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.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?
- 2.A ferris wheel's circular frame has centre O. A support strut runs from O to a point A on the rim, and a second strut runs from O to point B on the rim, with angle AOB = 76°. A cabin is mounted at point C on the major arc AB, and D is a separate point on the minor arc AB. Work out the difference between angle ACB and angle ADB.
- 3.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
- 4.Two points on a circle divide its circumference into two arcs of different lengths. Write down the name given to the shorter of the two arcs.
- 5.A roof has two triangular sections, P and Q. Section P has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them. Section Q has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them, arranged the same way round. A roofer wants to cut identical triangular felt panels for both sections without measuring Section Q separately. Which condition confirms that the two sections are congruent?
- 6.A right-angled triangle has two sides of length 5 cm and 12 cm, and the angle between those two sides is 90°. Work out the length of the hypotenuse.
- 7.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?
- 8.A solid is built from centimetre cubes standing on a table. Seen from the front, the left-hand column is 2 cubes high, the middle column is 2 cubes high and the right-hand column is 1 cube high. Work out how many squares make up the front elevation.
- 9.A chord divides a circle into two segments of different sizes. Write down the name given to the smaller of the two segments.
- 10.A footpath is 3750 m long. Work out this distance in kilometres.
- 11.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
- 12.Shape S is enlarged by a scale factor of 3, centre the origin. A vertex of S is at (2, 1). Work out the coordinates of the image of this vertex.
- 13.A circular table mat has diameter 20 cm. Yasmin has a 70 cm length of ribbon and sews it around the edge of the mat with no overlap. Using π = 3.14, work out how much ribbon is left over after going around the mat once.
- 14.In triangle ABC, AB = 6 cm, AC = 8 cm and angle BAC = 60°. Work out the exact area of triangle ABC, giving your answer in the form .
- 15.A tile manufacturer cuts a hexagonal tile so that all six sides measure exactly 4 cm, but a check shows only two pairs of the six interior angles are equal to each other, not all six angles. The customer's order specifies regular hexagonal tiles. Based on the check, is this tile a regular hexagon?
Answer key
- (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.
- (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
- (d) Minor arc — Method: compare the lengths of the two arcs formed by the two points, and recall the term for the shorter one. Working: the two points split the circumference into two arcs; the shorter one is the minor arc and the longer one is the major arc. A student who answers major arc has picked the longer arc by mistake. A student who answers minor segment has confused the curved boundary with the enclosed two-dimensional region. A student who answers chord has named the straight line joining the two points instead of the curved arc. Answer: minor arc.
- (d) SAS — two sides, included angle — Each section has two known sides, 3.6 m and 2.4 m, with the 70° angle between them, matching in both sections; this is exactly the SAS condition, so 'SAS — two sides, included angle' is correct. 'SSS — but only two sides given' is wrong because SSS requires three pairs of equal sides, but only two sides are given for each triangle here. 'ASA — angle between two sides' is wrong because ASA requires two angles with a side between them, but only one angle, 70°, is given, not two. 'Cannot prove — only one angle' is wrong because SAS is specifically designed to prove congruence from exactly two sides and the one angle between them, so no further angle is needed.
- (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
- (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.
- (c) 5 — The front elevation shows one square for every cube visible from the front, column by column: the left-hand column is 2 cubes high, so it contributes 2 squares; the middle column is 2 cubes high, so it contributes 2 more; the right-hand column is 1 cube high, so it contributes 1. The total is 2 + 2 + 1 = 5 squares. "6" comes from drawing a full 3 by 2 rectangle, treating every column as if it reached the greatest height. "4" comes from losing a square from one of the two tall columns, counting 2 + 1 + 1. "3" comes from counting one square per column — the width of the solid — and ignoring the heights altogether.
- (b) Minor segment — Method: compare the sizes of the two regions cut off by the chord, and recall the term used for the smaller one. Working: the chord creates two segments; the smaller region is called the minor segment and the larger one the major segment. A student who answers major segment has picked the larger region by mistake instead of the smaller one. A student who answers minor arc has named the curved boundary rather than the two-dimensional region it encloses. A student who answers semicircle has wrongly assumed the chord must pass through the centre. Answer: minor segment.
- (a) 3.75 km — To convert metres to kilometres, divide by 1000: 3750 ÷ 1000 = 3.75 km. Dividing by 100 instead of 1000 gives 37.5 km. Dividing by 10 instead of 1000 gives 375 km. Dividing by 10 000 instead of 1000 gives 0.375 km.
- (a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
- (b) (6, 3) — For an enlargement centred on the origin, multiply both coordinates by the scale factor: (2 × 3, 1 × 3) = (6, 3). A pupil who adds the scale factor to each coordinate instead of multiplying gets (2 + 3, 1 + 3) = (5, 4). A pupil who multiplies only the x-coordinate gets (6, 1). A pupil who multiplies only the y-coordinate gets (2, 3). The correct image is (6, 3).
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (c) $12\sqrt{3}$ cm² — Use Area = 1/2 × AB × AC × sin(angle BAC) = 1/2 × 6 × 8 × sin 60°. Since sin 60° = √3⁄2, this is 1/2 × 6 × 8 × √3⁄2 = 24 × √3⁄2 = 12√3 cm². Using AB × AB instead of AB × AC (taking 6 × 6 rather than 6 × 8) gives 1/2 × 6 × 6 × sin 60° = 18 × √3⁄2 = 9√3 cm². Using AC × AC instead of AB × AC (taking 8 × 8 rather than 6 × 8) gives 1/2 × 8 × 8 × sin 60° = 32 × √3⁄2 = 16√3 cm². Leaving out the 1/2 from the area formula gives 6 × 8 × sin 60° = 48 × √3⁄2 = 24√3 cm².
- (d) No — angles must be equal too — A regular polygon must have both all sides equal and all angles equal. This tile has all six sides equal, but its interior angles are not all equal, so it fails the angle condition and is not regular — 'No — angles must be equal too' is correct. 'Yes — all sides are equal' is wrong because equal sides alone are not enough; a shape can have equal sides but unequal angles, as here. 'Yes — six equal sides means regular' is wrong for the same reason: equal sides do not automatically guarantee equal angles. 'No — hexagons can't be regular' is wrong because regular hexagons certainly exist (six equal sides and six equal 120° angles); it is this particular tile that fails to be regular, not hexagons in general.
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