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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- (a) (−1, −0.5) — Method: for an enlargement centred at a point O that is not the origin, first find the point's position relative to O, multiply that by the scale factor, then add O's coordinates back on — image = O + k × (F − O). Working: F relative to O is (4, 2) − (2, 1) = (2, 1). Multiplying by the scale factor, −1.5 × (2, 1) = (−3, −1.5). Adding O back on: (2, 1) + (−3, −1.5) = (−1, −0.5). Answer: (−1, −0.5). Measure F's position from the aperture O, not from the origin, apply the full signed scale factor −1.5 to that displacement, and add O's own coordinates back on afterwards: measuring from the origin instead of O, treating the factor as +1.5, or adding O onto the wrong displacement all place the image mark somewhere else.
- (b) An enlargement, centre O, scale factor −1/2. — A single enlargement with a negative scale factor both changes the size (by the magnitude of the factor) and rotates the image 180° about the centre (giving the inverted orientation) in one transformation. Halving the size needs a magnitude of 1/2, and inverting needs a negative sign, so the scale factor is −1/2, centre O. 'Scale factor 1/2' gives the correct size but no inversion, since a positive scale factor keeps the same orientation. 'A 180° rotation with no change in size' gives the inversion but not the halving — K is described as smaller than J, so size must change too. 'Scale factor −2' inverts correctly but doubles the size instead of halving it.
- (a) 65° — OT and OC are both radii, so triangle OTC is isosceles with OT = OC, and its base angles are equal: angle OTC = angle OCT = (180° − 130°) ÷ 2 = 25°. A tangent is perpendicular to the radius at the point of contact, so angle OTP = 90°. Since C lies inside angle OTP, the chord TC splits this right angle into angle OTC and angle PTC, so angle PTC = angle OTP − angle OTC = 90° − 25° = 65°. Stopping after finding the base angle of the isosceles triangle, without subtracting it from the right angle at T, leaves 25° instead of the angle actually asked for. Adding the base angle to the right angle instead of subtracting it, 90° + 25° = 115°, reverses the direction the two angles combine in. Finding the sum of the two base angles of the isosceles triangle, 180° − 130° = 50°, and stopping there, gives another wrong value entirely.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (b) A semicircle of radius 4 m, away from the wall. — Every point the dog can reach is at most 4 m from the fixed ring, so without any wall the region would be a full circle of radius 4 m. The wall runs straight through the ring and blocks the dog from crossing it, and since the wall extends further than the lead in both directions, exactly half of that circle is cut off — leaving a semicircle of radius 4 m on the side of the wall the dog is tied on. (A full circle of radius 4 m ignores that the wall blocks half of the region; a quarter circle of radius 4 m would only be correct if the ring were fixed at a corner where two walls met, not along a single straight wall; a rectangle 4 m wide along the wall ignores that the lead lets the dog swing round in a curve, not stay a fixed distance out from the wall.)
- (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
- (a) 28.6 m — The perimeter of a sector is the two straight radii plus the curved arc. This sector's 90° angle is one quarter of a full turn, so its arc length is one quarter of the full circle's circumference. The full circumference is 2 × 3.14 × 8 = 50.24 m, and one quarter of that is 50.24 ÷ 4 = 12.56 m. Add the two 8 m radii: 12.56 + 8 + 8 = 28.56 m, which rounds to 28.6 m. Choosing 12.6 m gives the arc length alone (rounded), forgetting the two straight edges of the sector. Choosing 20.6 m adds only one radius to the arc length instead of two, missing one of the two straight sides. Choosing 41.1 m comes from using the diameter, 16 m, as if it were the radius when working out the arc length (2 × 3.14 × 16 = 100.48, one quarter of which is 25.12), then adding the two correct 8 m radii (25.12 + 8 + 8 = 41.12).
- (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) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (c) 10.4 cm — Method: rearrange Area = (1/2)ab sin C to make the unknown side the subject: b = 2 × Area ÷ (a × sin C). Working: b = 2 × 36 ÷ (9 × sin 50°) = 10.4 cm (1 d.p.). Answer: 10.4 cm. Forgetting to double the area before dividing gives b = 36 ÷ (9 × sin 50°) = 5.2 cm; using cos 50° instead of sin 50° gives b = 2 × 36 ÷ (9 × cos 50°) = 12.4 cm; and multiplying by sin 50° instead of dividing by it — inverting the rearrangement — gives b = 2 × 36 × sin 50° ÷ 9 = 6.1 cm. Always double the area before dividing, and check whether the unknown should be multiplied or divided by sin C once you've rearranged.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (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) 8.1 m — Multiply the length on the plan by the scale factor: 4.5 × 1.8 = 8.1, so the real length is 8.1 m. Choosing 6.3 m adds the two numbers instead of multiplying them (4.5 + 1.8 = 6.3). Choosing 2.5 m divides the plan length by the scale factor the wrong way round (4.5 ÷ 1.8 = 2.5) instead of multiplying. Choosing 9.0 m rounds the scale factor up to 2 before multiplying (4.5 × 2 = 9.0), losing the accuracy the 1.8 was giving.
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