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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- (c) 18 — Method: divide the real length by 5 to find how many 'units' of 5 m it contains, then multiply by 2 cm for each unit. Working: 45 ÷ 5 = 9, so the real bridge is 9 lots of 5 m; each lot is represented by 2 cm on the model, so the model length is 9 × 2 = 18 cm. Options: 9 comes from stopping after the division, without multiplying by the 2 cm per unit; 90 comes from multiplying the real length by 2 directly, without dividing by 5 first; 4.5 comes from dividing by 5 and then dividing by 2 again, instead of multiplying by 2. Answer: 18.
- (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°.
- (c) 2 — A point 4 cm from A lies on a circle of radius 4 cm centred at A; a point 3 cm from B lies on a circle of radius 3 cm centred at B. Since AB = 5 cm, and 4 + 3 = 7 is greater than 5 while 4 − 3 = 1 is less than 5, the two circles genuinely cross each other, at two separate points, one on each side of line AB. "1" comes from wrongly assuming the circles only touch rather than cross, which would need 4 + 3 to equal exactly 5. "0" comes from wrongly assuming the circles miss each other completely. "4" comes from counting where each circle crosses the line AB itself (two points each) instead of counting where the two circles cross each other.
- (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.
- (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.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (c) base angles of an isosceles triangle are equal — Because AB = AC, triangle ABC is isosceles, and the base angles of an isosceles triangle — the two angles opposite the equal sides — are always equal, which is why angle C equals angle B, 70°. Angles in a triangle adding up to 180° is a true fact about the triangle as a whole, but it is not the reason two specific angles are equal to each other. Corresponding angles are equal is a fact about parallel lines cut by a transversal, which does not apply inside a single triangle like this. Vertically opposite angles are equal is a fact about two lines crossing, not about a triangle's base angles.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (d) 49° — Method: first combine the two given angles at the centre to find the whole central angle FOM, then apply the angle-at-the-centre theorem, which halves the central angle to give the angle at the circumference on the major arc. Working: angle FOM = angle FOP + angle POM = 55 + 43 = 98 degrees. Since S is on the major arc FM, angle FSM = 98 ÷ 2 = 49 degrees. Answer: 49°. Add BOTH given angles to find the whole angle FOM before you halve anything: halving only one of them, doubling the total instead of halving it, or subtracting from 180° all give the wrong angle for the programme.
- (c) (4, 3) — Method: the midpoint of a segment is the mean of its two end points, so its x-coordinate is the mean of the two x-coordinates and its y-coordinate is the mean of the two y-coordinates. Working: for x, (1 + 7) ÷ 2 = 8 ÷ 2 = 4. For y, (3 + 3) ÷ 2 = 6 ÷ 2 = 3. The midpoint is therefore (4, 3). Answer: (4, 3). The distractors: (3, 3) comes from halving the difference of the x-coordinates, (7 − 1) ÷ 2 = 3, which measures half the distance instead of locating the point; (3.5, 3) comes from halving only the larger x-coordinate and leaving the smaller one out of the working; (4, 0) comes from averaging the x-coordinates correctly but then subtracting the y-coordinates, 3 − 3, rather than averaging them.
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (c) 6 squares — Method: the plan view shows the footprint of the solid; a second layer stacked on top of floor positions that are already covered does not create any new squares in the plan. Working: the row of 4 cubes and the row of 2 cubes attached at the end do not overlap, so the footprint has 4 + 2 = 6 distinct squares. Answer: 6 squares. The distractors: 12 squares comes from counting the total number of cubes used, including the second layer (6 floor positions × 2 layers = 12), instead of the footprint. 4 squares comes from counting only the row of four and forgetting the attached row of two. 5 squares comes from wrongly treating the corner square as shared between the two rows (4 + 1 instead of 4 + 2).
- (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.
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