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.
Answer key: Geometry and measures worksheet — GCSE Higher
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- (d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (a) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
- (c) (1/3)a + (2/3)b — Method: OP = OA + AP, and since AP is twice PB, AP is 2/3 of the whole of AB, with AB = b − a. Working: OP = a + 2/3(b − a) = a − (2/3)a + (2/3)b = (1/3)a + (2/3)b. Answer: OP = (1/3)a + (2/3)b. Measuring 2/3 of AB from B's end instead of A's swaps the fractions round, giving (2/3)a + (1/3)b; adding (2/3)b onto the whole of a without first subtracting a inside the bracket gives a + (2/3)b; and treating the ratio as though AP and PB were equal gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from the point named first in the ratio, subtract before you scale, and then add the result to OA.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
- (a) 7 — Rearranging F + V − E = 2 gives F = 2 − V + E = 2 − 10 + 15 = 7. A candidate who works out E − V without the +2, giving 15 − 10, answers 5. A candidate who rearranges with a sign error, working out 2 + V − E = 2 + 10 − 15 = −3 and then drops the negative sign, answers 3. A candidate who adds all three numbers together, V + E + 2 = 10 + 15 + 2, answers 27, having used the wrong operation entirely. The correct number of faces is 7.
- (a) (−1, 1) — Translating by $\binom{−2}{3}$ subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. This gives image vertices (−1, 4), (3, 4), (3, 7) and (−1, 7). The point (−1, 1) is not one of these: it has the correct new x-coordinate (1 − 2 = −1) but keeps the original y-coordinate (1) instead of adding 3, as if only the horizontal part of the vector had been applied.
- (c) 72° — Method: opposite angles in a cyclic quadrilateral always sum to 180°. Working: angle DAB and angle BCD are opposite angles of the cyclic quadrilateral ABCD, so angle BCD = 180 − 108 = 72 degrees. Answer: 72°. Opposite angles in a cyclic quadrilateral are SUPPLEMENTARY, not equal, so do not simply copy the given angle; and do not subtract from 360°, which is the rule for angles round a point, or confuse this with the angle-in-a-semicircle theorem, which gives a fixed 90° that has nothing to do with this quadrilateral.
- (a) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (b) Rhombus — A rhombus has exactly two lines of symmetry, formed by its two diagonals, and both pairs of opposite angles are equal, but its diagonals are unequal in length. A square also has opposite angles equal, but it has four lines of symmetry and its diagonals ARE equal, so it does not fit. A kite normally has only one line of symmetry and only one pair of opposite angles equal, so it does not fit. A general parallelogram has no lines of symmetry at all, so it does not fit. The correct answer is rhombus.
- (b) 80° — Method: opposite angles of a cyclic quadrilateral sum to 180°, so form and solve an equation for x, then substitute back to find angle DAB. Working: (2x + 10) + (3x − 5) = 180, so 5x + 5 = 180, x = 35, and angle DAB = 2(35) + 10 = 80°. Using 360° instead of 180° as the total (as if the two angles were the whole circle rather than a pair of opposite angles) gives x = 71 and angle DAB = 152°; solving correctly for x but reporting x itself instead of substituting it back into 2x + 10 gives 35°; and mishandling the signs when combining the two expressions, using (2x + 10) + (3x + 5) = 180 instead of (2x + 10) + (3x − 5) = 180, gives x = 33 and angle DAB = 76°. Opposite angles of a cyclic quadrilateral — not adjacent ones — are the pair that sums to 180°.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (d) 1687.5 ml — The length scale factor from the standard bottle to the giant bottle is 1.5, so the volume scale factor is 1.5³ = 3.375. The capacity of the giant bottle is 500 × 3.375 = 1687.5 ml. 750 ml comes from multiplying by the length factor 1.5 directly, without cubing it. 1125 ml comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 2250 ml comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 as the scale factor instead of 1.5³.
- (a) 25° — PA and PB are tangents from the same external point, so PA = PB and OAPB is a kite with right angles at A and B, by the tangent–radius theorem: angle OAP = angle OBP = 90°. The four angles of the kite sum to 360°, so angle AOB = 360° − 90° − 90° − 50° = 130°. Triangle OAB is isosceles because OA = OB, both radii, so its base angles are equal: angle OAB = (180° − 130°) ÷ 2 = 25°. Using angle APB itself as the apex angle of triangle OAB instead of angle AOB gives (180° − 50°) ÷ 2 = 65°. Using a triangle's angle sum on the four-sided kite, and so counting only one of its two right angles, treats angle AOB as 180° − 90° − 50° = 40°, and halving what is then left of triangle OAB gives (180° − 40°) ÷ 2 = 70°. Finding angle AOB = 130° correctly but forgetting to halve for the isosceles base angle gives 180° − 130° = 50°. Halve that remaining angle, and 25° is what's left for angle OAB.
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