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.
Geometry and measures worksheet — GCSE Higher
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- 1.In triangle ABC, AB = 8.6 cm, AC = 11.4 cm and angle BAC = 47°. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 2.In a right-angled triangle one of the other two angles is 45°, and the side opposite that 45° angle is 7 cm. Work out the length of the hypotenuse. Give your answer to 1 decimal place.
- 3.In triangle ABC, AB = 8 cm, AC = 7 cm and the area of the triangle is 24 cm². Given that angle BAC is acute, work out the size of angle BAC. Give your answer to 1 decimal place.
- 4.Line 1 passes through (0, 1) and (2, 5). Line 2 passes through (3, 2) and (5, 6). Work out the gradient of each line. Then write down whether the two lines are parallel.
- 5.A goat is tied by a rope 7 m long to a post at a corner of a rectangular field, where two fences meet at a right angle. The goat can reach anywhere inside the field that the rope allows. Using π = 3.14, work out the area the goat can graze, to the nearest square metre.
- 6.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 7.A transversal crosses a pair of parallel lines. At one line, the angle is (5x + 4)°. The corresponding angle at the other line is (3x + 24)°. Work out the value of x.
- 8.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
- 9.A, B and C are points on a circle with centre O. B is on the minor arc AC. Angle ABC = 100°. Work out the size of the non-reflex angle AOC.
- 10.A flower bed in the shape of a parallelogram has a base of 3.5 m and a perpendicular height of 2 m. Compost costs £4.80 per m². Work out the total cost of covering the flower bed with compost.
- 11.Triangle ABC is similar to triangle PQR. Side AB is 8 cm and corresponds to side PQ, which is 12 cm. Side AC is 6 cm and corresponds to side PR. Work out the length of PR.
- 12.A shape is reflected in the line y = −x. Which of these points is invariant under this reflection?y = −x
- 13.A tangent touches a circle with centre O at the point P. Q is a point on the tangent. Write down the circle fact that tells you the size of angle OPQ.
- 14.Triangle ABC has AB = 10 cm, BC = 7 cm and angle A = 40°. Triangle DEF has DE = 10 cm, EF = 7 cm and angle D = 40°. Sam says the two triangles are congruent by SAS. Which statement about Sam's claim is correct?
- 15.Lines AB and CD are parallel. A straight line EF crosses AB at point P and crosses CD at point Q. At P, angle APE = 65°. Angle APE and angle BPQ are vertically opposite. Angle BPQ and angle PQD are co-interior (allied) angles. Work out angle PQD.
Answer key
- (a) 35.9 cm² — Method: the area of any triangle is 1/2 × a × b × sin C, where a and b are two sides and C is the angle between them. Working: the 47° angle lies between AB = 8.6 cm and AC = 11.4 cm, so the area is 1/2 × 8.6 × 11.4 × sin 47° = 49.02 × 0.73135 = 35.851. Answer: the area is 35.9 cm² to 1 decimal place. The distractors: 71.7 cm² comes from leaving out the factor 1/2 and working out 8.6 × 11.4 × sin 47°; 33.4 cm² comes from pressing cos instead of sin, 49.02 × cos 47°, which is the same as using the complement 43° in place of 47°; 52.6 cm² comes from pressing tan instead of sin, 49.02 × tan 47°.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (a) 59.0° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject: sin C = 2 × Area ÷ (a × b). Working: sin C = 2 × 24 ÷ (8 × 7) = 0.857, so C = sin⁻¹(0.857) = 59.0° (1 d.p.), which is acute as the question requires. Answer: 59.0°. Taking the obtuse angle instead of the acute one asked for, 180 − 59.0 = 121.0°; forgetting to double the area before dividing gives sin C = 24 ÷ (8 × 7) = 0.4286, whose acute angle is 25.4°; and combining that same forgotten-doubling error with the obtuse branch gives 180 − 25.4 = 154.6°. Always double the area first, and then pick the acute branch, since that is what this question asks for.
- (a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
- (b) 160° — B is on the minor arc AC, so the angle at the centre theorem applies to the reflex angle AOC: reflex angle AOC = 2 × 100° = 200°. The non-reflex angle AOC is the rest of the full turn: 360° − 200° = 160°. Reporting the reflex angle itself, without subtracting it from 360°, gives 200°. Subtracting angle ABC from 180° instead, as if this were a cyclic quadrilateral, gives 80°. Halving angle ABC instead of doubling it gives 50°. Double first, then take the angle away from a full turn, and 160° is what's left.
- (b) £33.60 — The area of the parallelogram flower bed is base × height = 3.5 × 2 = 7 m². The cost is 7 × £4.80 = £33.60. £16.80 comes from using the triangle formula instead of the parallelogram formula: 3.5 × 2 = 7, and half of 7 is 3.5 m², then 3.5 × £4.80 = £16.80. £26.40 comes from adding the base and height, 3.5+2 = 5.5, instead of multiplying them, then multiplying by £4.80. £7.00 correctly finds the area, 7 m², but forgets to multiply it by the cost per m².
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (a) (4, −4) — A point is invariant under a reflection only if it lies exactly on the mirror line. The line y = −x consists of every point where the y-coordinate is the negative of the x-coordinate: (4, −4) satisfies this, since −4 = −(4), so it is invariant. (5, 5) lies on the line y = x, a different line altogether, not y = −x. (4, 4) has equal coordinates, but that alone does not put it on y = −x; it would need y = −4, not 4. (−4, −4) also has equal coordinates and lies on y = x, not y = −x — its coordinates would need opposite signs to sit on the given mirror line. Only a point whose coordinates are negatives of each other stays fixed under this reflection.
- (d) The angle between a tangent and a radius is 90° — OP is the radius drawn to the point of contact P, and a circle theorem states that a tangent always meets that radius at a right angle, so angle OPQ = 90°. A tangent is not parallel to the radius it touches — at the point of contact it is perpendicular to that radius, not parallel to it. A tangent does not pass through the centre — a straight line through the centre that also touches the circle at one point would have to be a diameter, which is a different line entirely. The angle-in-a-semicircle theorem needs a triangle drawn inside the circle with a diameter as its longest side; there is no such triangle here, just a tangent and a radius.
- (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.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
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