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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (d) 283 cm³ — Volume of a cylinder = πr²h. Using π = 3.14: V = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6, which rounds to 283 cm³. A pupil who uses the diameter, 6 cm, instead of the radius gets 3.14 × 6² × 10 = 1130.4 ≈ 1130 cm³. A pupil who forgets to square the radius gets 3.14 × 3 × 10 = 94.2 ≈ 94 cm³. A pupil who uses the circumference formula 2πr instead of πr² gets 2 × 3.14 × 3 × 10 = 188.4 ≈ 188 cm³. The correct volume is 283 cm³.
- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
- (a) 3/4 — 1 litre = 1000 ml, so 750 ml is 750/1000 of a litre. Dividing both the numerator and denominator by 250 simplifies this to 3/4. Writing the fraction upside down, as the litre out of the 750 ml, gives 4/3. Finding the fraction of the litre that is NOT filled, 250/1000, gives 1/4. Dividing the numerator by 250 but the denominator by only 100, an inconsistent simplification, gives 3/10.
- (a) 5.9 — The real length is 9.8 × 60 = 588 cm. Converting to metres, by dividing by 100, gives 5.88 m, which rounds to 5.9 m to 1 decimal place. A candidate who rounds 5.88 down instead of up gets 5.8 m. A candidate who forgets to convert from centimetres to metres gets 58.8. A candidate who divides by 60 instead of multiplying gets 0.16, to 2 decimal places. The real length, to 1 decimal place, is 5.9 m.
- (b) It is the longest of the three chords — Method: in any circle the length of a chord is decided by how far the chord lies from the centre, because a chord passing nearer the centre cuts further across the circle. Working: the chord through O lies at a distance of zero from the centre, and no chord can lie closer than that, so no chord of the circle can be longer than it; a chord through the centre is a diameter. The other two chords lie at some distance greater than zero, so each of them falls short of that maximum. Answer: It is the longest of the three chords. The distractors: It is the shortest of the three chords comes from reversing the rule and picturing a chord near the centre as a short line tucked inside; It is the same length as the other two chords comes from carrying the fact that all radii of a circle are equal across to chords, which are not all equal; It is half the length of each of the other two chords comes from confusing a chord through the centre with a radius, which really is half a diameter.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (b) A triangular prism standing on its triangular end — Method: a rectangular elevation with no sloping sides means the solid keeps the same cross-section all the way from the bottom to the top; work out which solid, standing the right way up, has a triangular cross-section that stays that shape as you go higher. Working: a triangular prism standing upright on its triangular end has a triangle as its plan view, and because the cross-section is constant all the way up, both the front and side elevations are plain rectangles. Answer: a triangular prism standing on its triangular end. The distractors: a triangle-based pyramid standing on its triangular base does give a triangle as its plan view, but its cross-section shrinks towards the apex, so its front and side elevations come to a point and are triangles, not rectangles. A cuboid standing on a rectangular face is wrong because its plan view is a rectangle, not a triangle. A triangular prism lying on one of its rectangular faces is wrong because it is then the triangular end that faces the side, so its plan view is a rectangle and one of its elevations is a triangle.
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (d) $\binom{−5}{−3}$ — Left is a negative horizontal move and down is a negative vertical move, so the vector is $\binom{−5}{−3}$. $\binom{5}{−3}$ comes from treating 'left' as a positive move. $\binom{−5}{3}$ comes from treating 'down' as a positive move. $\binom{−3}{−5}$ comes from swapping the horizontal and vertical components.
- (d) The centre — Method: sort the four names by what kind of object each one is, because only one of them names a single point rather than a line or a curve. Working: a chord is a straight line joining two points on the circle; a diameter is the special chord that runs right across the circle through the middle, so it too is a line; an arc is a piece of the circle's own curve. That leaves one name, and it belongs to the point from which every radius is drawn, which is why every point on the circle is the same distance from it. Answer: The centre. The distractors: The chord is picked by a candidate who reads 'the same distance' as a line of fixed length rather than as a point; The diameter is picked by a candidate who names the line that passes through the point instead of the point itself; The arc is picked by a candidate who hunts for a part of the circle's own curve rather than for a point inside it.
- (a) 105° — The interior angles of a pentagon add up to (5 − 2) × 180° = 540°. Subtracting the three known angles, 540 − 100 − 110 − 120 = 210°, and this 210° is shared equally between the two angles marked x°, so each one is 210 ÷ 2 = 105°. 210° stops one step early, giving the total of the two unknown angles instead of one of them. 108° is the interior angle of a regular pentagon, which does not apply here since this pentagon's angles are not all equal. 55° comes from halving one of the given angles, 110°, instead of halving the remaining total.
- (a) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (b) a rectangle — Lying on its side, the cylinder's curved surface touches the table along a straight line, and the two flat circular ends face sideways rather than up or down; viewed from directly above, the outline traced is a rectangle — as long as the cylinder and as wide as its diameter. "a circle" would be correct if the cylinder stood upright on one of its circular ends instead of lying on its side. "a triangle" belongs to a cone lying or standing so that it narrows to a point in that view, which a cylinder never does. "an oval" is a common guess from picturing the round ends, but from directly above those ends are edge-on and contribute to the rectangle's short sides, not a curved outline.
- (b) $\binom{-6}{10}$ — Translating twice by the same vector doubles both components: 2 × $\binom{-3}{5}$ = $\binom{-6}{10}$. $\binom{-3}{5}$ forgets to double the vector at all, giving only one translation's worth. $\binom{-9}{15}$ trebles the vector instead of doubling it. $\binom{-6}{5}$ doubles only the top number and forgets to double the bottom number.
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