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) sin 30°, tan 30°, cos 30° — sin 30° = 1/2 = 0.5, tan 30° = √3/3 ≈ 0.577 and cos 30° = √3/2 ≈ 0.866, so the correct order from smallest to largest is sin 30°, tan 30°, cos 30°. 'sin 30°, cos 30°, tan 30°' swaps the last two, wrongly putting cos 30° before tan 30°. 'cos 30°, tan 30°, sin 30°' is the correct list written backwards, from largest to smallest. 'tan 30°, sin 30°, cos 30°' wrongly swaps sin 30° and tan 30° at the start.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (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.
- (c) AB and CD are equal in length — AB = CD states that the line segments AB and CD are equal in length; it says nothing about their direction or position. 'AB is parallel to CD' would be written AB ∥ CD, not AB = CD. 'A, B, C and D all lie on one line' is not what an equals sign between two segment names states at all. 'AB is perpendicular to CD' would be written AB ⊥ CD, not AB = CD.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (c) (−35, −55) — Add the three stages component by component to find the drone's position relative to base: (30+(−10)+15, 40+20+(−5)) = (35, 55). The flight back to base is the negative of this vector, reversing both numbers: (−35, −55). (35, 55) is the vector from base to the drone's position — it forgets to reverse direction for the return flight. (−35, 55) only reverses the top number. (35, −55) only reverses the bottom number.
- (c) 1 — cos 45° = √2/2 and sin 45° = √2/2. Squaring each gives (√2/2)² = 2/4 = 1/2, so (cos 45°)² + (sin 45°)² = 1/2 + 1/2 = 1. The distractor √2 comes from adding cos 45° + sin 45° directly without squaring first (√2/2 + √2/2 = √2). The distractor 2 comes from squaring the top of the fraction, (√2)² = 2, but then dividing by 2 instead of 4 for each term, giving 1 + 1 = 2. The distractor 1/2 comes from squaring only cos 45° and forgetting to add the sin 45° term.
- (b) (5, 1) — First scale a by 2: 2a = (2×3, 2×(−2)) = (6, −4). Then add b component by component: (6+(−1), −4+5) = (5, 1). (2, 3) is a + b without doubling a first. (4, 6) doubles both a and b instead of only a. (7, −9) subtracts b from 2a instead of adding it.
- (b) 30 cm — The perpendicular from the centre of a circle to a chord bisects the chord, so this line, half the chord and the radius form a right-angled triangle. Using Pythagoras' theorem, half the chord = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. The full chord AB is twice this length: AB = 2 × 15 = 30 cm. Stopping after finding the half-chord, without doubling it for the whole chord, gives 15 cm. Adding the radius and the perpendicular distance directly, 17 + 8 = 25 cm, ignores that these two lengths are the two shorter sides of a right-angled triangle, not parts of a straight line. Subtracting instead, 17 − 8 = 9 cm, makes the same mistake in the other direction.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
- (b) Minor segment — Method: compare the sizes of the two regions cut off by the chord, and recall the term used for the smaller one. Working: the chord creates two segments; the smaller region is called the minor segment and the larger one the major segment. A student who answers major segment has picked the larger region by mistake instead of the smaller one. A student who answers minor arc has named the curved boundary rather than the two-dimensional region it encloses. A student who answers semicircle has wrongly assumed the chord must pass through the centre. Answer: minor segment.
- (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.
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