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.Write down the name given to a quadrilateral with exactly one pair of parallel sides.
- 2.A picture frame is a rhombus with a diagonal of 16 cm and a diagonal of 12 cm. The two diagonals meet at right angles at their midpoints. Work out the length of one side of the rhombus.
- 3.How many vertices does a cuboid have?
- 4.A sector of a circle has angle 120° and an arc length of 31.4 cm. Using π = 3.14, work out the radius of the circle.
- 5.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 6.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
- 7.A straight line is drawn from the centre of a circle to a point on the circle. Write down the name of this line.
- 8.Here are four trigonometric ratios of special angles. Write down the one that does not have a value.
- 9.A zip-wire is fixed at an angle of 60° to the horizontal ground. The zip-wire is 12 m long. Using the exact value of cos 60°, work out the exact horizontal distance it covers.
- 10.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 11.Two straight lines AB and CD cross at point O. Angle AOC = 74°. Work out angle AOD.
- 12.a is the column vector with top number 3 and bottom number −2. b is the column vector with top number −1 and bottom number 5. Work out 2a + b, giving your answer as a column vector in the form (top, bottom).
- 13.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 14.In triangle ABC and triangle DEF, the angle at A equals the angle at D, and AB ÷ DE = AC ÷ DF. Write down what this proves about the two triangles.
- 15.A ship's radar can detect other ships up to 20 km away, over a sector-shaped area with angle 90°. A monitoring company charges £0.50 per square kilometre covered by one radar sweep. Using π = 3.14, work out the total charge for one sweep.
Answer key
- (a) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs, and a kite and rhombus are defined by side lengths, not by having only one pair of parallel sides.
- (c) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (d) 8 — A cuboid is a 3D shape with 6 rectangular faces, 12 edges and 8 vertices (corners). Counting the corners of a cuboid gives 8. A candidate who mixes up vertices with faces answers 6. A candidate who mixes up vertices with edges answers 12. A candidate who counts only the 4 corners of the top face, forgetting the 4 corners of the bottom face, answers 4. The correct number of vertices is 8.
- (d) 15 cm — Arc length = (angle ÷ 360) × 2 × π × r. Here 120 ÷ 360 = 1/3, and 2 × 3.14 = 6.28, so 31.4 = (1/3) × 6.28 × r. Multiplying both sides by 3 gives 6.28 × r = 94.2, so r = 94.2 ÷ 6.28 = 15 cm. (5 cm comes from forgetting the angle fraction altogether and dividing the arc length by 2 × π alone: 31.4 ÷ 6.28 = 5; 30 cm comes from leaving out the factor of 2, dividing by (1/3) × 3.14 = 1.0467 instead of (1/3) × 6.28: 31.4 ÷ 1.0467 = 30; 7.5 cm comes from correctly finding a radius of 15 cm but then treating that 15 cm as a diameter and halving it.)
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (b) The radius — Method: each of the four names describes a line in a fixed position relative to the circle, so match the description in the question against those positions. Working: the line described has one end at the centre and its other end on the circle. A chord joins two points that both lie on the circle, so it is ruled out. A diameter joins two points on the circle by passing through the centre, so it is twice as long as the line described. A tangent touches the circle at exactly one point and never reaches the centre. The line with one end at the centre and one end on the circle is a radius. Answer: The radius. The distractors: The diameter comes from remembering only that the line involves the centre, and not noticing that it stops there instead of carrying on to the far side; The chord comes from seeing one end on the circle and assuming both ends are; The tangent comes from matching the phrase 'a point on the circle' to the line that touches at exactly one point, ignoring that a tangent never reaches the centre.
- (b) tan 90° — Method: write each ratio as one side of a right-angled triangle divided by another, and look for the one whose bottom line can be zero. Working: sine and cosine are both a side divided by the hypotenuse, and a hypotenuse is never zero, so sin 90° = 1 and cos 90° = 0 are both perfectly good values — a value of zero is still a value. Tangent is the opposite side divided by the adjacent side. As the angle opens out towards 90° the adjacent side shrinks to nothing, and a division by zero has no result, while at the other end of the row the adjacent side is the long one and tan 0° = 0. Answer: tan 90°.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
- (a) $\binom{4}{4}$ — A movement to the right is a positive top number and a movement up is a positive bottom number, so 4 right and 4 up gives $\binom{4}{4}$. $\binom{4}{-4}$ wrongly gives the vertical movement a negative sign, as if it were downward. $\binom{-4}{4}$ wrongly gives the horizontal movement a negative sign, as if it were leftward. $\binom{-4}{-4}$ gets both signs wrong, as if the translation were left and down.
- (a) 106° — Angle AOC and angle AOD share the arm OA and together they make up the straight line CD, so they add up to 180°: angle AOD = 180° − 74° = 106°. 74° comes from treating AOC and AOD as vertically opposite angles, which are equal — but they are adjacent angles on a straight line, not vertically opposite. 16° comes from subtracting 74° from 90°, treating the angles as complementary. 148° comes from doubling 74° instead of subtracting it from 180°.
- (b) (5, 1) — First scale a by 2: 2a = (2×3, 2×(−2)) = (6, −4). Then add b component by component: (6+(−1), −4+5) = (5, 1). (2, 3) is a + b without doubling a first. (4, 6) doubles both a and b instead of only a. (7, −9) subtracts b from 2a instead of adding it.
- (c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (d) Similar, by SAS — Method: read off whether the given sides are equal or only in the same ratio, and where the given angle sits. Sides in proportion point to similarity; sides equal in length would be needed for congruence. Working: AB ÷ DE = AC ÷ DF gives two pairs of corresponding sides in the same ratio, and the equal angle at A and D lies between AB and AC in the first triangle and between DE and DF in the second, so it is the included angle. Two pairs of sides in proportion with the included angle equal is the SAS condition for similarity. Answer: similar, by SAS. The distractors: 'Similar, by AA' quotes a condition that needs two pairs of equal angles, while only one pair is given here; 'Congruent, by SAS' reads AB ÷ DE = AC ÷ DF as AB = DE and AC = DF, turning a statement about proportion into one about equal lengths; 'Congruent, by SSS' makes that same misreading and adds an assumption that the third pair of sides is equal too, which nothing in the question says.
- (a) £157.00 — Area covered = (angle ÷ 360) × π × r² = (90 ÷ 360) × 3.14 × 400 = 0.25 × 1256 = 314 km². Charge = 314 × £0.50 = £157.00. (£7.85 comes from forgetting to square the radius, using 0.25 × 3.14 × 20 = 15.7 km² and then charging that; £628.00 comes from finding the area of a full circle, 3.14 × 400 = 1256 km², and forgetting the angle fraction before charging; £15.70 comes from using the arc length formula, 0.25 × 2 × 3.14 × 20 = 31.4, in place of the area, and charging that.)
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