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 designer enlarges a drawing of a model car for a poster. She first enlarges the drawing by a scale factor of 1.5, and then enlarges that result by a further scale factor of 2. On the original drawing, the position of a wheel relative to the front bumper is given by the column vector with top number 4 and bottom number −3, in centimetres. What is the corresponding column vector on the poster, in centimetres?
- 2.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 3.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 4.A garden water trough is a prism whose cross-section is a right-angled triangle with base 40 cm and height 30 cm. The trough is 120 cm long. Work out the volume of water needed to fill the trough completely.
- 5.A sector of a circle is bounded by two straight edges and one curved edge. Write down the mathematical term for each of the two straight edges.
- 6.a is the column vector with top number 3 and bottom number 2. b is the column vector with top number −1 and bottom number 4. Work out a + b, giving your answer as a column vector in the form (top, bottom).
- 7.Compasses are opened to exactly half the length of AB, and arcs are drawn centred at A and at B, to construct the perpendicular bisector of AB. Why does this fail?
- 8.Write down the exact value of sin 60°.
- 9.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.
- 10.A sign-maker is told a metal bracket outline must have two pairs of adjacent sides equal in length (not opposite pairs), exactly one pair of opposite angles equal, and exactly one line of symmetry. Which quadrilateral matches this description?
- 11.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 12.Quadrilateral ABCD has vertices A(1, 1), B(5, 1), C(5, 4) and D(1, 4). The quadrilateral is translated by the vector . Which of these points is NOT a vertex of the image?
- 13.p is the column vector with top number 5 and bottom number 2. q is the column vector with top number 1 and bottom number 6. Work out p + q, giving your answer as a column vector in the form (top, bottom).
- 14.A map has a scale of 1 cm : 5 km. A road is drawn 3 cm long on the map. What is the real length of the road?
- 15.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
Answer key
- (b) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
- (c) Radius — Method: recall that a sector is formed using two straight lines drawn from the centre out to the circle's edge. Working: each straight edge of a sector runs from the centre of the circle to a point on the circumference. A student who answers chord has confused a straight edge from the centre with one joining two points on the circumference. A student who answers diameter has wrongly assumed the two straight edges must form a single full diameter. A student who answers arc has named the curved edge instead of the straight edges. Answer: radius (radii).
- (b) (2, 6) — To add column vectors, add the top numbers together and add the bottom numbers together, separately: top = 3 + (−1) = 2, bottom = 2 + 4 = 6, giving (2, 6). A candidate who writes the top and bottom numbers of the correct answer in the wrong order gives (6, 2). A candidate who subtracts instead of adds, working out 3 − (−1) = 4 and 2 − 4 = −2, gives (4, −2). A candidate who adds the top numbers correctly but subtracts the bottom numbers by mistake, working out 2 − 4 = −2, gives (2, −2). The correct answer is (2, 6).
- (b) The arcs only meet exactly on AB, not above or below it. — With a radius of exactly half of AB, the arc centred at A and the arc centred at B each reach precisely to the midpoint of AB, so they meet only at that one point, on the line AB itself — there are no two intersection points above and below the line to join, so the perpendicular bisector cannot be drawn. (The claim that this radius is always too short to draw any arc is wrong, since a radius equal to half of AB is a perfectly valid, positive length for a compass arc; the claim that the radius must equal the full length of AB is wrong — a longer radius than half of AB would also work, it does not have to equal AB exactly; the claim that the arcs would not cross line AB at all is wrong, since they do cross it — that is exactly the problem, they meet only on it.)
- (c) √3/2 — Method: sin 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2 and base 1, and Pythagoras gives its height as √3 because 2² − 1² = 3. Working: sine is the opposite side over the hypotenuse, the side opposite the 60° angle is the height √3, and the hypotenuse is 2, so sin 60° = √3 ÷ 2. Answer: sin 60° = √3/2. Swapping sine for cosine at this angle gives cos 60° = 1/2; taking the value from the other special triangle, the right-angled isosceles one, gives sin 45° = √2/2; and reading the tangent row instead of the sine row gives tan 60° = √3.
- (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.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (a) (−1, 1) — Translating by $\binom{−2}{3}$ subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. This gives image vertices (−1, 4), (3, 4), (3, 7) and (−1, 7). The point (−1, 1) is not one of these: it has the correct new x-coordinate (1 − 2 = −1) but keeps the original y-coordinate (1) instead of adding 3, as if only the horizontal part of the vector had been applied.
- (d) (6, 8) — Method: add column vectors by adding the top numbers together and the bottom numbers together. Working: top numbers 5 + 1 = 6; bottom numbers 2 + 6 = 8. Answer: p + q = (6, 8). A candidate who subtracts instead of adds gets (4, −4). A candidate who swaps the top and bottom of the correct answer gets (8, 6). A candidate who only adds the top numbers and leaves the bottom number of p unchanged gets (6, 2).
- (a) 15 km — Multiply the length drawn on the map by the scale: 3 × 5 = 15 km. Choosing 8 km adds the two scale numbers together (3 + 5 = 8) instead of multiplying them. Choosing 30 km comes from misreading the scale as 1 cm : 10 km and doubling the correct answer. Choosing 5 km simply repeats the scale's distance figure and ignores that the road is drawn 3 cm long, not 1 cm.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
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