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 ramp rises at 30° to the horizontal. Its sloping surface is 4.8 m long. Safety rules say the vertical rise of a ramp must be no more than 2.5 m. Work out how far below that limit the rise of this ramp is.
- 2.Write down which one of these quadrilaterals always has diagonals that are equal in length and that cross at right angles.
- 3.The three angles of a triangle are 10°, 58° and 112°. Write down the name that describes this triangle.
- 4.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.
- 5.A kite string makes an angle of 30° with the ground. The kite is flying at a height of 6 m directly above a point on the ground. Using the exact value of sin 30°, work out the exact length of the kite string.
- 6.Shape S has an area of 48 cm². It is enlarged by a scale factor of 1/4 to give shape T. Work out the area of shape T.
- 7.Triangle JKL and triangle MNP are similar, with JK ÷ MN = JL ÷ MP = 2. A student says triangle JKL must be congruent to triangle MNP. Write down why the student is wrong.
- 8.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 9.A right-angled triangle has a hypotenuse of 29 cm and one shorter side of 20 cm. Work out the length of the other shorter side.
- 10.Quadrilateral WXYZ has WX parallel to ZY, and WZ = XY (the two non-parallel sides are equal in length). Give a reason why angle W = angle X.
- 11.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.
- 12.Triangle PQR has vertices P(0, 0), Q(6, 0) and R(0, 8). Work out the perimeter of the triangle.
- 13.A flower bed in the shape of a parallelogram has a base of 3.5 m and a perpendicular height of 2 m. Compost costs £4.80 per m². Work out the total cost of covering the flower bed with compost.
- 14.The perimeter of a square is 20 cm. Work out the area of the square.
- 15.In triangle ABC, point D lies on side AB and point E lies on side AC, so that DE is parallel to BC. AD = 3 cm, DB = 6 cm and DE = 4 cm. Work out the length of BC.
Answer key
- (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
- (a) Square — Method: two conditions are being asked for at once, so test each shape against both — the two diagonals must always be the same length as each other, and they must always meet at 90°. Working: in a rectangle the diagonals are equal but they meet at 90° only in the special case where the rectangle is also a rhombus; in a rhombus the diagonals do meet at 90° but they are of different lengths unless the rhombus is also a rectangle; the shape that satisfies both conditions for every example of it is the one that is both, and its diagonals are equal and perpendicular. Answer: the square. The distractors: the rectangle is where a candidate stops who tests only the equal-length condition and never checks the angle at the crossing; the rhombus is where a candidate stops who tests only the right-angle condition and never checks the two lengths; the parallelogram is chosen by a candidate who remembers that the diagonals of a parallelogram bisect each other and treats bisecting each other as being equal to each other, which is a different property.
- (c) Scalene — Method: equal angles in a triangle sit opposite equal sides, so compare the three angles with each other. Working: 10°, 58° and 112° are all different, so no two sides of the triangle are equal either, and a triangle with no equal sides is scalene. Answer: scalene. The distractors: isosceles is chosen by candidates who pair up the two acute angles, 10° and 58°, as base angles without checking that they are actually equal; equilateral is chosen by candidates who check that the three angles add to 180° and take that as meaning the triangle is regular; right-angled is chosen by candidates who see that 112° is larger than 90° and classify the triangle as containing a right angle, when in fact none of the three angles is 90°.
- (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.
- (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
- (a) 3 cm² — Area scale factor = (linear scale factor)² = (1/4)² = 1/16. Area of T = 48 × 1/16 = 3 cm². (12 cm² comes from multiplying by the linear scale factor 1/4 directly, without squaring it; 24 cm² comes from taking the square root of the scale factor instead of squaring it; 768 cm² comes from squaring the reciprocal of the scale factor, 4, instead of the scale factor itself.)
- (c) The scale factor is 2, not 1, so the sides are not equal — Congruent shapes must be exactly the same size as well as the same shape, which means a scale factor of 1. Here the scale factor between the triangles is 2, so the sides are different lengths and the triangles cannot be congruent, even though they are similar. 'Similar triangles are never congruent' is too strong — a scale factor of exactly 1 would make them both similar and congruent. 'The angles are not necessarily equal' is wrong, since similar triangles always have equal matching angles. 'Congruent triangles must have a right angle' is an unrelated, false fact about congruence.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (a) Isosceles trapezium: base angles are equal — WX is parallel to ZY and the two non-parallel sides WZ and XY are equal in length, so WXYZ is an isosceles trapezium. In an isosceles trapezium the two angles at each of the parallel sides are equal, so angle W = angle X (and angle Z = angle Y). A student who answers with the parallelogram property has quoted a fact that is true of a parallelogram, but WZ and XY are given as non-parallel so WXYZ is not a parallelogram — and in a parallelogram "opposite angles" pairs W with Y, not W with X. A student who answers with the kite property has again taken a true fact about the wrong shape: a kite's equal sides are two pairs of adjacent sides, not the two non-parallel sides of a trapezium. A student who answers with the rhombus property has used something no rhombus has — a rhombus has two pairs of parallel sides and only two pairs of equal angles; all four angles are equal only in a square.
- (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.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (b) £33.60 — The area of the parallelogram flower bed is base × height = 3.5 × 2 = 7 m². The cost is 7 × £4.80 = £33.60. £16.80 comes from using the triangle formula instead of the parallelogram formula: 3.5 × 2 = 7, and half of 7 is 3.5 m², then 3.5 × £4.80 = £16.80. £26.40 comes from adding the base and height, 3.5+2 = 5.5, instead of multiplying them, then multiplying by £4.80. £7.00 correctly finds the area, 7 m², but forgets to multiply it by the cost per m².
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (d) 12 cm — Since DE is parallel to BC, triangle ADE is similar to triangle ABC (the two triangles share angle A, and the parallel lines make the angles at D and E equal to the angles at B and C). The whole side AB = AD + DB = 3 + 6 = 9 cm. The scale factor from the small triangle to the large triangle is AB ÷ AD = 9 ÷ 3 = 3, so BC = DE × 3 = 4 × 3 = 12 cm. Scaling by DB ÷ AD instead of AB ÷ AD gives 4 × (6 ÷ 3) = 8 cm. Adding DB directly onto DE, instead of scaling, gives 4 + 6 = 10 cm. Treating the triangles as congruent rather than similar, so assuming corresponding sides are simply equal, gives BC = DE = 4 cm.
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