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.
Geometry and measures worksheet — GCSE Foundation
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- 1.A straight line crosses a pair of parallel lines. One of the co-interior (allied) angles is 118°. Work out the size of the other co-interior angle.
- 2.A tile is described as having two pairs of parallel sides, four equal sides, and diagonals that cross at right angles but are not equal in length. Give a reason why this tile cannot be a square.
- 3.A roof truss is shaped like an isosceles triangle resting on a horizontal ceiling joist. The truss's sloping edges are equal in length, and the angle at its apex is 40°. The ceiling joist continues in a straight line beyond the foot of the truss. Work out the angle between the truss's sloping edge and the extended joist, on the outside of the truss.
- 4.A parallelogram has an area of 136 cm² and a base of 17 cm. Work out the perpendicular height of the parallelogram.
- 5.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 6.The point (4, −3) is reflected in the y-axis. Work out the coordinates of the image point.
- 7.A solid has 5 faces, 9 edges and 6 vertices. Two of its faces are triangles and the other three are rectangles. Which solid is it?
- 8.A children's slide is built as a right-angled triangle: the vertical ladder side is 2.4 m, the horizontal base along the ground is 3.2 m, and the sloping slide is the third side. Work out the exact length of the sloping slide.
- 9.A circle has diameter 10 cm. Using π = 3.14, work out the circumference of the circle.
- 10.Write down the name given to a triangle whose three sides are all the same length.
- 11.A square has sides of length 7 cm. Work out its area.
- 12.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 13.A ship sails 12 km on a bearing of 070° from a harbour to a buoy. It then sails a further 9 km, still on a bearing of 070°, from the buoy to a lighthouse. How far is the lighthouse from the harbour, in kilometres?
- 14.Triangle PQR has vertices P(0, 0), Q(6, 0) and R(0, 8). Work out the perimeter of the triangle.
- 15.To construct a perpendicular to a line at a given point P on the line, compasses are first opened to draw two arcs crossing the line, one on each side of P. What is the correct next step?
Answer key
- (b) 62 — Method: co-interior (allied) angles between parallel lines add up to 180°. Working: 180 − 118 = 62. Answer: 62°. A candidate who treats co-interior angles as equal, like corresponding angles, gives 118. A candidate who uses 360° instead of 180°, working out 360 − 118, gets 242. A candidate who subtracts as if the angles were complementary, working out 118 − 90, gets 28.
- (a) A square's diagonals are equal in length — A square has two pairs of parallel sides, four equal sides, and diagonals that are equal in length and cross at right angles. This tile's diagonals are not equal, so it is a rhombus that is not a square.
- (a) 110° — The truss is isosceles, so its two base angles are equal; since the three angles of the triangle add up to 180° and the apex is 40°, each base angle is (180 − 40) ÷ 2 = 70°. The angle between the sloping edge and the joist, extended beyond the base of the truss, sits on a straight line with that 70° base angle, and angles on a straight line add up to 180°, so the required angle is 180 − 70 = 110°. "70°" comes from stopping after finding the base angle and giving it directly, without also using the straight-line fact to find the angle on the OUTSIDE of the truss. "140°" comes from doubling the base angle instead of using the straight-line fact. "20°" comes from taking half of the apex angle (40° ÷ 2), mistaking it for the required angle instead of properly using the triangle's angle sum.
- (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.)
- (a) $\binom{4}{4}$ — A movement to the right is a positive top number and a movement up is a positive bottom number, so 4 right and 4 up gives $\binom{4}{4}$. $\binom{4}{-4}$ wrongly gives the vertical movement a negative sign, as if it were downward. $\binom{-4}{4}$ wrongly gives the horizontal movement a negative sign, as if it were leftward. $\binom{-4}{-4}$ gets both signs wrong, as if the translation were left and down.
- (d) (−4, −3) — Reflecting in the y-axis keeps the y-coordinate the same and changes the sign of the x-coordinate, so (4, −3) maps to (−4, −3). "(4, 3)" changes the sign of the y-coordinate instead, which is what happens when reflecting in the x-axis. "(−4, 3)" changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the y-axis. "(3, −4)" swaps the two coordinates around instead of reflecting either of them.
- (a) Triangular prism — Count the given properties against each solid. A triangular prism has 5 faces (2 triangular ends + 3 rectangular sides), 9 edges and 6 vertices — this matches exactly, so the solid is a triangular prism. A square-based pyramid also has 5 faces, but 8 edges and 5 vertices, and only one face is a square rather than three rectangles — the edge and vertex counts don't match. A cuboid has 6 faces, 12 edges and 8 vertices, none of which match the numbers given. A tetrahedron has 4 faces, all triangles, with 6 edges and 4 vertices — too few faces, and no rectangular faces at all.
- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
- (d) Equilateral — Method: triangles are named by their sides — three equal sides, exactly two equal sides, or no equal sides — or by their angles. Working: the triangle described has three equal sides, which is the equilateral case, and each of its angles is 60°. Answer: equilateral. The distractors: isosceles is the name for a triangle with exactly two equal sides, and is chosen by candidates who remember only that it is the 'equal sides' word; scalene is the name for a triangle whose sides are all different, so it is the opposite of what is described; right-angled classifies a triangle by an angle of 90° rather than by its sides, and is chosen by candidates who recall that the angles of this triangle are all equal and take 'equal angles' to mean 'right angles'.
- (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.
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (b) Draw wider arcs from each crossing point, meeting above. — Once the two arcs cross the line at points either side of P, compasses are opened to a radius greater than before and arcs are drawn from each of those two points so that they meet above (or below) the line; joining that meeting point to P gives the perpendicular. (Joining the two crossing points with a straight line only retraces part of the original line, since both points already lie on it; drawing a circle centred at P through both crossing points does not locate any new point needed for the perpendicular; drawing an arc centred at P through only one crossing point repeats the first step instead of moving on to the second pair of arcs.)
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