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) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (c) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (c) Rectangle — A rectangle has two pairs of parallel sides and four right angles, but does not require all sides to be equal — this matches exactly, so Rectangle is correct. A square also has four right angles and parallel sides, but additionally requires all four sides to be equal, which contradicts 'not all the same length', so it is wrong. A rhombus has two pairs of parallel sides and all four sides equal, but its angles are not generally 90° unless it is also a square, so it does not match the right-angle condition here. A kite has no pairs of parallel sides at all, so it does not match the first condition given.
- (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.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (b) P, since OP = 5 and OQ = 6 — Using the distance formula, OP = √(3² + 4²) = √(9 + 16) = √25 = 5, and OQ = √(6² + 0²) = √36 = 6. Since 5 is less than 6, P is closer to the origin. Naming Q as closer, with OQ = 5 and OP = 6, has the two distances swapped around the wrong point. Naming P as closer but with OP = 6 and OQ = 5 also has the two values swapped, even though it names the right point. The distances are not equal, since 5 is not the same as 6, so P and Q are not equally distant from the origin.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (c) 140° — Method: OAPB is a quadrilateral whose angles sum to 360°; the tangent-radius angles at A and B are each 90°, and angle APB is given, so angle AOB is whatever is left. Working: angle OAP = angle OBP = 90°, so angle AOB = 360 − 90 − 90 − 40 = 140 degrees. Answer: 140°. Both tangent-radius angles are 90° each and must both be subtracted, not just one, and the quadrilateral's angles sum to 360°, not 180°: do not assume angle AOB simply matches angle APB, which is a different angle in a different part of the figure.
- (c) 5√3 m — The cable, the pole and the ground form a right-angled triangle: the ground distance (5 m) is adjacent to the 60° angle, and the height of the pole is opposite it, so height = 5 × tan 60° = 5 × √3 = 5√3 m. 5√3/2 m comes from using sin 60° = √3/2 instead of tan 60°. 5/√3 m comes from using tan 30° = 1/√3, the reciprocal-angle value, instead of tan 60°. 10√3 m comes from doubling the correct height by mistake.
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