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 tile is described as having two pairs of parallel sides, four equal sides, and diagonals that cross at right angles but are not equal in length. Give a reason why this tile cannot be a square.
- 2.A recipe uses 0.75 kg of sugar. Work out how many grams of sugar this is.
- 3.The midpoint of the line segment AB is (3, 5). A is the point (1, 3). Work out the coordinates of B.
- 4.A kitchen scale is marked in equal divisions of 20 g. The pointer rests three divisions past the 400 g mark. Work out the reading on the scale.
- 5.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?
- 6.A regular nonagon has 9 sides. Work out the sum of its interior angles.
- 7.A garden designer describes a flower bed as a shape with four straight sides: exactly one pair of sides is parallel, those two parallel sides are different lengths, and the other two sides are equal in length to each other but not parallel to each other. Which shape should the designer draw?
- 8.Work out the distance between the points (−3, 4) and (5, −2).
- 9.Priya says: 'To find the arc length of a sector, you divide the circumference by the angle in degrees.' Which statement about Priya's method is correct?
- 10.In triangle ABC the angle at A is 90° and BC = 10 cm. For the angle at B, sin B = 3/5. Work out the length of AC.
- 11.A sector of a circle has radius 6 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector.
- 12.Work out the exact value of tan 30° + tan 30°.
- 13.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 14.A cylindrical drum and a cuboid box are being wrapped for a display. A shop assistant wraps only the flat faces of each item, not any curved surface. How many flat faces in total does the assistant wrap across both items?
- 15.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
Answer key
- (a) A square's diagonals are equal in length — A square has two pairs of parallel sides, four equal sides, and diagonals that are equal in length and cross at right angles. This tile's diagonals are not equal, so it is a rhombus that is not a square.
- (a) 750 — Method: to change kilograms to grams, multiply by 1000. Working: 0.75 × 1000 = 750. Answer: 750 g. A candidate who multiplies by 100 instead of 1000 gets 75. A candidate who multiplies by 10000 instead of 1000 gets 7500. A candidate who divides by 1000 instead of multiplying gets 0.00075.
- (a) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (c) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (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.
- (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
- (a) isosceles trapezium — One pair of parallel sides, plus a separate pair of equal non-parallel sides, is exactly the definition of an isosceles trapezium — the shape the designer should draw. Parallelogram is wrong because a parallelogram needs BOTH pairs of opposite sides parallel, but only one pair is parallel here. Kite is wrong because a kite has two separate pairs of adjacent equal sides and no requirement for any sides to be parallel, a different combination of properties. Rhombus is wrong because a rhombus needs all four sides equal, but the description only makes two of the four sides equal to each other.
- (d) 10 — Method: the distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical gaps, so work out both gaps first, handling the negative coordinates carefully, and then apply Pythagoras' theorem. Working: the horizontal gap is 5 − (−3) = 5 + 3 = 8 and the vertical gap is 4 − (−2) = 4 + 2 = 6. Then d² = 8² + 6² = 64 + 36 = 100, so d = √100 = 10. Answer: 10. The distractors: 14 comes from adding the two gaps, 8 + 6, instead of adding their squares and taking the root; 100 comes from stopping at the sum of the squares and never taking the square root; 50 comes from reaching 100 correctly and then halving it instead of taking its square root, a candidate who has read the last step as “halve” rather than “root”.
- (a) Wrong: multiply by angle ÷ 360, not divide by it. — The arc length is the fraction angle ÷ 360 of the circumference, and that fraction is MULTIPLIED by the circumference — it is never divided by the angle itself. So Priya's method, dividing by the angle, is wrong, and the correct statement names both the fraction and the multiplication. Saying she is correct accepts her wrong division. Saying the arc length is the full circumference multiplied by the angle in degrees, with no division at all, drops the 360 that turns the angle into a fraction. Saying the arc length is the radius multiplied by the angle in degrees swaps the circumference for the radius, which is a different and, in degrees, incorrect formula.
- (b) 6 cm — Method: the right angle is at A, so BC is the hypotenuse and AC is the side opposite the angle at B; sin B = opposite ÷ hypotenuse therefore gives AC ÷ BC = 3/5. Working: AC ÷ 10 = 3/5, so AC = 10 × 3 ÷ 5 = 6. Answer: 6 cm. The distractors: 8 cm is AB, the side next to the angle at B, which is what cos B = 4/5 produces — the right method used on the wrong side; 3 cm comes from reading the 3 in the ratio as a length and never scaling it up to the 10 cm hypotenuse; 30 cm comes from multiplying by 3 and forgetting to divide by 5.
- (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.)
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
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