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) 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.
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (a) 12π + 16 cm — A three-quarter sector's perimeter is the curved arc plus the two straight radii that close the shape. The full circumference is 2 × π × 8 = 16π cm, and three-quarters of that is 12π cm. Adding the two straight radii, 8 cm each, gives 12π + 16 cm. Leaving out the straight edges gives just 12π cm. Using one-quarter of the circumference, the piece left over rather than the piece asked for, gives 4π + 16 cm. Adding only one radius instead of two gives 12π + 8 cm.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (d) 16 — Method: use the scale to find the real length and width separately, then use the perimeter formula. Working: real length = 4 cm × 125 = 500 cm = 5 m; real width = 2.4 cm × 125 = 300 cm = 3 m; perimeter = 2 × (5 + 3) = 16 m. A student who answers 8 has added the real length and width but forgotten to double the total for the perimeter. A student who answers 1600 has correctly worked out the perimeter in centimetres but forgotten to convert it to metres. A student who answers 500 has only converted the length to real centimetres and stopped there, ignoring the width and the perimeter step. Answer: 16 m.
- (d) SAS — Method: a congruence condition is named by the parts that are given equal and the order in which they sit round the triangle, so count the sides and the angles first. Working: AB = DE and AC = DF are two pairs of equal sides, and the equal angle at A and D lies between AB and AC, so the given parts read side, included angle, side. Answer: SAS. The distractors: SSS needs three pairs of equal sides, and the third pair, BC and EF, is not given — it follows from the proof rather than being part of it; ASA reads the two equal sides as two equal angles, swapping which facts are which; RHS applies only when the triangles contain a right angle and the equal pair includes the hypotenuse, and nothing here says the angle at A is 90°.
- (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
- (d) No transformation — every point stays exactly where it was — The two vectors (5, −3) and (−5, 3) are opposites, so adding them gives (0, 0): every point ends up exactly where it started, and there is no transformation at all. Misreading the second vector's signs and effectively adding (5, −3) to itself instead of to its opposite gives a translation by the vector (10, −6). Assuming two translations must combine into a reflection gives a reflection in the x-axis — but a reflection reverses orientation, and translations never do. Assuming that two opposite vectors must mean a half turn gives a rotation of 180° about the origin — but a 180° rotation moves every point except its own centre, whereas this pair of translations leaves every single point exactly where it was. Two translations by opposite vectors always cancel exactly, leaving every point unmoved.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (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.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (b) (2, 3) — Applying A first: reflecting (2, 1) in the x-axis gives (2, −1). Applying B to that image: translating (2, −1) by (0, 4) gives (2, −1 + 4) = (2, 3). Applying the transformations in the opposite order — B first, then A — gives a different result: (2, 1) translates to (2, 5), which then reflects to (2, −5); this shows that the order genuinely matters here. Applying only A and stopping there, without the translation, gives (2, −1). Applying only B and stopping there, without the reflection, gives (2, 5). Do both transformations, in the order A then B, and the image of P is (2, 3).
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