE circle, ignoring the 90° fraction, gives radius² = 78.5 ÷ 3.14 = 25, so a radius of 5 cm. Correctly finding a radius of 10 cm but then doubling it, mistaking the question for asking the diameter, gives 20 cm.
- (a) 151 cm³ — Volume of the cone = 1/3 × 3.14 × 3² × 10 = 94.2 cm³. Volume of the hemisphere = 1/2 × (4/3 × 3.14 × 3³) = 1/2 × 113.04 = 56.52 cm³. Total volume = 94.2 + 56.52 = 150.72 cm³, which rounds to 151 cm³. A student who uses a full sphere instead of a hemisphere gets 94.2 + 113.04 = 207.24 cm³, rounding to 207. A student who uses a cylinder instead of a cone for the base, forgetting the 1/3, gets 3.14 × 3² × 10 = 282.6 cm³, plus the hemisphere's 56.52 cm³, totalling 339.12 cm³, rounding to 339.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (d) 18.8 cm — Arc length is a fraction of the full circumference. Here 120 ÷ 360 = 1/3, and the full circumference is 2 × 3.14 × 9 = 56.52 cm, so the arc is 56.52 ÷ 3 = 18.84 cm, which rounds to 18.8 cm. Forgetting the fraction and giving the full circumference rounds to 56.5 cm. Using 9 cm as a diameter instead of a radius gives an arc of 9.4 cm. Using 120 ÷ 180 instead of 120 ÷ 360 for the fraction gives an arc of 37.7 cm.
- (b) 8.5 m — First apply the scale to convert the plan length to a real length in centimetres: 3.4 × 250 = 850 cm. Then convert centimetres to metres by dividing by 100: 850 ÷ 100 = 8.5, so the wall is 8.5 m long. Choosing 850 m applies the scale correctly but forgets to convert the answer from centimetres into metres. Choosing 0.85 m divides by 1000 instead of 100, confusing the centimetre-to-metre conversion with a metre-to-kilometre one. Choosing 3.4 m ignores the scale factor completely and just restates the plan length as if it were already the real length.
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