Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Geometry and measures worksheet — GCSE Higher
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- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (b) trapezium — A quadrilateral with exactly one pair of parallel sides is a trapezium; field ABCD has AB parallel to DC and no other pair of parallel sides, so trapezium is correct. Parallelogram requires BOTH pairs of opposite sides to be parallel, but the plan says only AB and DC are parallel. Rhombus requires all four sides to be equal in length, which is not stated here. A kite is defined by two pairs of adjacent equal sides, not by having a pair of parallel sides, so it does not match this description either.
- (a) opposite angles of a parallelogram are equal — P and R are opposite vertices of the parallelogram, and opposite angles of a parallelogram are always equal, which is why angle R equals angle P, 65°. Co-interior angles adding up to 180° is the correct fact for angle Q or angle S, the angles adjacent to P along a side, not for the opposite angle R. Alternate angles are equal is a fact about a transversal crossing two parallel lines, which explains other angle relationships in the parallelogram, not the one between opposite angles P and R directly. Angles on a straight line adding up to 180° applies to two angles that sit together on one straight line, which P and R do not.
- (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (c) 58° — In an isosceles triangle, the base angles opposite the equal sides are equal. Since DE = DF, angle F is the base angle equal to angle E, so angle F = 58°.
- (d) y = 2 — Two reflections in perpendicular lines that cross at a point combine to a 180° rotation about that point. The line x = 3 is vertical, so the second line must be horizontal, and it must pass through the centre of rotation (3, 2) — that line is y = 2. Taking the y-coordinate of the centre but writing it against the wrong letter gives y = 3. Assuming the second line must also be vertical, like the first one, and just swapping in the other coordinate gives x = 2. Reaching for the standard mirror line y = x without checking that it actually passes through (3, 2) gives y = x — it does not pass through that point at all. The line that is both perpendicular to x = 3 and through (3, 2) is y = 2.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
- (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.
- (a) 2 — Gradient = (change in y) ÷ (change in x) = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. "0.5" comes from dividing the change in x by the change in y the wrong way round: 4 ÷ 8. "8" is only the change in y, forgetting to divide by the change in x at all. "−2" comes from a sign error, as if the y-coordinate had decreased rather than increased.
- (a) 9 — AB is a vertical segment, since A and B share the x-coordinate 1, and its length is the difference in y-coordinates: 5 − 1 = 4. BC is a horizontal segment, since B and C share the y-coordinate 5, and its length is the difference in x-coordinates: 6 − 1 = 5. The total path length is 4 + 5 = 9. 20 comes from multiplying the two lengths, 4 × 5, instead of adding them. 5 is only the length of BC, forgetting to include AB. 4 is only the length of AB, forgetting to include BC.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (c) A rotation of 180° about the origin — An enlargement by scale factor −1 sends every point (x, y) to (−x, −y) — both coordinates change sign. A rotation of 180° about the origin does exactly the same thing to every point, so the two transformations have identical effect. A reflection in the x-axis only changes the sign of the y-coordinate, sending (x, y) to (x, −y), leaving the x-coordinate untouched. A reflection in the y-axis only changes the sign of the x-coordinate, sending (x, y) to (−x, y), leaving the y-coordinate untouched. Treating a negative scale factor as though it behaves like a positive one gives no transformation at all, but the minus sign is not decorative — it reverses both coordinates. Both signs flip together, which is exactly what a 180° rotation about the origin does.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
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