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.An enlargement has a scale factor of −2. Priti says the image must be a mirror image of the original shape. Is Priti correct?
- 2.In quadrilateral ABCD the two sides AB and BC are each 6 cm long and meet each other at B. The two sides CD and DA are each 9 cm long and meet each other at D. No side of ABCD is parallel to any other side. Write down the mathematical name of this quadrilateral.
- 3.A roof truss is shaped like an isosceles triangle resting on a horizontal ceiling joist. The truss's sloping edges are equal in length, and the angle at its apex is 40°. The ceiling joist continues in a straight line beyond the foot of the truss. Work out the angle between the truss's sloping edge and the extended joist, on the outside of the truss.
- 4.In triangle OAB, OA = a and OB = b. P lies on AB such that AP is twice PB. Express the vector OP in terms of a and b.
- 5.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?
- 6.A treasure map has a scale of 1 cm to 4 m. A path from the start to a rock is drawn as two straight sections, measuring 3 cm and 2.5 cm. Work out the total real distance from the start to the rock, in metres.
- 7.Point Q has coordinates (3, 5). It is translated by the vector (1, −2), and the image is then enlarged by scale factor 3, centre the origin. Work out the coordinates of the final image of Q.
- 8.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.
- 9.Each of these has an exact value. Write down the one whose value is the greatest.
- 10.Point A has coordinates (2, 5). Point A is translated to point B using the column vector with top number 6 and bottom number −3. Write down the coordinates of point B.
- 11.A textbook contains the instruction 'Draw AB' and, on the next line, the statement 'AB = 5 cm'. Which statement about the notation AB is correct?
- 12.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).
- 13.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 14.In triangle ABC and triangle DEF, the angle at A and the angle at D are both 40°, and the angle at B and the angle at E are both 70°. No side lengths are given. Write down whether the two triangles must be similar, with the reason.
- 15.A raised garden bed is a prism whose cross-section is a triangle with a base of 1.2 m and a perpendicular height of 0.8 m. The bed is 3 m long. Work out the volume of soil needed to fill it, in m³.
Answer key
- (a) No — it's a 180° rotation, not a reflection. — A negative scale factor scales the shape by the size of the factor and turns it through 180° about the centre of enlargement — every point of the image lands on the opposite side of the centre from the object. A 180° turn is a rotation, and a rotation never reverses a shape's handedness (its sense), whereas a reflection always does, so the image is not a mirror image: Priti is wrong. The option claiming negative numbers always mean a reflection confuses the sign of a scale factor with the transformation of reflection — they are not the same thing. The option saying only the size changes ignores the sign altogether: the 180° turn does change the way the shape faces, even though it leaves the handedness unchanged. The option equating 'upside down' with 'mirrored' assumes the two always mean the same thing, but a 180° rotation and a reflection in a horizontal line can both look 'upside down' while being different transformations — only the reflection reverses the shape's handedness.
- (b) Kite — Method: name a quadrilateral by matching what is given — which sides are equal, whether those equal sides lie next to each other or opposite each other, and whether any sides are parallel — against the definitions of the special quadrilaterals. Working: the two 6 cm sides meet at B and the two 9 cm sides meet at D, so each pair of equal sides is a pair of neighbours rather than a pair of opposites, and the stem rules out any parallel sides. The quadrilateral with two pairs of equal adjacent sides and no parallel sides is a kite. Answer: kite. The distractors: a rhombus is chosen by candidates who see two pairs of equal sides and read that as all four sides being equal, which the two different lengths of 6 cm and 9 cm rule out; a parallelogram is chosen by candidates who remember that a parallelogram has two pairs of equal sides but not that in a parallelogram the equal sides are the opposite ones, and who pass over the statement that nothing is parallel; an isosceles trapezium is chosen by candidates who notice that the shape is symmetrical about the line BD and treat symmetry on its own as the mark of a trapezium, when a trapezium needs a pair of parallel sides.
- (a) 110° — The truss is isosceles, so its two base angles are equal; since the three angles of the triangle add up to 180° and the apex is 40°, each base angle is (180 − 40) ÷ 2 = 70°. The angle between the sloping edge and the joist, extended beyond the base of the truss, sits on a straight line with that 70° base angle, and angles on a straight line add up to 180°, so the required angle is 180 − 70 = 110°. "70°" comes from stopping after finding the base angle and giving it directly, without also using the straight-line fact to find the angle on the OUTSIDE of the truss. "140°" comes from doubling the base angle instead of using the straight-line fact. "20°" comes from taking half of the apex angle (40° ÷ 2), mistaking it for the required angle instead of properly using the triangle's angle sum.
- (c) (1/3)a + (2/3)b — Method: OP = OA + AP, and since AP is twice PB, AP is 2/3 of the whole of AB, with AB = b − a. Working: OP = a + 2/3(b − a) = a − (2/3)a + (2/3)b = (1/3)a + (2/3)b. Answer: OP = (1/3)a + (2/3)b. Measuring 2/3 of AB from B's end instead of A's swaps the fractions round, giving (2/3)a + (1/3)b; adding (2/3)b onto the whole of a without first subtracting a inside the bracket gives a + (2/3)b; and treating the ratio as though AP and PB were equal gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from the point named first in the ratio, subtract before you scale, and then add the result to OA.
- (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.
- (b) 22 — Method: add the two drawn lengths together first, then apply the scale to the total. Working: 3 cm + 2.5 cm = 5.5 cm; 5.5 cm × 4 = 22 m. A student who answers 10 has only converted one of the two sections (2.5 cm × 4) and forgotten the other. A student who answers 5.5 has added the two drawn lengths but forgotten to apply the scale at all. A student who answers 44 has doubled the correct answer, effectively applying the scale twice. Answer: 22 m.
- (b) (12, 9) — Apply the transformations in the order given: first translate, then enlarge. Translating (3, 5) by the vector (1, −2) gives (3 + 1, 5 − 2) = (4, 3). Enlarging this by scale factor 3 about the origin multiplies both coordinates by 3: (4 × 3, 3 × 3) = (12, 9). Enlarging first and translating afterwards reverses the order and gives (3 × 3 + 1, 5 × 3 − 2) = (10, 13), a different point because the two transformations do not commute. Enlarging the original point by scale factor 3 while forgetting to translate it at all gives (3 × 3, 5 × 3) = (9, 15). Reversing the signs of the translation vector before applying it gives (3 − 1, 5 + 2) = (2, 7), which then enlarges to (2 × 3, 7 × 3) = (6, 21).
- (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.
- (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
- (a) (8, 2) — Method: translating a point by a column vector means adding the vector's top number to the x-coordinate and its bottom number to the y-coordinate. Working: x-coordinate 2 + 6 = 8; y-coordinate 5 + (−3) = 2. Answer: B = (8, 2). A candidate who ignores the negative sign and adds 3 instead of −3 gets (8, 8). A candidate who translates in the reverse direction, subtracting the vector from A instead of adding it, gets (−4, 8). A candidate who swaps the vector's top and bottom numbers before adding gets (−1, 11).
- (c) AB means the segment or its length, shown by context — Method: recall the standard convention for a two-letter label such as AB. Working: AB names the segment AND its length; the sentence around it shows which is meant — 'draw AB' means the segment, 'AB = 5 cm' means the length. Options: 'a different symbol is needed' and 'must be written as |AB|' both describe stricter rules than the convention GCSE actually uses; 'AB can only mean the segment' ignores that the same label is also used for the length. Answer: AB means the segment or its length, shown by context.
- (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.
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (d) Yes, by the AA condition — Method: similarity is decided by the angles, and because the three angles of a triangle add up to 180°, two matching pairs force the third pair to match as well. Working: the angles at A and D are both 40° and the angles at B and E are both 70°, so the third angles are both 180° − 40° − 70° = 70° and all three pairs are equal. Two pairs were enough, and that is the AA condition. Answer: yes, by the AA condition. The distractors: 'Yes, by the SSS condition' names a condition about three pairs of sides in proportion, and the question gives no side lengths at all; 'No, because no side lengths are given' treats side information as necessary, which it is for congruence but not for similarity; 'No, because the triangles may be different sizes' turns the definition of similarity into an objection, since similar figures are allowed to differ in size and only their shape must match.
- (c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
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