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
MathsUKwww.geekhero.co.uk
- 1.A hexagonal prism has 8 faces and 12 vertices. Using F + V − E = 2, work out how many edges it has.
- 2.A, B, C and D are points on a circle, and ABCD is a cyclic quadrilateral with the vertices in that order around the circle. Angle ABC = 108° and angle ADC = 72°. Which circle theorem is the reason that angle ABC + angle ADC = 180°?
- 3.Two coastguard stations A and B are 18 km apart, with B due east of A. A boat at C is on a bearing of 062° from A and on a bearing of 315° from B. Work out the distance of the boat from station B. Give your answer to 1 decimal place.
- 4.A solid pyramid has a square base of side 6 cm. Its apex is directly above the centre of the base, at a vertical height of 8 cm. Work out the volume of the pyramid.
- 5.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
- 6.A quadrilateral has vertices A(1, 1), B(5, 1), C(6, 4) and D(2, 4). By comparing the y-coordinates of A, B and of D, C, write down whether the sides AB and DC are parallel, and give a reason for your answer.
- 7.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 8.In triangle ABC the angle at C is 90°, AC = 8 cm and the angle at A is 30°. Work out the length of BC. Give your answer to 1 decimal place.
- 9.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 10.A solid cone has a base radius of 6 cm and a vertical height of 10 cm. Work out the volume of the cone. Use π = 3.14 and give your answer correct to 1 decimal place.
- 11.O is the centre of a circle, and PT is a tangent to the circle at the point T. A is a point on the circle, with OA and OT both radii and angle AOT = 122°. At the point T, the chord TA lies between the radius TO and the tangent TP. Work out the size of angle ATP, the angle between the chord and the tangent.
- 12.A landscaper marks out two triangular flower beds using stakes and a tape measure. Bed 1 has two edges of 5 m and 7 m, meeting at a corner with a marked angle of 60° between them. Bed 2 has edges of 5 m and 7 m, meeting at a corner also marked at 60° between them. The landscaper wants to check the two beds will be exactly the same size and shape using only these three measurements. Which condition proves the two beds are congruent?
- 13.Two triangles are similar. The sides of the first triangle are 3 cm, 4 cm and 5 cm. The shortest side of the second triangle is 9 cm. Work out the scale factor from the first triangle to the second.
- 14.A packaging designer is told a box has 6 faces, 12 edges and 8 vertices, and every face is a rectangle. To fit exactly onto a shelf, the designer needs to know whether the box must be a cube. Based only on this information, which solid must the box be?
- 15.The diagram shows a solid made from two cuboids joined together. Which shape is the front elevation of this solid?
Answer key
- (c) 18 — Rearranging F + V − E = 2 gives E = F + V − 2. Substitute F = 8 and V = 12: 8 + 12 − 2 = 18 edges. Choosing 20 comes from adding the faces and vertices but forgetting to subtract the 2 (8 + 12 = 20). Choosing 22 comes from adding the 2 instead of subtracting it (8 + 12 + 2 = 22). Choosing 16 comes from subtracting 2 twice by mistake (8 + 12 − 2 − 2 = 16).
- (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
- (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.
- (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.
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (d) 4.6 cm — Method: BC is opposite the 30° angle and AC is next to it, so the ratio that links the two is tan θ = opposite ÷ adjacent. Working: tan 30° = BC ÷ 8, so BC = 8 × tan 30° = 4.6188…, which is 4.6 to 1 decimal place. Answer: 4.6 cm. The distractors: 13.9 cm comes from dividing by tan 30° instead of multiplying by it; 4.0 cm comes from using sin 30°, which treats the 8 cm side as the hypotenuse when it is the side next to the 30° angle; 6.9 cm comes from using cos 30° in place of tan 30°, which gives the wrong pair of sides.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (b) 61° — Triangle OAT is isosceles because OA = OT, both radii, so its base angles are equal: angle OTA = angle OAT = (180° − 122°) ÷ 2 = 29°. The tangent PT meets the radius OT at 90°, by the tangent–radius theorem, so angle OTP = 90°. Since TA lies between TO and TP, angle OTP splits into angle OTA and angle ATP: angle ATP = 90° − 29° = 61°. Stopping after the isosceles step and reporting the base angle itself gives 29°. Finding the base angles as 180° − 122° without halving, then subtracting from 90°, gives 90° − 58° = 32°. Adding the base angle to 90° instead of subtracting it gives 90° + 29° = 119°. Subtract the base angle from the right angle, and 61° is what's left.
- (b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (b) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (c) An L-shape — Method: to find the front elevation, trace the outline of the solid seen from directly in front. Working: the low, wide cuboid gives a wide rectangle across the bottom, and the taller, narrower cuboid sitting on one end adds a narrower rectangle rising above only that end of the base, so the outline steps up on one side only. Answer: an L-shape. The distractors: a rectangle comes from taking the outline of the box the whole solid would just fit inside, ignoring the step created by the taller block. A T-shape comes from placing the taller block in the middle of the base instead of at one end, so that the base would show on both sides of it. A parallelogram comes from copying a face as it is drawn in the sketch — the top of the taller block is drawn as a sloping parallelogram because the solid is drawn at an angle — instead of drawing the true outline seen looking straight at the front.
Build your own mix at the worksheet builder.