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) One third of the cylinder's volume — Volume of a cylinder = base area × height. Volume of a cone = 1/3 × base area × height. For the same base radius and height, the cone's volume is exactly one third of the cylinder's, so it uses less wax. A student who thinks the cone is half the cylinder's volume has confused it with a different solid's ratio. A student who thinks the two volumes are the same has ignored the 1/3 factor in the cone formula entirely. A student who thinks the cone is two thirds of the cylinder's volume has the right idea that it is a fraction, but the wrong fraction.
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
- (c) (6, −5) — Method: the translation has already happened, so it must be undone: reverse the vector and apply the reverse to the point that is given. Working: reversing $\binom{-2}{6}$ gives $\binom{2}{-6}$, so the x-coordinate is 4 + 2 = 6 and the y-coordinate is 1 − 6 = −5. Check: from (6, −5) the given vector gives 6 − 2 = 4 and −5 + 6 = 1, which is the point named in the question. Answer: (6, −5). Applying the vector forwards instead of backwards gives (2, 7). Reversing the horizontal movement but not the vertical one gives (6, 7), and reversing the vertical movement but not the horizontal one gives (2, −5).
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (b) 96 cm³ — Method: the volume of a pyramid is one third of the base area multiplied by the vertical height. Work out the area of the square base, multiply by the height, then divide by 3. Working: the base area is 6 × 6 = 36 cm², then 36 × 8 = 288, and 288 ÷ 3 = 96. Answer: 96 cm³. The distractors: 288 cm³ comes from multiplying the base area by the height and forgetting the one third, which is the volume of a cuboid with the same base and height; 144 cm³ comes from halving that 288 instead of taking a third of it; 16 cm³ comes from using the base edge of 6 cm in place of the base area, (6 × 8) ÷ 3.
- (a) Similar, but AAA alone does not prove congruence — Three equal corresponding angles (AAA) show that the two triangles are similar — the same shape — but says nothing about their size, so it does not prove congruence. They could be congruent, or one could simply be an enlargement of the other; without matching side lengths, congruence is not established, so 'congruent because AAA proves congruence' is wrong. Equal angles do not force equal sides — a triangle can be enlarged to any size while keeping the same angles, so that option is also wrong. Something CAN be said here — that the triangles are similar — so 'no relationship can be determined' is wrong too.
- (b) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
- (b) Statement 3 — a triangle's angles are said to sum to 360° — The angles of any triangle sum to 180°, not 360° — Statement 3 uses the wrong total, and that error is what sends its final line to the impossible claim that angle ACB = 180°. The correct working is angle OAC + angle OBC + angle ACB = 180°, and since angle ACB = angle OCA + angle OCB, this gives 2 × angle ACB = 180°, so angle ACB = 90°, which is the actual theorem. Statement 1 correctly identifies OA and OC as equal radii, making triangle OAC isosceles — nothing wrong there. Statement 2 correctly does the same for triangle OBC. Statement 4 correctly states that A, O and B are collinear, since a diameter passes through the centre — also nothing wrong there. Statement 3 is the one to flag: it is the angle sum it quotes that is wrong, not the diagram or the radii.
- (b) 23 — If every position were filled to the full height of 3, the total would be 2 × 4 × 3 = 24 crates. One corner position has only 2 crates instead of 3, one crate short of full height there, so the actual total is 24 − 1 = 23. "24" comes from using the full height everywhere and forgetting the one incomplete corner. "22" comes from removing 2 crates for the incomplete corner instead of the 1 that is actually missing (3 − 2 = 1, not 2). "21" comes from removing all 3 crates at that corner, as though the position were completely empty rather than 2 crates short.
- (d) 2 — Method: if f is 3 times e, then each part of f equals 3 times the matching part of e. Working: using the bottom numbers, 6 = 3 × k, so k = 2. Answer: k = 2. A candidate who multiplies instead of dividing, working out 6 × 3, gets 18. A candidate who uses the top numbers' ratio instead, 15 ÷ 5, and gives that ratio as k gets 3. A candidate who adds instead of using the multiple relationship, working out 6 + 3, gets 9.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (a) Isosceles trapezium: base angles are equal — WX is parallel to ZY and the two non-parallel sides WZ and XY are equal in length, so WXYZ is an isosceles trapezium. In an isosceles trapezium the two angles at each of the parallel sides are equal, so angle W = angle X (and angle Z = angle Y). A student who answers with the parallelogram property has quoted a fact that is true of a parallelogram, but WZ and XY are given as non-parallel so WXYZ is not a parallelogram — and in a parallelogram "opposite angles" pairs W with Y, not W with X. A student who answers with the kite property has again taken a true fact about the wrong shape: a kite's equal sides are two pairs of adjacent sides, not the two non-parallel sides of a trapezium. A student who answers with the rhombus property has used something no rhombus has — a rhombus has two pairs of parallel sides and only two pairs of equal angles; all four angles are equal only in a square.
- (c) 68° — Since PA and PB are both tangents from the external point P, PA = PB, so triangle PAB is isosceles with equal base angles at A and B. The angles of triangle PAB sum to 180°, so angle PAB + angle PBA = 180° − 44° = 136°, and since the two base angles are equal, angle PAB = 136° ÷ 2 = 68°. Angle PAB is the angle between the tangent at A and the chord AB, so by the alternate segment theorem it equals the angle in the alternate segment, angle ACB = 68°. Using angle APB directly as angle ACB, without using the isosceles triangle to find angle PAB first, gives 44°. Halving angle APB directly, rather than subtracting it from 180° before halving, gives 22°. Finding 180° − 44° = 136° correctly but forgetting to divide by 2 for one base angle leaves 136°.
- (a) 46° — Method: use the angle at the centre, then the cyclic quadrilateral, then the angle sum of a triangle. Working: angle ABC = 152° ÷ 2 = 76° (angle at the centre is twice the angle at the circumference); angle ADC = 180° − 76° = 104° (opposite angles of a cyclic quadrilateral sum to 180°); angle DAC = 180° − 104° − 30° = 46° (angle sum of triangle ACD). Skipping the halving step and taking angle ABC = 152° gives angle ADC = 180° − 152° = 28° and then angle DAC = 122°; using angle ABC = 76° directly as angle ADC, skipping the cyclic quadrilateral step, gives 74°; and finding angle DAC as 180° − 104° without subtracting the given 30° gives 76°. Each of the three steps must be carried through in order — halve, then subtract from 180°, then subtract from 180° again with the extra angle.
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