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.
Non-calculator
Answer key: Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (b) (6, 3) — For an enlargement centred on the origin, multiply both coordinates by the scale factor: (2 × 3, 1 × 3) = (6, 3). A pupil who adds the scale factor to each coordinate instead of multiplying gets (2 + 3, 1 + 3) = (5, 4). A pupil who multiplies only the x-coordinate gets (6, 1). A pupil who multiplies only the y-coordinate gets (2, 3). The correct image is (6, 3).
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (d) Cylinder — Method: check how many faces are flat and how many are curved, and how many flat faces there are. Working: this solid has a curved surface joining two flat circular faces of equal size, which is exactly a cylinder. A student who answers cone has forgotten a cone has only one flat circular face, not two. A student who answers sphere has forgotten a sphere has no flat faces at all. A student who answers triangular prism has spotted that the solid has two identical end faces but has taken those ends to be triangles rather than circles, and a prism's side faces are flat rectangles, not one curved surface. Answer: cylinder.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (c) 5 cm — Diameter = circumference ÷ π, so 31.4 ÷ 3.14 = 10 cm, and the radius is half the diameter, so 10 ÷ 2 = 5 cm. 10 cm is the diameter itself, given as the radius by forgetting the final halving step. 15.7 cm comes from halving the circumference, 31.4 ÷ 2 = 15.7, and stopping there, treating half the circumference as the radius without ever dividing by π. 2.5 cm comes from halving the correct radius again, effectively dividing by 2 twice instead of once.
- (a) 49 cm² — The area of a square is side × side = 7 × 7 = 49 cm². 28 cm² is the perimeter of the square (4 × 7), not its area. 14 cm² adds just two of the sides together. 21 cm² multiplies the side length by 3 instead of squaring it.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (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.
- (b) 20 cm — Method: corresponding sides of similar triangles are in the same ratio, and the longest side of one triangle corresponds to the longest side of the other; a ratio of 2 : 5 means each length is multiplied by 5 ÷ 2 = 2.5 going from the smaller triangle to the larger one. Working: the longest side of the smaller triangle is 8 cm, so the matching side of the larger triangle is 8 × 2.5 = 20. Answer: 20 cm. The distractors: 10 cm comes from scaling the shortest side, 4 cm, instead of the longest; 40 cm comes from multiplying by 5 and forgetting to divide by 2; 3.2 cm comes from multiplying by 2 ÷ 5 instead of 5 ÷ 2, which scales from the larger triangle down to the smaller one.
- (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.
- (b) 28 — Method: total bricks = (number of squares in the footprint) × (the height of the wall in bricks). Working: the footprint has the row of 5 squares plus the 2 squares in the arm that stands out from the middle of that row, and they do not overlap, giving 5 + 2 = 7 squares; multiplying by the height of 4 bricks gives 7 × 4 = 28. Answer: 28. The distractors: 20 comes from using only the row of 5 and ignoring the arm (5 × 4). 24 comes from treating the bottom square of the arm as if it were one of the row's own squares, so the arm is counted as adding only 1 new square instead of 2 (5 + 1 = 6, then 6 × 4). 32 comes from counting the square of the row directly below the arm a second time as part of the arm (5 + 3 = 8, then 8 × 4).
- (d) All angles equal and all sides equal — Congruent triangles are identical in both shape and size, so every corresponding angle is equal and every corresponding side is equal, so 'all angles equal and all sides equal' is correct. 'All angles equal, sides may differ' describes 'similar' triangles, which have equal angles but sides in the same ratio rather than necessarily equal, not congruent ones. 'All sides equal, angles may differ' is not geometrically possible for triangles, since equal sides throughout force the angles to match too. 'Same area, angles and sides may differ' is wrong because equal area alone does not guarantee congruence; two triangles can share an area with completely different shapes.
- (b) circumference — The distance all the way around the outside edge of a circle is called the circumference. The diameter is a straight line across the circle through the centre, so it is a length through the circle, not around it. The radius is a straight line from the centre to the edge, again a length across, not around. The area is the amount of surface inside the circle, a region, not a length at all.
- (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.)
Build your own mix at the worksheet builder.