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
Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.A trapezium-shaped allotment plot has two parallel sides that face each other, and two sloping sides that are equal in length (an isosceles trapezium). One of the angles next to the shorter parallel side is 118°. Work out the angle next to the other end of the same shorter parallel side.
- 2.In triangle ABC and triangle DEF, the angle at A equals the angle at D, and AB ÷ DE = AC ÷ DF. Write down what this proves about the two triangles.
- 3.A straight line crosses two parallel lines. One of the angles formed is (2x + 10)°, and the angle alternate to it is 74°. Work out the value of x.
- 4.Two similar flags have widths in the ratio 2 : 5. The width of the smaller flag is 8 cm. Work out the width of the larger flag.
- 5.In kite WXYZ, WX = WZ and XY = ZY (so at each of X and Z, one side from each of the two unequal pairs meets). Angle X = 100°. Work out angle Z.
- 6.A triangle has vertices A(1, 1), B(5, 1) and C(5, 4). Work out the lengths of AB, BC and CA, and write down whether triangle ABC is right-angled.
- 7.In triangle ABC, angle ABC = 90° and angle BAC = 30°. The hypotenuse AC = 12 cm. Work out the exact length of AB.
- 8.A square has sides of length 7 cm. Work out its area.
- 9.Point C is at (5, 2). It is reflected in the x-axis. Work out the coordinates of the image of point C.
- 10.A ladder rests against a wall, making an angle of 60° with the ground. The ladder is 2 m long. Using the exact value of sin 60°, work out how high up the wall the ladder reaches, giving your answer in centimetres.
- 11.Triangle ABC has a right angle at B and hypotenuse AC = 17 cm, with AB = 8 cm. Triangle DEF has a right angle at E and hypotenuse DF = 17 cm. Which extra fact about DEF would complete the RHS condition for proving that ABC ≅ DEF, with the vertices matching in that order?
- 12.Two triangles are congruent to each other. What does this tell you about their angles and sides?
- 13.In triangle ABC and triangle DEF, the angle at A and the angle at D are both 40°, and the angle at B and the angle at E are both 70°. No side lengths are given. Write down whether the two triangles must be similar, with the reason.
- 14.A solid pyramid has a square base of side 6 cm. Its apex is directly above the centre of the base, at a vertical height of 8 cm. Work out the volume of the pyramid.
- 15.Triangle XYZ has angle X = 90°. Angle Y = 34°. Work out angle Z.
Answer key
- (d) 118° — In an isosceles trapezium, the two angles next to the same parallel side are equal, because the sloping sides are equal in length. So the angle at the other end of the shorter parallel side also equals 118°.
- (d) Similar, by SAS — Method: read off whether the given sides are equal or only in the same ratio, and where the given angle sits. Sides in proportion point to similarity; sides equal in length would be needed for congruence. Working: AB ÷ DE = AC ÷ DF gives two pairs of corresponding sides in the same ratio, and the equal angle at A and D lies between AB and AC in the first triangle and between DE and DF in the second, so it is the included angle. Two pairs of sides in proportion with the included angle equal is the SAS condition for similarity. Answer: similar, by SAS. The distractors: 'Similar, by AA' quotes a condition that needs two pairs of equal angles, while only one pair is given here; 'Congruent, by SAS' reads AB ÷ DE = AC ÷ DF as AB = DE and AC = DF, turning a statement about proportion into one about equal lengths; 'Congruent, by SSS' makes that same misreading and adds an assumption that the third pair of sides is equal too, which nothing in the question says.
- (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
- (a) 20 cm — The scale factor from the smaller flag to the larger flag is 5 ÷ 2 = 2.5, so the larger width is 8 × 2.5 = 20 cm. The distractor 3.2 cm comes from using the ratio the wrong way round, 8 × 2 ÷ 5 = 3.2. The distractor 11 cm comes from adding the difference between the ratio numbers (5 − 2 = 3) to the given width, 8 + 3 = 11. The distractor 40 cm comes from multiplying by 5 without dividing by 2 first, 8 × 5 = 40.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (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) (5, −2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, 2) → (5, −2). ((−5, 2) comes from reflecting in the y-axis instead, changing the sign of the x-coordinate; (−5, −2) comes from changing the sign of both coordinates, as for a reflection in the origin; (2, 5) comes from swapping the coordinates, which is what happens for a reflection in the line y = x.)
- (c) 100√3 cm — The height is opposite the 60° angle, so height = 2 × sin 60° = 2 × √3/2 = √3 m. Converting to centimetres: √3 m = 100√3 cm. '√3 cm' forgets to convert the answer from metres to centimetres. '200√3 cm' comes from mis-recalling sin 60° as √3 instead of √3/2, dropping the denominator of the exact value: 2 × √3 = 2√3 m = 200√3 cm. '50√3 cm' comes from halving the ladder's length before multiplying by sin 60°, instead of using the full 2 m.
- (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.
- (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.
- (d) Yes, by the AA condition — Method: similarity is decided by the angles, and because the three angles of a triangle add up to 180°, two matching pairs force the third pair to match as well. Working: the angles at A and D are both 40° and the angles at B and E are both 70°, so the third angles are both 180° − 40° − 70° = 70° and all three pairs are equal. Two pairs were enough, and that is the AA condition. Answer: yes, by the AA condition. The distractors: 'Yes, by the SSS condition' names a condition about three pairs of sides in proportion, and the question gives no side lengths at all; 'No, because no side lengths are given' treats side information as necessary, which it is for congruence but not for similarity; 'No, because the triangles may be different sizes' turns the definition of similarity into an objection, since similar figures are allowed to differ in size and only their shape must match.
- (b) 96 cm³ — Method: the volume of a pyramid is one third of the base area multiplied by the vertical height. Work out the area of the square base, multiply by the height, then divide by 3. Working: the base area is 6 × 6 = 36 cm², then 36 × 8 = 288, and 288 ÷ 3 = 96. Answer: 96 cm³. The distractors: 288 cm³ comes from multiplying the base area by the height and forgetting the one third, which is the volume of a cuboid with the same base and height; 144 cm³ comes from halving that 288 instead of taking a third of it; 16 cm³ comes from using the base edge of 6 cm in place of the base area, (6 × 8) ÷ 3.
- (a) 56° — The angles of a triangle add up to 180°. Add the two known angles: 90° + 34° = 124°. Subtract from 180°: 180° − 124° = 56°.
Build your own mix at the worksheet builder.