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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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (c) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
- (b) AB = DE — RHS needs a right angle, the hypotenuse, and one OTHER side to be equal; the right angles and hypotenuses are already equal, so a matching pair of the remaining sides, AB = DE, completes RHS. Angle A = angle D is an extra ANGLE fact, not the extra SIDE fact that RHS specifically requires. AC being parallel to DF says nothing about either triangle's side lengths, so it cannot complete a congruence condition. Being drawn the same way up is about orientation on the page, not about any measurement, so it proves nothing about congruence.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (c) tan θ = opposite/adjacent — The tangent ratio is defined as tan θ = opposite/adjacent, so this is the ratio that connects the opposite and adjacent sides. "sin θ = opposite/adjacent" wrongly labels this ratio as sine, when sine actually connects the opposite side and the hypotenuse. "cos θ = opposite/adjacent" wrongly labels it as cosine, when cosine connects the adjacent side and the hypotenuse. "tan θ = adjacent/opposite" uses the correct ratio name but has the opposite and adjacent sides the wrong way round.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
- (b) Acute — An acute angle is any angle less than 90°. Since 47° is less than 90°, it is acute. An obtuse angle is between 90° and 180°, which 47° is not. A reflex angle is greater than 180°, which is far too big for 47°. A right angle is exactly 90°, not 47°. The angle is acute.
- (b) 50° — Method: the three angles of a triangle add up to 180°, and the two angles opposite the equal sides are equal, so subtract the given angle from 180° and halve the remainder. Working: 180° − 80° = 100°, and 100° ÷ 2 = 50°. Answer: 50°. The distractors: 100° comes from subtracting from 180° and forgetting to halve, so it is the two equal angles together; 40° comes from halving the 80° that is given rather than halving what is left of the 180°; 80° comes from assuming that the two base angles must match the angle between the equal sides.
- (d) 2π cm — The arc is a fraction of the whole circumference. The fraction is 72 ÷ 360 = 1/5 of the circle, and the full circumference is 2 × π × 5 = 10π cm. So the arc length is 1/5 × 10π = 2π cm. Taking the whole circumference and forgetting the fraction gives 10π cm. Using 72 ÷ 180 instead of 72 ÷ 360 gives 4π cm. Treating the 5 cm as a diameter instead of a radius gives π cm.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (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°.
- (d) 68 cm² — Method: the area of a parallelogram is base × perpendicular height. Working: 4 × 17 = 68. Answer: 68 cm². The distractors: 34 cm² comes from halving the product, which is the rule for a triangle and not for a parallelogram; 42 cm comes from treating the two given lengths as the sides of the shape and working out a perimeter, 2 × (4 + 17), which is a length and not an area; 21 cm comes from adding the base and the height, 4 + 17, instead of multiplying them.
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