Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Foundation
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- 1.A tent is shaped like a triangular prism: two triangular end faces and three rectangular faces. One of the rectangular faces lies flat on the ground as the groundsheet. How many of the tent's faces are above the ground, not touching the ground?
- 2.The diagram shows a solid made from two cuboids joined together. Which shape is the front elevation of this solid?
- 3.A solid cube has a volume of 8 cm³. Work out the length of one edge of the cube.
- 4.A regular polygon has an exterior angle of 45°. Work out the number of sides of the polygon.
- 5.A sector of a circle has radius 8 cm and takes up three-quarters of the full circle. Work out the perimeter of the sector, in terms of π.
- 6.A roof truss has two horizontal parallel rafters, one above the other. A straight strut crosses both rafters. Where the strut crosses the lower rafter, the angle above the rafter and to the left of the strut is 65°. Work out the size of the angle above the upper rafter and to the right of the strut, where the strut crosses it.
- 7.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 8.A survey team measures the angle between two fences with a protractor and gets a reading of 254°. What type of angle is this?
- 9.A solid has 6 faces and 8 vertices. Using the formula F + V − E = 2, work out how many edges it has.
- 10.A carpenter cuts two triangular metal brackets for a shelf. Bracket P has sides of 12 cm, 16 cm and 20 cm. Bracket Q has sides of 16 cm, 20 cm and 12 cm, listed in a different order. Before fitting them as a matching pair, the carpenter wants to check they are exactly the same shape and size. Which of the following correctly justifies this?
- 11.A rectangular notice board measures 150 cm by 80 cm. Work out its area in m².
- 12.A shopkeeper builds a display from two cuboid boxes, shown in the diagram. She wants to cover the front of the display with coloured paper. Work out the total area of the front elevation, in square centimetres.
- 13.A ramp rises at 30° to the horizontal. Its sloping surface is 4.8 m long. Safety rules say the vertical rise of a ramp must be no more than 2.5 m. Work out how far below that limit the rise of this ramp is.
- 14.u is the column vector with top number 5 and bottom number 1. v is the column vector with top number 2 and bottom number 4. Work out u − v, giving your answer as a column vector in the form (top, bottom).
- 15.A carpenter measures two triangular braces. Brace 1 has a 55° angle, a 65° angle, and, opposite the 55° angle, a side of 8 cm. Brace 2 has matching 55° and 65° angles and, opposite its 55° angle, a side of 8 cm. The carpenter says that because the equal 8 cm side is not between the two equal angles in either brace, he cannot be sure the two braces are the same size. Is the carpenter correct?
Answer key
- (b) 4 — A triangular prism has 5 faces in total: 2 triangular ends and 3 rectangular faces. One rectangular face is the groundsheet, lying on the ground, so the other 5 − 1 = 4 faces are above the ground. A candidate who forgets one of the triangular ends when counting the remaining faces answers 3. A candidate who forgets to subtract the groundsheet at all answers 5. A candidate who only counts the two sloped rectangular faces, forgetting the two triangular ends, answers 2. The number of faces above the ground is 4.
- (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.
- (b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
- (c) 8 — Method: the exterior angles of a polygon add up to 360°, so divide 360° by the size of one exterior angle. Working: 360 ÷ 45 = 8. Answer: 8 sides. A candidate who divides into a half turn instead of a full turn, working out 180 ÷ 45, gets 4. A candidate who reads off the given exterior angle as if it were the number of sides gets 45. A candidate who subtracts instead of dividing, working out 360 − 45, gets 315.
- (a) 12π + 16 cm — A three-quarter sector's perimeter is the curved arc plus the two straight radii that close the shape. The full circumference is 2 × π × 8 = 16π cm, and three-quarters of that is 12π cm. Adding the two straight radii, 8 cm each, gives 12π + 16 cm. Leaving out the straight edges gives just 12π cm. Using one-quarter of the circumference, the piece left over rather than the piece asked for, gives 4π + 16 cm. Adding only one radius instead of two gives 12π + 8 cm.
- (c) 115 — Method: use corresponding angles to carry the 65° angle from the lower rafter up to the upper rafter, then use angles on a straight line to move to the other side of the strut. Working: the angle above the upper rafter and to the left of the strut corresponds to the given angle, so it is 65°; the angle above the upper rafter and to the right of the strut lies on a straight line with it, so it is 180 − 65 = 115. Answer: 115°. A candidate who assumes the angle stays 65° without allowing for the move from the left of the strut to the right of it gives 65. A candidate who uses 90° instead of 180°, working out 90 − 65, gets 25. A candidate who adds instead of subtracting, working out 180 + 65, gets 245.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (c) Reflex — Method: compare the measured angle to the key boundaries of 90 degrees, 180 degrees and 360 degrees. Working: 254 degrees is greater than 180 degrees but less than 360 degrees, so it lies in the reflex range. A student who answers obtuse knows the angle is bigger than 90 degrees but has not realised it is also bigger than 180 degrees; they may have worked with the smaller angle at the same point, 360 - 254 = 106 degrees, instead of the reading given. A student who answers acute has misread the scale and taken the reading as 54 degrees rather than 254 degrees. A student who answers right has guessed a commonly recognised angle type without comparing the size properly. Answer: reflex.
- (d) 12 — Method: rearrange F + V − E = 2 so that E is on its own: E = F + V − 2. Working: E = 6 + 8 − 2 = 12. A student who answers 14 has added F and V but forgotten to subtract 2 at all. A student who answers 16 has added 2 instead of subtracting it. A student who answers 10 has subtracted 2 twice by mistake. Answer: 12 edges.
- (a) Yes - SSS, the three side lengths all match — Bracket P's sides (12 cm, 16 cm, 20 cm) can each be matched to one of Bracket Q's sides (16 cm, 20 cm, 12 cm) — the same three lengths, just listed differently — so the brackets are congruent by SSS. SAS is wrong here because no angle is stated for either bracket, only three sides. The order the sides are listed in does not matter for SSS — only whether the SET of three lengths matches, and it does, so 'listed in a different order' is not a reason to say no. Nothing extra is needed: SSS proves congruence from side lengths alone, without any angles, so it can be determined.
- (d) 1.2 m² — Method: an area in square metres needs lengths in metres, so convert first and then multiply. Working: 100 cm = 1 m, so 150 cm = 1.5 m and 80 cm = 0.8 m, and the area = 1.5 × 0.8 = 1.2 m². Answer: 1.2 m². The same result comes from working in centimetres: 150 × 80 = 12 000 cm², and a square metre is a square of side 100 cm, so 100 × 100 = 10 000 cm² make one square metre and 12 000 ÷ 10 000 = 1.2. Dividing the 12 000 cm² by 100 instead, as though a square metre held only 100 square centimetres, gives 120 m²; dividing by 1000 gives 12 m². Working out the perimeter rather than the area gives 1.5 + 0.8 + 1.5 + 0.8 = 4.6, which is a length and not an area.
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (d) No — third angle is also fixed — Since both braces have angles of 55° and 65°, their third angles must both be 60°, because angles in a triangle sum to 180°. All three angles now match, so the braces have the same shape. Both 8 cm sides lie in the same position relative to those angles — opposite the 55° angle in each brace — so one matching pair of corresponding sides fixes the size as well, exactly as ASA or AAS would. The braces are therefore guaranteed to be congruent and the carpenter is incorrect: 'No — third angle is also fixed' is correct. 'Yes — side must be included' is wrong because the side does not have to lie physically between the two named angles; once the third angle is fixed, a corresponding equal side anywhere is enough. 'No — any two angles enough alone' is wrong because two equal angles with no side length at all would only show the triangles are similar, not congruent. 'Yes — third angle may differ' is wrong because the third angle is fixed at 60° by the angle sum and cannot vary.
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