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.
Non-calculator
Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.A scale drawing of a rectangular lawn uses a scale of 1 : 125. On the drawing, the lawn measures 4 cm by 2.4 cm. Work out the perimeter of the real lawn, in metres.
- 2.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 3.A straight transversal crosses a pair of parallel lines. At one crossing point, the angle between the transversal and a parallel line is 118°. Work out the size of the co-interior (allied) angle at the other crossing point.
- 4.p is the column vector with top number 2 and bottom number 3. q is the column vector with top number −1 and bottom number 4. Work out 2p + q, giving your answer as a column vector in the form (top, bottom).
- 5.The column vector r has top number 2 and bottom number 6. Which of these column vectors is parallel to r?
- 6.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 7.Point C is at (5, 2). It is reflected in the x-axis. Work out the coordinates of the image of point C.
- 8.Work out the distance between the points (−3, 4) and (5, −2).
- 9.Write down the exact value of sin 0°.
- 10.An isosceles trapezium has exactly one pair of parallel sides, and its two non-parallel sides are equal in length. How many lines of symmetry does it have?
- 11.A solid is built from centimetre cubes: a base layer of 2 rows of 3 cubes each (a 3 by 2 rectangle of cubes), with one extra cube placed on top of one corner cube of that base. Looking down from directly above (the plan view), how many squares are visible?
- 12.An isosceles triangle ABC has AB equal to AC. AM is a line from vertex A, perpendicular to BC, meeting BC at point M. Which condition proves that triangle ABM is congruent to triangle ACM?
- 13.ABCD is an isosceles trapezium. The sides AB and DC are parallel, and the two sloping sides AD and BC are equal in length. Angle D is 112°. Work out the size of angle B.
- 14.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 15.Two angles are supplementary. One of them is twice the size of the other. Work out the size of the smaller angle.
Answer key
- (d) 16 — Method: use the scale to find the real length and width separately, then use the perimeter formula. Working: real length = 4 cm × 125 = 500 cm = 5 m; real width = 2.4 cm × 125 = 300 cm = 3 m; perimeter = 2 × (5 + 3) = 16 m. A student who answers 8 has added the real length and width but forgotten to double the total for the perimeter. A student who answers 1600 has correctly worked out the perimeter in centimetres but forgotten to convert it to metres. A student who answers 500 has only converted the length to real centimetres and stopped there, ignoring the width and the perimeter step. Answer: 16 m.
- (c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
- (c) 62° — Co-interior (allied) angles on parallel lines sum to 180°, so the co-interior angle is 180° − 118° = 62°. 118° comes from treating the angles as corresponding angles, which are equal, instead of co-interior angles, which sum to 180°. 59° comes from halving the given angle. 236° comes from doubling the given angle.
- (b) (3, 10) — Method: multiply every part of p by 2, then add the matching parts of q. Working: 2p = (4, 6); adding q gives top 4 + (−1) = 3 and bottom 6 + 4 = 10. Answer: 2p + q = (3, 10). A candidate who forgets to double p first, working out p + q instead, gets (1, 7). A candidate who doubles q instead of p, working out p + 2q, gets (0, 11). A candidate who subtracts q instead of adding it, working out 2p − q, gets (5, 2).
- (a) (3, 9) — Two column vectors are parallel only when one is the other multiplied by a single scalar — both numbers scaled by the same factor with the same sign. r has top:bottom ratio 2:6, which simplifies to 1:3. (3, 9) has ratio 3:9, also 1:3, so (3, 9) = 1.5 × r: it is parallel to r. (6, 2) swaps the top and bottom numbers of r, giving ratio 6:2 (3:1), which is not the same ratio. (2, 3) halves only the bottom number of r, giving ratio 2:3, not 1:3. (−2, 6) changes the sign of only the top number of r, giving ratio −2:6, which no longer matches 2:6 throughout.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (b) (5, −2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, 2) → (5, −2). ((−5, 2) comes from reflecting in the y-axis instead, changing the sign of the x-coordinate; (−5, −2) comes from changing the sign of both coordinates, as for a reflection in the origin; (2, 5) comes from swapping the coordinates, which is what happens for a reflection in the line y = x.)
- (d) 10 — Method: the distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical gaps, so work out both gaps first, handling the negative coordinates carefully, and then apply Pythagoras' theorem. Working: the horizontal gap is 5 − (−3) = 5 + 3 = 8 and the vertical gap is 4 − (−2) = 4 + 2 = 6. Then d² = 8² + 6² = 64 + 36 = 100, so d = √100 = 10. Answer: 10. The distractors: 14 comes from adding the two gaps, 8 + 6, instead of adding their squares and taking the root; 100 comes from stopping at the sum of the squares and never taking the square root; 50 comes from reaching 100 correctly and then halving it instead of taking its square root, a candidate who has read the last step as “halve” rather than “root”.
- (b) 0 — At 0°, the opposite side of the right-angled triangle has shrunk to zero length, so sin 0° = 0. 1 is the value of sin 90° (mixing up the two angles). 1/2 is the value of sin 30°. 'Undefined' is what happens for tan 90°, not sin 0° — sin 0° has a perfectly good exact value.
- (d) 1 — An isosceles trapezium has one line of symmetry, running through the midpoints of the two parallel sides. 0 would be true for a scalene trapezium, whose non-parallel sides are unequal, but this trapezium's non-parallel sides are equal, so it is symmetrical. 2 is the number of lines of symmetry of a rectangle, not a trapezium. 4 comes from wrongly applying the 'number of sides equals number of lines of symmetry' rule, which only holds for REGULAR polygons — a trapezium is not regular.
- (d) 6 — The plan view shows every square of the base footprint, whether or not there is a taller stack above it — the base layer alone already covers a 3 by 2 rectangle of cubes, which is 6 squares. The extra cube on top of a corner cube sits directly above a square that is already counted, so it adds no NEW square to the plan — height does not show up in a plan view, only footprint does. "7" comes from wrongly counting the extra cube as an additional square. "5" comes from missing one square of the base rectangle, perhaps forgetting a corner. "3" comes from counting only one row of the base rectangle and forgetting that the base is two rows deep.
- (d) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (d) 60 — Method: write the two supplementary angles as x and 2x, since one is twice the other, and solve x + 2x = 180. Working: 3x = 180, so x = 60. Answer: the smaller angle is 60°. A candidate who gives the larger angle, 2x, instead of the smaller angle gets 120. A candidate who uses a complementary sum of 90° instead of a supplementary sum of 180° gets 30. A candidate who divides 180 by 2 instead of by 3 gets 90.
Build your own mix at the worksheet builder.