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.A square tile has sides of length 4.5 cm. Work out the perimeter of the tile.
- 2.In triangle ABC the base BC is 10 cm long and the sloping side AB is 13 cm long. The perpendicular height from A down to BC is 12 cm. Work out the area of triangle ABC.
- 3.A ship's radio can be heard up to 30 km from the ship. A lighthouse's light can be seen up to 20 km from the lighthouse. The ship and the lighthouse are 40 km apart along the coast. Describe the region where BOTH the radio can be heard AND the light can be seen.
- 4.A hexagonal prism has 8 faces and 12 vertices. Using F + V − E = 2, work out how many edges it has.
- 5.A solid has 10 vertices and 15 edges. Using the formula F + V − E = 2, work out how many faces it has.
- 6.A landscaper marks out two triangular flower beds using stakes and a tape measure. Bed 1 has two edges of 5 m and 7 m, meeting at a corner with a marked angle of 60° between them. Bed 2 has edges of 5 m and 7 m, meeting at a corner also marked at 60° between them. The landscaper wants to check the two beds will be exactly the same size and shape using only these three measurements. Which condition proves the two beds are congruent?
- 7.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 8.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 9.A roof has two triangular sections, P and Q. Section P has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them. Section Q has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them, arranged the same way round. A roofer wants to cut identical triangular felt panels for both sections without measuring Section Q separately. Which condition confirms that the two sections are congruent?
- 10.Write down the mathematical name for a quadrilateral that has exactly one pair of parallel sides.
- 11.A scale drawing of a bridge uses a scale of 1 : 25. A support beam on the drawing measures 4.4 cm. What is the real length of the beam, in metres?
- 12.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?
- 13.A cylindrical water tank has a solid hemispherical dome on top, both with radius 2 m. Work out the volume of the hemispherical dome alone. Use π = 3.14 and give your answer correct to 3 decimal places.
- 14.Triangle DEF has a right angle at E. Angle DFE = 60° and EF = 9 cm. Work out the exact length of DE.
- 15.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.
Answer key
- (c) 18 cm — Method: a square has four equal sides, so the perimeter is 4 × the side length. Working: 4 × 4.5 = 4 × 4 + 4 × 0.5 = 16 + 2 = 18. Answer: 18 cm. The distractors: 9 cm comes from 2 × 4.5, adding only one pair of sides; 16 cm comes from rounding the side length down to 4 cm before multiplying, so the 0.5 cm on each side is lost; 20.25 cm² comes from working out 4.5 × 4.5, which is the area of the tile and carries a squared unit.
- (b) 60 cm² — Method: the area of a triangle is half the base multiplied by the perpendicular height, and here the perpendicular height is the 12 cm, not the sloping side. Working: 10 × 12 = 120, then 120 ÷ 2 = 60. Answer: 60 cm². The distractors: 65 cm² comes from using the sloping side of 13 cm as the height, (10 × 13) ÷ 2; 120 cm² comes from using the right two lengths but forgetting to halve, 10 × 12; 78 cm² comes from taking 13 cm and 12 cm as the base and the height and ignoring BC altogether, (13 × 12) ÷ 2.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (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).
- (a) 7 — Rearranging F + V − E = 2 gives F = 2 − V + E = 2 − 10 + 15 = 7. A candidate who works out E − V without the +2, giving 15 − 10, answers 5. A candidate who rearranges with a sign error, working out 2 + V − E = 2 + 10 − 15 = −3 and then drops the negative sign, answers 3. A candidate who adds all three numbers together, V + E + 2 = 10 + 15 + 2, answers 27, having used the wrong operation entirely. The correct number of faces is 7.
- (b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (a) $\binom{4}{4}$ — A movement to the right is a positive top number and a movement up is a positive bottom number, so 4 right and 4 up gives $\binom{4}{4}$. $\binom{4}{-4}$ wrongly gives the vertical movement a negative sign, as if it were downward. $\binom{-4}{4}$ wrongly gives the horizontal movement a negative sign, as if it were leftward. $\binom{-4}{-4}$ gets both signs wrong, as if the translation were left and down.
- (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.
- (d) SAS — two sides, included angle — Each section has two known sides, 3.6 m and 2.4 m, with the 70° angle between them, matching in both sections; this is exactly the SAS condition, so 'SAS — two sides, included angle' is correct. 'SSS — but only two sides given' is wrong because SSS requires three pairs of equal sides, but only two sides are given for each triangle here. 'ASA — angle between two sides' is wrong because ASA requires two angles with a side between them, but only one angle, 70°, is given, not two. 'Cannot prove — only one angle' is wrong because SAS is specifically designed to prove congruence from exactly two sides and the one angle between them, so no further angle is needed.
- (b) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs of parallel sides, not one, so it does not fit the definition. A kite has no parallel sides at all. A rhombus, like a parallelogram, has two pairs of parallel sides. The correct name is trapezium.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (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°.
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (d) 9√3 cm — DE is opposite the 60° angle at F, and EF is adjacent to it, so DE = EF × tan 60° = 9 × √3 = 9√3 cm. 9√3/2 cm comes from using sin 60° = √3/2 instead of tan 60°. 3√3 cm comes from using tan 30° = 1/√3 instead of tan 60° (9 × 1/√3 = 9/√3 = 3√3). 18 cm is the hypotenuse DF, not DE: it comes from using cos 60° = 1/2 and working out 9 ÷ 1/2 = 18, which finds the wrong side of the triangle.
- (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.
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