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.Triangle ABC is similar to triangle PQR. Side AB is 8 cm and corresponds to side PQ, which is 12 cm. Side AC is 6 cm and corresponds to side PR. Work out the length of PR.
- 2.A rectangular garden gate is 3 m wide and 4 m high. A straight brace runs from one corner of the gate to the opposite corner. Work out the length of the brace.
- 3.In one circle, chord PQ is 6 cm long, chord RS is 10 cm long, chord TU is 14 cm long and chord VW is 15 cm long. Write down which of these chords lies closest to the centre of the circle.
- 4.A translation is described by the column vector . Write down a description of this translation in words, stating the horizontal and vertical movement.
- 5.The three angles of a triangle are 10°, 58° and 112°. Write down the name that describes this triangle.
- 6.Two similar flags have widths in the ratio 2 : 5. The width of the smaller flag is 8 cm. Work out the width of the larger flag.
- 7.Triangle A has a vertex at (1, 2). Triangle A is enlarged, centre (1, 1), to give triangle B, whose corresponding vertex is at (1, 5). Work out the scale factor of the enlargement.
- 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.The point A(2, 5) is translated to the point B(7, 3). Write down the column vector that describes this translation.
- 10.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 11.A sector of a circle has radius 5 cm and angle 72°. Work out the arc length of the sector, in terms of π.
- 12.Triangle ABC has AB = (2x + 3) cm, BC = 11 cm and angle B = 90°. Triangle DEF has DE = 13 cm, EF = 11 cm and angle E = 90°. Given that the two triangles are congruent by SAS, work out the value of x.
- 13.A right-angled triangle has a hypotenuse of 29 cm and one shorter side of 20 cm. Work out the length of the other shorter side.
- 14.A landscaper marks out two triangular flower beds using stakes and a tape measure. Bed 1 has two edges of 5 m and 7 m, meeting at a corner with a marked angle of 60° between them. Bed 2 has edges of 5 m and 7 m, meeting at a corner also marked at 60° between them. The landscaper wants to check the two beds will be exactly the same size and shape using only these three measurements. Which condition proves the two beds are congruent?
- 15.A trapezium-shaped garden bed has parallel sides of 2.5 m and 4.5 m, and a perpendicular width of 3 m. A gardener wants to know the area of the bed before ordering topsoil. Work out the area of the garden bed.
Answer key
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (b) 5 m — Method: the brace, the width and the height form a right-angled triangle in which the brace faces the right angle, so it is the hypotenuse and Pythagoras' theorem applies, a² + b² = c². Working: c² = 3² + 4² = 9 + 16 = 25, so c = √25 = 5. Answer: 5 m. The distractors: 7 m comes from adding the two sides, 3 + 4, instead of adding their squares; 25 m comes from stopping at c² = 25 and forgetting to take the square root; 12 m comes from multiplying 3 × 4, which gives the area of the gate in square metres and not a length across it.
- (c) VW — Method: within one circle a chord's distance from the centre is fixed by its length, because a chord passing nearer the centre cuts further across the circle; so the chord that lies closest to the centre is simply the longest one listed. Working: the four lengths are 6 cm, 10 cm, 14 cm and 15 cm. Placing them in order, the greatest is 15 cm, and that length belongs to VW, so VW lies closest to the centre. Answer: VW. The distractors: PQ comes from reversing the rule and taking the shortest chord to be the one tucked nearest the centre; TU comes from knowing that the very longest chord is a diameter, deciding that such a chord passes through the centre rather than lying close to it, ruling the 15 cm chord out on that ground and taking the next longest; RS comes from reading 'closest to the centre' as 'nearest the middle of the list of lengths' and picking a middling value.
- (b) 7 units right and 2 units down — In the column vector $\binom{7}{−2}$, the top number gives the horizontal movement and the bottom number gives the vertical movement. A positive top number (7) means 7 units to the right, and a negative bottom number (−2) means 2 units down. The distractor '7 units left and 2 units up' reverses both signs, as if the vector were $\binom{−7}{2}$. The distractor '2 units right and 7 units down' swaps the horizontal and vertical numbers around. The distractor '7 units right and 2 units up' gets the horizontal movement correct but reverses the sign of the vertical movement.
- (c) Scalene — Method: equal angles in a triangle sit opposite equal sides, so compare the three angles with each other. Working: 10°, 58° and 112° are all different, so no two sides of the triangle are equal either, and a triangle with no equal sides is scalene. Answer: scalene. The distractors: isosceles is chosen by candidates who pair up the two acute angles, 10° and 58°, as base angles without checking that they are actually equal; equilateral is chosen by candidates who check that the three angles add to 180° and take that as meaning the triangle is regular; right-angled is chosen by candidates who see that 112° is larger than 90° and classify the triangle as containing a right angle, when in fact none of the three angles is 90°.
- (a) 20 cm — The scale factor from the smaller flag to the larger flag is 5 ÷ 2 = 2.5, so the larger width is 8 × 2.5 = 20 cm. The distractor 3.2 cm comes from using the ratio the wrong way round, 8 × 2 ÷ 5 = 3.2. The distractor 11 cm comes from adding the difference between the ratio numbers (5 − 2 = 3) to the given width, 8 + 3 = 11. The distractor 40 cm comes from multiplying by 5 without dividing by 2 first, 8 × 5 = 40.
- (b) 4 — The scale factor is the distance from the centre to the image, divided by the distance from the centre to the object: (5 − 1) ÷ (2 − 1) = 4 ÷ 1 = 4. (2.5 comes from dividing the raw y-coordinates, 5 ÷ 2, without first subtracting the centre's coordinate; 3 comes from subtracting the two distances instead of dividing them; 0.25 comes from dividing the distances the wrong way round.)
- (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.
- (c) $\binom{5}{-2}$ — Method: the top number of a column vector is the change in the x-coordinate and the bottom number is the change in the y-coordinate, each worked out as image minus object. Working: across, 7 − 2 = 5; up, 3 − 5 = −2. So the point moves 5 to the right and 2 down. Answer: $\binom{5}{-2}$. Subtracting the other way round, object minus image, gives $\binom{-5}{2}$, which is the journey from B back to A. Recording the vertical change as 2 because the gap between 3 and 5 is 2, without noting the direction, gives $\binom{5}{2}$, a movement 2 upwards. Adding the coordinates instead of subtracting them gives $\binom{9}{8}$.
- (a) $\binom{4}{4}$ — A movement to the right is a positive top number and a movement up is a positive bottom number, so 4 right and 4 up gives $\binom{4}{4}$. $\binom{4}{-4}$ wrongly gives the vertical movement a negative sign, as if it were downward. $\binom{-4}{4}$ wrongly gives the horizontal movement a negative sign, as if it were leftward. $\binom{-4}{-4}$ gets both signs wrong, as if the translation were left and down.
- (d) 2π cm — The arc is a fraction of the whole circumference. The fraction is 72 ÷ 360 = 1/5 of the circle, and the full circumference is 2 × π × 5 = 10π cm. So the arc length is 1/5 × 10π = 2π cm. Taking the whole circumference and forgetting the fraction gives 10π cm. Using 72 ÷ 180 instead of 72 ÷ 360 gives 4π cm. Treating the 5 cm as a diameter instead of a radius gives π cm.
- (d) 5 — Method: since the triangles are congruent by SAS, the corresponding sides AB and DE must be equal, because both are the side next to the given right angle that is not BC or EF. Working: AB = DE gives 2x + 3 = 13, so 2x = 10, so x = 5. Options: 10 comes from dropping the coefficient of x and solving x + 3 = 13 instead of 2x + 3 = 13; 4 comes from matching AB to the wrong side, EF, giving 2x + 3 = 11, so 2x = 8, so x = 4; 8 comes from a sign error, solving 2x − 3 = 13 instead of 2x + 3 = 13, giving 2x = 16, so x = 8. Answer: 5.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
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