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 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?
- 2.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.
- 3.A triangle has vertices X, Y and Z, with a right angle at Y. Side XY is twice as long as side YZ. Which of the following correctly describes this triangle?
- 4.A trapezium has vertices (0, 2), (4, 2), (3, 5) and (1, 5). It is rotated 180° about the point (2, 2), and the image is then translated by the vector (1, −5). Work out the coordinates of the image of (0, 2).
- 5.A hexagonal prism lies on a table. A flat cut is made straight across it, at right angles to its length (so the cut is parallel to the two identical end faces). What shape is the cross-section?
- 6.Two triangles are proved congruent using the ASA condition. What can be concluded about the two remaining pairs of corresponding sides that were not part of the original ASA facts?
- 7.The perimeter of a square is 20 cm. Work out the area of the square.
- 8.A(1, 2), B(5, 2) and C(5, 6) are three of the four vertices of a square ABCD. Work out the coordinates of D.
- 9.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.
- 10.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.
- 11.A solid has 6 faces: one pentagon and five triangles that all meet at a single point above the pentagon. What is the name of this solid?
- 12.Shape S has an area of 48 cm². It is enlarged by a scale factor of 1/4 to give shape T. Work out the area of shape T.
- 13.A right-angled triangle has a hypotenuse of 14 cm. One of the other angles is 30°. Work out the exact length of the side opposite the 30° angle.
- 14.A gardener wants to plant a tree so that it is the same distance from two straight garden walls that meet at a corner, and also the same distance from two ornamental posts standing 4 m apart. Which construction locates this point?
- 15.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
Answer key
- (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) 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'.
- (b) A right-angled triangle — The right angle at Y makes this a right-angled triangle, so that description is correct. Since XY is twice YZ, those two sides cannot be equal. The third side XZ is opposite the right angle, so it is the hypotenuse and is longer than either XY or YZ, so it cannot equal either of them. No two sides are equal, which rules out 'an isosceles triangle' and 'a right-angled isosceles triangle', both of which wrongly assume two equal sides. 'An equilateral triangle' would need all three sides equal, which contradicts XY being twice YZ, so it is wrong too.
- (b) (5, −3) — Method: rotate the point about the given centre first, then translate the image, in the stated order. Working: rotating (0, 2) by 180° about (2, 2) uses the rule (x, y) → (4 − x, 4 − y), since the centre doubles in each coordinate. This gives 4 − 0 = 4 and 4 − 2 = 2, so (0, 2) maps to (4, 2). Translating (4, 2) by the vector (1, −5) gives 4 + 1 = 5 and 2 − 5 = −3, so the final image is (5, −3). Answer: (5, −3). Rotate about the centre GIVEN in the question, (2, 2), not about the origin, and translate the rotated image afterwards, in that order: rotating about the wrong centre, swapping the order, or stopping after one step all give a different point.
- (b) Hexagon — Cutting straight across a prism, at right angles to its length, always gives a cross-section that is the same shape as its end faces. The end faces of a hexagonal prism are hexagons (6-sided), so the cross-section is a hexagon. Choosing Pentagon comes from miscounting the sides of the hexagonal end as five instead of six. Choosing Rectangle comes from cutting along the LENGTH of the prism instead of across it, which gives a rectangular face, not the cross-section asked for. Choosing Triangle comes from confusing a hexagonal prism with a triangular prism.
- (c) They must also be equal — Once two triangles are proved congruent by any condition, including ASA, they are identical in every respect: every pair of corresponding sides and every pair of corresponding angles must be equal, not just the ones originally used to prove the congruence. So the two remaining pairs of corresponding sides must also be equal, making 'they must also be equal' correct. 'They might be equal or not' and 'not enough information to say' both wrongly suggest that congruence only guarantees the specific facts used to prove it, when congruence actually guarantees the triangles are identical overall. 'They must be different' is backwards: the triangles being identical is the entire point of proving congruence, not a reason for a side to differ.
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (b) (1, 6) — Method: when a square is set square-on to the grid, so that its sides run parallel to the axes, each vertex shares its x-coordinate with one neighbour and its y-coordinate with the other, and the missing vertex then borrows one coordinate from each of the two vertices it is joined to; so the first job is to check from the given points that the sides really do run parallel to the axes. Working: A(1, 2) and B(5, 2) share y = 2, so AB is a horizontal side; B(5, 2) and C(5, 6) share x = 5, so BC is a vertical side, which confirms that this square lies square-on to the axes and that the rule may be used. In square ABCD the vertex D is joined to C and to A. DC must be horizontal like AB, so D takes the y-coordinate of C, which is 6; DA must be vertical like CB, so D takes the x-coordinate of A, which is 1. D is therefore (1, 6), and checking confirms every side is 4 long. Answer: (1, 6). The distractors: (1, 5) comes from lifting the first number out of each of A and C, pairing the x-coordinate of A with the x-coordinate of C; (6, 1) comes from finding the right two numbers but writing them the wrong way round, height before sideways position; (9, 6) comes from stepping a further 4 to the right from C instead of closing the square back to the column A stands in.
- (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.
- (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.
- (c) Pentagonal pyramid — Method: a pyramid has one base and triangular faces that all meet at a single apex; the base shape gives the pyramid its name. Working: the base is a pentagon and the other five faces are triangles meeting at one point, so this is a pyramid with a pentagon base. A student who answers pentagonal prism has confused a pyramid, whose sloping faces meet at an apex, with a prism, which has two identical parallel faces. A student who answers hexagonal pyramid has miscounted the base as having 6 sides instead of 5. A student who answers triangular pyramid has misread the five triangular side faces as meaning the base itself is a triangle. Answer: pentagonal pyramid.
- (a) 3 cm² — Area scale factor = (linear scale factor)² = (1/4)² = 1/16. Area of T = 48 × 1/16 = 3 cm². (12 cm² comes from multiplying by the linear scale factor 1/4 directly, without squaring it; 24 cm² comes from taking the square root of the scale factor instead of squaring it; 768 cm² comes from squaring the reciprocal of the scale factor, 4, instead of the scale factor itself.)
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (a) where the angle bisector meets the posts' perpendicular bisector — Being equidistant from the two walls means lying on the angle bisector of the corner; being equidistant from the two posts means lying on the perpendicular bisector of the 4 m segment joining them. A single point satisfying both conditions is wherever those two loci cross. "where the angle bisector meets the line joining the posts" uses the straight line between the posts instead of its perpendicular bisector — a point on that line is not generally equidistant from both posts. "the perpendicular bisector of the posts, alone" satisfies only the posts condition, ignoring the walls entirely. "the angle bisector of the corner, alone" satisfies only the walls condition, ignoring the posts entirely.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
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