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 shopkeeper builds a display from two cuboid boxes, shown in the diagram. She wants to cover the front of the display with coloured paper. Work out the total area of the front elevation, in square centimetres.
- 2.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.
- 3.A rectangular garden bed measures 5 m by 3 m. A semicircular flower border is attached along one of the 3 m-wide ends, on the outside of the rectangle, so that the 3 m side is the diameter of the semicircle. Work out the total perimeter of the combined shape (the two long sides, the one remaining short side, and the curved edge). Use π = 3.14.
- 4.A raised garden bed is a prism whose cross-section is a triangle with a base of 1.2 m and a perpendicular height of 0.8 m. The bed is 3 m long. Work out the volume of soil needed to fill it, in m³.
- 5.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 6.A cylinder has a radius of 5 cm. Its volume is 471 cm³. Using π = 3.14, work out the height of the cylinder.
- 7.A sector of a circle has radius 12 cm and angle 45°. Using π = 3.14, work out the area of the sector.
- 8.A garden bed is designed as a trapezium. The two parallel sides are 8 m and 12 m, and the perpendicular distance between them is 5 m. Grass seed covers 4 m² per bag, sold only in whole bags. Work out how many bags of grass seed are needed.
- 9.A circular coaster has a radius of 4 cm. Work out the circumference of the coaster. Give your answer in terms of π.
- 10.A sector of a circle has radius 10 cm and angle 72°. Using π = 3.14, work out the arc length of the sector.
- 11.A circle has a diameter of 24 cm. A sector of this circle has an angle of 90°. Using π = 3.14, work out the area of the sector.
- 12.A trapezium has parallel sides of 5.6 cm and 8.4 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 13.A birthday cake is a cylinder of radius 10 cm and height 8 cm, with a thin ribbon fixed exactly once around the curved side, at half the height, and joined with no overlap. Work out the length of ribbon needed. Use π = 3.14.
- 14.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 15.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
Answer key
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (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) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (c) 56.52 cm² — Sector area = (45 ÷ 360) × 3.14 × 12² = 0.125 × 3.14 × 144 = 56.52 cm². (4.71 cm² comes from forgetting to square the radius; 452.16 cm² comes from finding the area of the whole circle and forgetting the angle fraction; 9.42 cm² comes from using the arc length formula instead of the sector area formula.)
- (c) 13 — Area of the trapezium = 1/2 × (8 + 12) × 5 = 1/2 × 100 = 50 m². Number of bags = 50 ÷ 4 = 12.5, which rounds up to 13 bags since seed is sold only in whole bags. A student who mistakenly uses 2 m² of coverage per bag instead of 4 m² finds 50 ÷ 2 = 25 bags.
- (a) 8π cm — Circumference = 2πr. Substitute r = 4: circumference = 2 × π × 4 = 8π cm. Using r in place of 2r (halving the formula) gives 4π cm. Using the area formula πr² in place of the circumference formula gives π × 4² = 16π cm. Multiplying 2 × 4 without including π at all gives 8 cm.
- (d) 12.56 cm — Arc length = (angle ÷ 360) × 2 × π × r = (72 ÷ 360) × 2 × 3.14 × 10 = 0.2 × 62.8 = 12.56 cm. (62.8 cm comes from finding the full circumference and forgetting to take the angle fraction; 6.28 cm comes from leaving out the factor of 2, using πr instead of 2πr; 25.12 cm comes from using the diameter, 20 cm, in place of the radius.)
- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (b) 28 cm² — Area of a trapezium = (sum of parallel sides) ÷ 2 × height. Sum of parallel sides = 5.6 + 8.4 = 14 cm. Half of that is 14 ÷ 2 = 7 cm. Area = 7 × 4 = 28 cm². A pupil who forgets to halve gets 14 × 4 = 56 cm². A pupil who uses only the longer parallel side, as if this were a rectangle, gets 8.4 × 4 = 33.6 cm². A pupil who subtracts the parallel sides instead of adding them gets (8.4 − 5.6) ÷ 2 × 4 = 1.4 × 4 = 5.6 cm². The correct area is 28 cm².
- (a) 62.8 cm — The ribbon goes once around the circular cross-section, so its length equals the circumference: 2πr = 2 × 3.14 × 10 = 62.8 cm.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
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