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.
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Geometry and measures worksheet — GCSE Higher
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- 1.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 2.A straight line touches the edge of a circle at exactly one point and does not cross into the circle at all. Write down the term for this line.
- 3.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?
- 4.A surveyor marks two fixed points A and B, with position vectors OA = a and OB = b (in km) from a base station O. A relay mast P is to be placed on the line AB such that AP : PB = 3 : 2. Express the vector OP in terms of a and b.
- 5.A line passes through the points (1, 2) and (3, 6). The line is reflected in the x-axis, and the image is then translated by the vector (2, 0). Work out the gradient of the image line after both transformations.
- 6.The sum of the interior angles of a polygon is 1980°. Work out the number of sides of the polygon.
- 7.A shape is reflected in the x-axis and the image is then reflected in the y-axis. Which single transformation is equivalent to this combination for every point?
- 8.Triangle T has an area of 6 cm². It is enlarged by scale factor −2 about the origin, and the image is then rotated 90° about the origin. Work out the area of the final image, and state whether it is the same way round as T or a mirror image of it.
- 9.A cylinder has a radius of 5 cm. Its volume is 471 cm³. Using π = 3.14, work out the height of the cylinder.
- 10.A bead starts at position (2, −1) on a grid, in centimetres. It is moved by the column vector u, with top number 3 and bottom number 5, and then moved by the column vector v, with top number −7 and bottom number 2. Work out the coordinates of the bead's final position.
- 11.In triangle ABC the angle at C is 90° and the hypotenuse AB is 12 cm. H is the point on AB for which CH is perpendicular to AB, and AH = 3 cm. Work out the length of AC.
- 12.Rectangle ABCD has vertices A(−1, 2), B(4, 2), C(4, 5) and D(−1, 5). The rectangle is translated so that C moves to (0, −1). Write down the coordinates of the image of A.
- 13.A square-based pyramid is cut by a flat plane parallel to its base, partway up between the base and the apex. What shape is the cross-section?
- 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.Two similar shapes have lengths in the ratio 1 : 3. A side of the smaller shape is 4 cm long. Work out the length of the corresponding side of the larger shape.
Answer key
- (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
- (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
- (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.
- (d) (2/5)a + (3/5)b — Method: OP = OA + AP, and since AP : PB = 3 : 2 splits AB into 5 equal parts, AP is 3/5 of the whole of AB, with AB = b − a. Working: OP = a + 3/5(b − a) = a − (3/5)a + (3/5)b = (2/5)a + (3/5)b. Answer: OP = (2/5)a + (3/5)b. Using the ratio the wrong way round, as though it read AP : PB = 2 : 3, gives (3/5)a + (2/5)b; adding (3/5)b onto the whole of a without subtracting a inside the bracket first gives a + (3/5)b; and treating the ratio as 1 : 1 gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from A, matching the ORDER the ratio is stated in, and subtract before you scale.
- (b) −2 — The gradient of the original line is (6 − 2) ÷ (3 − 1) = 4 ÷ 2 = 2. Reflecting in the x-axis sends every y-coordinate to its negative, which flips the sign of the gradient: the image line has gradient −2. Translating by (2, 0) is a horizontal shift, which does not change the line's steepness or direction at all, so the gradient stays at −2. Assuming the gradient is unaffected by the reflection gives 2, the original gradient carried straight through. Thinking a reflection in the x-axis turns a gradient into its positive reciprocal gives 1/2. Combining that same wrong idea with the sign flip from the reflection gives −1/2. Only the sign flips, from the reflection, and translating never changes a gradient at all, so the answer is −2.
- (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
- (a) A rotation of 180° about the origin — Method: composing two reflections in lines that cross is always a single rotation about the point where the lines meet, through twice the angle between them. Working: the x-axis and y-axis meet at the origin at an angle of 90°, so the combined transformation is a rotation about the origin through 2 × 90 = 180 degrees. Answer: a rotation of 180° about the origin. The rotation angle is TWICE the angle between the mirror lines, not the angle itself, and the centre is always where the two lines cross, not some other point, and the result of two reflections in intersecting lines is a rotation, never another reflection.
- (d) 24 cm²; same way round as T — An enlargement scales area by the square of the scale factor, whatever its sign: area factor = (−2)² = 4, so the image's area is 6 × 4 = 24 cm². A negative scale factor is equivalent to an enlargement by the positive scale factor together with a rotation of 180°, so it does not create a mirror image — it is still a direct transformation, and combining it with a further rotation cannot create one either, since rotations never change a shape's orientation. Using the scale factor itself rather than its square gives area 6 × 2 = 12 cm², which is too small. Believing that a negative scale factor reflects the shape gives 'a mirror image of T' alongside the correct area of 24 cm², and combining both mistakes gives 12 cm² together with 'a mirror image of T'.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (d) 6 cm — Method: CH cuts the triangle into two smaller right-angled triangles, so Pythagoras' theorem can be written in each of the three right-angled triangles and the results combined. Working: HB = 12 − 3 = 9 cm. In triangle ACH, CH² = AC² − 3² = AC² − 9; in triangle CHB, CH² = CB² − 9² = CB² − 81. Setting those equal gives AC² − 9 = CB² − 81. In triangle ABC, AC² + CB² = 12² = 144, so CB² = 144 − AC². Substituting gives AC² − 9 = 144 − AC² − 81, so 2 × AC² = 72 and AC² = 36, giving AC = 6. Answer: 6 cm. The distractors: 36 cm comes from stopping at AC² = 36 and never taking the square root; 9 cm is HB, the other part of the hypotenuse, written down in place of AC; 4 cm comes from working out 12 ÷ 3, treating AH as a scale factor between the two triangles rather than as a length.
- (d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
- (c) Square — Because the cut is parallel to the square base, it produces a shape that is similar to the base — the same shape, scaled down — so it is also a square, with four equal sides, smaller than the base. A candidate who looks at the sloped triangular faces instead of at the horizontal cut answers triangle. A candidate who pictures a slanted cut, which would shorten one pair of sides, answers trapezium, but a cut parallel to the base scales all four sides by the same amount. A candidate who counts an extra side, perhaps including an edge of the base, answers pentagon. The cross-section is a square.
- (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.
- (b) 12 cm — Method: in similar shapes every length is multiplied by the same scale factor, and a ratio of 1 : 3 means that factor is 3 going from the smaller shape to the larger one. Working: 4 × 3 = 12. Answer: 12 cm. The distractors: 7 cm comes from adding 3 to the side instead of multiplying by it, which is what happens when a ratio is read as a difference; 36 cm comes from multiplying by 3² = 9, the factor that scales areas, and applying it to a length; 4 cm comes from treating the two shapes as congruent, so that corresponding sides stay equal — similar shapes have equal angles, but their sides are in proportion.
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