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 solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 2.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?
- 3.A parallelogram has a base of 9 cm and a perpendicular height of 4 cm. Work out its area.
- 4.A surveyor is marking out field ABCD for a new crop. The plan states that side AB is parallel to side DC, but AB and DC are not equal in length, and sides BC and AD are not parallel to each other. Which term correctly describes the shape of field ABCD?
- 5.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?
- 6.A goat is tied by a rope 7 m long to a post at a corner of a rectangular field, where two fences meet at a right angle. The goat can reach anywhere inside the field that the rope allows. Using π = 3.14, work out the area the goat can graze, to the nearest square metre.
- 7.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
- 8.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.
- 9.In a right-angled triangle, the two shorter sides are a and b, and the hypotenuse is c. Write down the correct statement of Pythagoras' theorem.
- 10.In triangle ABC the angle at C is 90°, AC = 8 cm and the angle at A is 30°. Work out the length of BC. Give your answer to 1 decimal place.
- 11.In triangle ABC, angle ABC = 90° and angle BAC = 30°. The hypotenuse AC = 12 cm. Work out the exact length of AB.
- 12.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
- 13.A circular pizza base has a radius of 12 cm. Work out the area of the pizza base. Use π = 3.14.
- 14.A window is made from a rectangle with a semicircle on top. The rectangle is 40 cm wide and 60 cm tall. The semicircle has the same width as the rectangle. Work out the total area of the window. Use π = 3.14. Give your answer to the nearest whole number.
- 15.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?
Answer key
- (a) 615.44 cm² — Method: the surface area of a sphere is 4πr². Square the radius, multiply by π, then multiply by 4. Working: r² = 7² = 49, then 3.14 × 49 = 153.86, then 4 × 153.86 = 615.44. Answer: 615.44 cm². The distractors: 153.86 cm² comes from stopping at πr², which is the area of a flat circle of radius 7 cm and leaves out the factor of 4 that a curved surface needs; 87.92 cm² comes from 4 × 3.14 × 7, using the radius itself where the formula asks for its square; 1436.03 cm³ comes from working out the volume of the ball with (4 ÷ 3) × π × r³ instead of its surface area, which is a different measure and carries a cubic unit.
- (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.
- (c) 36 cm² — The area of a parallelogram is base × height = 9 × 4 = 36 cm². 26 cm² is 2 × (9 + 4), the perimeter you would get by treating the perpendicular height as though it were a side length — and a perimeter is a length, not an area. 18 cm² uses the triangle formula, half of base × height, by mistake. 72 cm² doubles the correct area.
- (b) trapezium — A quadrilateral with exactly one pair of parallel sides is a trapezium; field ABCD has AB parallel to DC and no other pair of parallel sides, so trapezium is correct. Parallelogram requires BOTH pairs of opposite sides to be parallel, but the plan says only AB and DC are parallel. Rhombus requires all four sides to be equal in length, which is not stated here. A kite is defined by two pairs of adjacent equal sides, not by having a pair of parallel sides, so it does not match this description either.
- (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.
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
- (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).
- (d) a² + b² = c² — Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides, so a² + b² = c². "a + b = c" adds the sides directly without squaring them at all. "a² − b² = c²" subtracts the squares instead of adding them. "a² + b² = c" adds the squares correctly but forgets to square the hypotenuse on the other side of the equation.
- (d) 4.6 cm — Method: BC is opposite the 30° angle and AC is next to it, so the ratio that links the two is tan θ = opposite ÷ adjacent. Working: tan 30° = BC ÷ 8, so BC = 8 × tan 30° = 4.6188…, which is 4.6 to 1 decimal place. Answer: 4.6 cm. The distractors: 13.9 cm comes from dividing by tan 30° instead of multiplying by it; 4.0 cm comes from using sin 30°, which treats the 8 cm side as the hypotenuse when it is the side next to the 30° angle; 6.9 cm comes from using cos 30° in place of tan 30°, which gives the wrong pair of sides.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (c) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 cm².
- (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°.
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