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.
Geometry and measures worksheet — GCSE Foundation
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- 1.A straight line crosses a pair of parallel lines. At one of the crossing points, one of the four angles measures 55°. Work out the size of the angle next to it on a straight line at that crossing point.
- 2.How many edges does a cube have?
- 3.A translation is described by the column vector . Write down the column vector of the translation that undoes it.
- 4.Put sin 30°, tan 30° and cos 30° in order of size, starting with the smallest.
- 5.Loose sweets are sold at £1.20 per 100 g. Work out the cost of 350 g of sweets.
- 6.A line segment joins two points on the circumference of a circle, without necessarily passing through the centre. Write down the general term for this line segment.
- 7.A stained-glass window panel is designed as a kite. Two adjacent sides are each 25 cm, and the other two adjacent sides are each 40 cm. A frame is fitted around the whole panel. The frame costs £2.50 per metre, sold only in whole metres. Work out the total cost of the frame.
- 8.A drone starts at the point (3.5, −2) on a grid measured in metres. It flies by the vector to check a first sensor. The drone then needs to fly in a straight line to reach the point (−1.7, 9) to check a second sensor. Work out the column vector of this second flight.
- 9.A quadrilateral has angles of 100°, 85° and 95° at three of its vertices. Work out the fourth angle.
- 10.In pentagon PQRST, which of the following correctly names the interior angle at vertex R, using standard three-letter angle notation?
- 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 circle has a diameter of 24 cm. A sector of this circle has an angle of 90°. Using π = 3.14, work out the area of the sector.
- 13.A trapezium has parallel sides of 5.6 cm and 8.4 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 14.A student draws a net using 5 identical squares arranged in a row of four with one extra square attached to the side of one of them. Can this net be folded to make a closed cube?
- 15.In a right-angled triangle, the two shorter sides are a and b, and the hypotenuse is c. Write down the correct statement of Pythagoras' theorem.
Answer key
- (d) 125° — Method: two angles that sit next to each other at a crossing point lie on a straight line, so they add up to 180°; the angle that faces the given one across the point is the equal one, and that is not the angle asked for here. Working: 180° − 55° = 125°. Answer: 125°. The distractors: 55° is the angle vertically opposite the given one, taken by a candidate who reads “next to” as the facing angle and applies the equal-angles rule to the wrong pair; 35° comes from subtracting from 90°, treating the pair as complementary instead of as angles on a straight line; 90° comes from assuming that a line crossing a pair of parallel lines must meet them at right angles, which the question never says.
- (a) 12 — A cube has 6 square faces, 12 edges and 8 vertices. The edges are the straight lines where two faces meet: 4 edges around the top face, 4 edges around the bottom face, and 4 vertical edges joining them, giving 4 + 4 + 4 = 12. A candidate who counts the vertices instead of the edges answers 8. A candidate who counts the faces instead answers 6. A candidate who counts only the edges around the top face, forgetting the bottom face and the vertical edges, answers 4. The correct number of edges is 12.
- (c) $\binom{-6}{5}$ — Method: to undo a translation, travel the same journey backwards. A move of 6 to the right is undone by a move of 6 to the left, and a move of 5 down is undone by a move of 5 up, so both numbers change sign. Working: the top number 6 becomes −6 and the bottom number −5 becomes 5. Check by combining the two: 6 − 6 = 0 across and −5 + 5 = 0 up, so the shape finishes where it started. Answer: $\binom{-6}{5}$. Reversing only the horizontal movement gives $\binom{-6}{-5}$ and reversing only the vertical movement gives $\binom{6}{5}$, and each of those leaves the shape displaced. Swapping the two entries instead of changing their signs gives $\binom{-5}{6}$.
- (d) sin 30°, tan 30°, cos 30° — sin 30° = 1/2 = 0.5, tan 30° = √3/3 ≈ 0.577 and cos 30° = √3/2 ≈ 0.866, so the correct order from smallest to largest is sin 30°, tan 30°, cos 30°. 'sin 30°, cos 30°, tan 30°' swaps the last two, wrongly putting cos 30° before tan 30°. 'cos 30°, tan 30°, sin 30°' is the correct list written backwards, from largest to smallest. 'tan 30°, sin 30°, cos 30°' wrongly swaps sin 30° and tan 30° at the start.
- (a) £4.20 — 350 g is 3.5 lots of 100 g, since 350 ÷ 100 = 3.5, so the cost is 3.5 × £1.20 = £4.20. Multiplying the mass in grams directly by the price, without dividing by 100 first, gives £420.00. Working out 100 ÷ 350 instead of 350 ÷ 100 inverts the ratio and gives about £0.34. Rounding 350 g down to 300 g gives 3 × £1.20 = £3.60.
- (a) Chord — Method: identify what the two endpoints of the line segment are, and whether it must pass through the centre. Working: a line joining any two points on the circumference, whether or not it passes through the centre, is a chord, and the diameter is just a special case of it. A student who answers diameter has wrongly assumed the line must pass through the centre. A student who answers radius has confused joining two circumference points with joining the centre to the circumference. A student who answers tangent has confused a line crossing through the circle with one that only touches its outside. Answer: chord.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
- (d) 80° — The four angles of any quadrilateral add up to 360°. Three of the angles add up to 100 + 85 + 95 = 280°, so the fourth angle is 360 − 280 = 80°. "70°" comes from using 350° as the total instead of 360°, an easy slip on the quadrilateral angle sum. "280°" comes from stopping at the sum of the three given angles and writing that total down as the answer, instead of subtracting it from 360°. "260°" comes from subtracting only the one angle 100° from 360°, instead of subtracting all three given angles.
- (a) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (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.)
- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (b) 28 cm² — Area of a trapezium = (sum of parallel sides) ÷ 2 × height. Sum of parallel sides = 5.6 + 8.4 = 14 cm. Half of that is 14 ÷ 2 = 7 cm. Area = 7 × 4 = 28 cm². A pupil who forgets to halve gets 14 × 4 = 56 cm². A pupil who uses only the longer parallel side, as if this were a rectangle, gets 8.4 × 4 = 33.6 cm². A pupil who subtracts the parallel sides instead of adding them gets (8.4 − 5.6) ÷ 2 × 4 = 1.4 × 4 = 5.6 cm². The correct area is 28 cm².
- (b) No, it needs one more square — A closed cube has exactly 6 faces, so its net must be made of exactly 6 identical squares, arranged so each one unfolds to a separate face with none overlapping. This net has only 5 squares, so it is one square short and cannot be folded into a closed cube. Choosing 'Yes, it folds into a cube' ignores that a cube needs 6 faces, not 5. Choosing 'No, it has one square too many' miscounts in the wrong direction — 5 is one too FEW, not one too many. Choosing 'Yes, but only if two squares overlap' is not a valid net: a net's faces must not overlap when folded.
- (d) a² + b² = c² — Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides, so a² + b² = c². "a + b = c" adds the sides directly without squaring them at all. "a² − b² = c²" subtracts the squares instead of adding them. "a² + b² = c" adds the squares correctly but forgets to square the hypotenuse on the other side of the equation.
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