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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- (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.
- (a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
- (d) Triangle 2, by 3.6 cm² — Method: find both areas with 1/2ab sin C, then compare them. Working: area of triangle 1 = 1/2 × 10 × 13 × sin 64° = 58.4 cm²; area of triangle 2 = 1/2 × 11 × 12 × sin 70° = 62.0 cm²; triangle 2 is larger, by 62.0 − 58.4 = 3.6 cm². Getting the right difference but naming triangle 1 as the larger one, the subtraction done the wrong way round, gives 'Triangle 1, by 3.6 cm²'; leaving out the 1/2 when finding triangle 1's area (giving 116.8 cm² instead of 58.4 cm²) and then subtracting gives 'Triangle 1, by 54.8 cm²'; and using cos 64° instead of sin 64° for triangle 1 (giving 28.5 cm² instead of 58.4 cm²) gives 'Triangle 2, by 33.5 cm²'. Work out both areas fully and correctly before comparing which is bigger.
- (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
- (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.
- (c) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (c) 5 hours — The distance from the origin to (7, 24) is √(7² + 24²) = √(49 + 576) = √625 = 25 km. Travelling at 5 km per hour, the time taken is 25 ÷ 5 = 5 hours. 25 hours mistakes the distance itself for the time, forgetting to divide by the speed. 125 hours comes from multiplying the distance by the speed, 25 × 5, instead of dividing. 0.2 hours comes from dividing the speed by the distance, 5 ÷ 25, the wrong way round.
- (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.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (a) 56° — The angles of a triangle add up to 180°. Add the two known angles: 90° + 34° = 124°. Subtract from 180°: 180° − 124° = 56°.
- (c) 56.52 cm² — Sector area = (45 ÷ 360) × 3.14 × 12² = 0.125 × 3.14 × 144 = 56.52 cm². (4.71 cm² comes from forgetting to square the radius; 452.16 cm² comes from finding the area of the whole circle and forgetting the angle fraction; 9.42 cm² comes from using the arc length formula instead of the sector area formula.)
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (d) A rectangle — Method: to find the plan view, work out the outline traced when looking straight down onto the solid from directly above, not the outline shown in the angled sketch. Working: this solid has two identical flat round ends joined by one curved surface, and it is lying on its side rather than standing upright; viewed from above, the curved surface gives two straight edges running the full length of the solid, the width of the round ends apart, and each flat round end — seen edge-on from directly above — also becomes a straight edge of that same width, with no curve remaining. Four straight edges, with opposite sides equal and meeting at right angles, form a rectangle. Answer: a rectangle. The distractors: a circle comes from picturing the solid as if it were standing upright on one of its flat ends, giving the plan of an upright version instead of working out the plan of the solid as it actually lies. An oval comes from copying the foreshortened shape of a round end as it is drawn in the angled sketch, instead of working out the true shape seen from directly above, which has no such foreshortening. A rectangle with rounded ends comes from carrying the curve of the round ends over into the plan view, when in fact a flat round end viewed edge-on from directly above shows no curve at all, only a straight edge.
- (d) 8.8 km — Method: turn each bearing into an angle of triangle ABC, find the third angle from the angle sum, then use the sine rule. Working: B is due east of A, so AB itself lies on a bearing of 090°, and the angle at A between AB and AC is 090° − 062° = 28°. From B, station A lies due west on a bearing of 270°, and C lies on 315°, so the angle at B is 315° − 270° = 45°. The third angle is 180° − 28° − 45° = 107°. The side BC faces the 28° angle and AB = 18 km faces the 107° angle, so BC = 18 × sin 28° ÷ sin 107° = 8.4505 ÷ 0.95630 = 8.8366. Answer: the boat is 8.8 km from B, to 1 decimal place. The distractors: 13.3 km comes from pairing BC with the 45° angle at B instead of the 28° angle it faces, which gives the distance AC; 12.0 km comes from never working out the third angle and dividing by sin 45° instead of sin 107°; 16.6 km comes from using the bearing 062° itself as the angle at A, instead of the 28° between AC and AB.
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