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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- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (b) (1, 6) — Method: when a square is set square-on to the grid, so that its sides run parallel to the axes, each vertex shares its x-coordinate with one neighbour and its y-coordinate with the other, and the missing vertex then borrows one coordinate from each of the two vertices it is joined to; so the first job is to check from the given points that the sides really do run parallel to the axes. Working: A(1, 2) and B(5, 2) share y = 2, so AB is a horizontal side; B(5, 2) and C(5, 6) share x = 5, so BC is a vertical side, which confirms that this square lies square-on to the axes and that the rule may be used. In square ABCD the vertex D is joined to C and to A. DC must be horizontal like AB, so D takes the y-coordinate of C, which is 6; DA must be vertical like CB, so D takes the x-coordinate of A, which is 1. D is therefore (1, 6), and checking confirms every side is 4 long. Answer: (1, 6). The distractors: (1, 5) comes from lifting the first number out of each of A and C, pairing the x-coordinate of A with the x-coordinate of C; (6, 1) comes from finding the right two numbers but writing them the wrong way round, height before sideways position; (9, 6) comes from stepping a further 4 to the right from C instead of closing the square back to the column A stands in.
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
- (b) 2 500 000 cm³ — Method: a volume conversion uses the length factor three times, once for each dimension. Working: 1 m = 100 cm, so a cube of side 1 m is a cube of side 100 cm, and 100 × 100 × 100 = 1 000 000, giving 1 m³ = 1 000 000 cm³. Then 2.5 × 1 000 000 = 2 500 000. Answer: 2 500 000 cm³. Using the plain length factor 100 gives 250 cm³. Using 1000, which is the factor that turns cubic metres into litres, gives 2500 cm³. Using 10 000, which is the factor that belongs to square metres and square centimetres, gives 25 000 cm³.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (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) SAS — Two pairs of matching sides are equal (AB = DE and BC = EF) and the angle between them is equal (angle ABC = angle DEF), which is exactly the Side-Angle-Side congruence test: SAS. SSS is wrong because only two pairs of sides are given, not three. ASA is wrong because only one angle is given, and it sits between the two sides rather than the side sitting between two angles. RHS is wrong because no right angle is mentioned in either triangle.
- (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) Yes - SSS, the three side lengths all match — Bracket P's sides (12 cm, 16 cm, 20 cm) can each be matched to one of Bracket Q's sides (16 cm, 20 cm, 12 cm) — the same three lengths, just listed differently — so the brackets are congruent by SSS. SAS is wrong here because no angle is stated for either bracket, only three sides. The order the sides are listed in does not matter for SSS — only whether the SET of three lengths matches, and it does, so 'listed in a different order' is not a reason to say no. Nothing extra is needed: SSS proves congruence from side lengths alone, without any angles, so it can be determined.
- (d) 12.56 cm — Arc length = (angle ÷ 360) × 2 × π × r = (72 ÷ 360) × 2 × 3.14 × 10 = 0.2 × 62.8 = 12.56 cm. (62.8 cm comes from finding the full circumference and forgetting to take the angle fraction; 6.28 cm comes from leaving out the factor of 2, using πr instead of 2πr; 25.12 cm comes from using the diameter, 20 cm, in place of the radius.)
- (c) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (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.)
- (d) 1.2 m² — Method: an area in square metres needs lengths in metres, so convert first and then multiply. Working: 100 cm = 1 m, so 150 cm = 1.5 m and 80 cm = 0.8 m, and the area = 1.5 × 0.8 = 1.2 m². Answer: 1.2 m². The same result comes from working in centimetres: 150 × 80 = 12 000 cm², and a square metre is a square of side 100 cm, so 100 × 100 = 10 000 cm² make one square metre and 12 000 ÷ 10 000 = 1.2. Dividing the 12 000 cm² by 100 instead, as though a square metre held only 100 square centimetres, gives 120 m²; dividing by 1000 gives 12 m². Working out the perimeter rather than the area gives 1.5 + 0.8 + 1.5 + 0.8 = 4.6, which is a length and not an area.
- (a) 110° — Method: in a parallelogram the two angles at the ends of one side are co-interior angles between a pair of parallel sides, so they add up to 180°. Working: angle A + angle B = 180°, so angle B = 180° − 70° = 110°. Answer: 110°. The distractors: 70° comes from applying the rule for opposite angles of a parallelogram, which are equal, to two angles that are next to each other instead; 20° comes from treating the two angles as complementary and working out 90° − 70°; 290° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given, 360° − 70°.
- (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.
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