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.
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Geometry and measures worksheet — GCSE Foundation
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- 1.A builder props a straight plank against a vertical wall to reach a window ledge. The foot of the plank is 2.1 m from the base of the wall, and the plank is 3.5 m long. Work out how high up the wall the plank reaches.
- 2.State whether the line segment joining A(−4, 3) and B(2, 6) is horizontal, vertical or neither, and give a reason for your answer.
- 3.A solid has a plan view that is a circle, and a front elevation that is a triangle. Which solid is this?
- 4.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.
- 5.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 6.Point B is at (1, 6). It is reflected in the x-axis. Work out the coordinates of the image of point B.
- 7.m is the column vector with top number 4 and bottom number 6. n is the column vector with top number −6 and bottom number −9. Given that n = k × m for some number k, work out the value of k.
- 8.In triangle ABC the angle at A is 90° and BC = 10 cm. For the angle at B, sin B = 3/5. Work out the length of AC.
- 9.A shape is translated by the vector , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector . Work out the coordinates of the image of that same vertex under this second translation.
- 10.Two similar triangles have lengths in the ratio 3 : 4. The perimeter of the smaller triangle is 24 cm. Work out the perimeter of the larger triangle.
- 11.A furniture catalogue drawing uses a scale of 1 cm to 1.5 m. A sofa in the drawing is measured as 7 cm long. What is the real length of the sofa, in metres?
- 12.The column vector r has top number 2 and bottom number 6. Which of these column vectors is parallel to r?
- 13.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.
- 14.A postcard measuring 10 cm by 15 cm is enlarged by a scale factor of 2 to make a poster. Work out the area of the poster, in cm².
- 15.Which pair of points lies on a horizontal line when the points are plotted on a coordinate grid?
Answer key
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (a) a cone — A cone, viewed from directly above, shows its circular base as a circle; viewed from the front, its curved surface narrows to a point at the apex, giving a triangle outline — so a circle plan with a triangle elevation identifies a cone. "a cylinder" has a circular plan too, but its front elevation is a rectangle, not a triangle, since its sides run straight up rather than narrowing to a point. "a sphere" gives a circle from every direction, plan and every elevation alike, not a triangle from the front. "a square-based pyramid" gives a triangle from the front, but its plan view is a square, not a circle.
- (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.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (d) (1, −6) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: (1, 6) → (1, −6). A pupil who reflects in the y-axis instead gets (−1, 6). A pupil who changes the sign of both coordinates gets (−1, −6). A pupil who forgets to change any sign leaves the point unmoved at (1, 6). The correct image is (1, −6).
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (b) 6 cm — Method: the right angle is at A, so BC is the hypotenuse and AC is the side opposite the angle at B; sin B = opposite ÷ hypotenuse therefore gives AC ÷ BC = 3/5. Working: AC ÷ 10 = 3/5, so AC = 10 × 3 ÷ 5 = 6. Answer: 6 cm. The distractors: 8 cm is AB, the side next to the angle at B, which is what cos B = 4/5 produces — the right method used on the wrong side; 3 cm comes from reading the 3 in the ratio as a length and never scaling it up to the 10 cm hypotenuse; 30 cm comes from multiplying by 3 and forgetting to divide by 5.
- (d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
- (c) 32 cm — Method: a perimeter is lengths added together, so it scales by the length scale factor itself, which is 4 ÷ 3 going from the smaller triangle to the larger one — not by its square. Working: 24 ÷ 3 = 8, and 8 × 4 = 32. Answer: 32 cm. The distractors: 18 cm comes from multiplying by 3 ÷ 4, scaling from the larger triangle down to the smaller one; 25 cm comes from adding the difference between the parts of the ratio, 4 − 3 = 1, to the perimeter; 8 cm comes from dividing by 3 and stopping there, before multiplying by 4.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
- (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.
- (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.)
- (d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
- (c) (2, 4) and (6, 4) — Method: two points lie on the same horizontal line when they stand at the same height above the x-axis, and a point's height is given by its y-coordinate, the second number in the bracket. Working: compare the second numbers in each pair. For (3, 5) and (3, 8) they are 5 and 8, which differ. For (1, 2) and (2, 1) they are 2 and 1, which differ. For (0, 0) and (1, 1) they are 0 and 1, which differ. For (2, 4) and (6, 4) they are 4 and 4, which agree, so only that pair lies on a horizontal line. Answer: (2, 4) and (6, 4). The distractors: (3, 5) and (3, 8) comes from swapping horizontal for vertical, since equal x-coordinates place two points one above the other; (1, 2) and (2, 1) comes from looking for the same pair of numbers rather than for the same y-coordinate; (0, 0) and (1, 1) comes from reading 'horizontal' as simply 'in a straight line', which any two points are.
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