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.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.
- 2.The point R(−6, 9) is translated by the vector . Write down the coordinates of the image of R.
- 3.Work out the exact value of sin 45° × cos 45°.
- 4.A translation moves the point (1, 1) to the point (9, 4). Write down the column vector of this translation.
- 5.A trapezium has parallel sides of length 6 cm and 10 cm, and a perpendicular height of 4 cm. Work out its area.
- 6.In triangle ABC and triangle DEF, AB = DE, angle ABC = angle DEF and BC = EF. Write down the congruence condition that proves the two triangles are congruent.
- 7.How many vertices does a cuboid have?
- 8.A sector of a circle has radius 5 cm and angle 72°. Work out the arc length of the sector, in terms of π.
- 9.A translation is described by the column vector . Write down a description of this translation in words, stating the horizontal and vertical movement.
- 10.Two circles are drawn with the same centre but different radii. Write down the term used to describe this pair of circles.
- 11.A is the point (1, 3) and B is the point (7, 3). Work out the coordinates of the midpoint of AB.
- 12.A circle has diameter 10 cm. Using π = 3.14, work out the circumference of the circle.
- 13.A vertical line passes through the point (5, 2). Write down the equation of this line.
- 14.A submarine travels due south. What is the three-figure bearing for this direction?
- 15.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?
Answer key
- (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).
- (b) (−1, −2) — To translate R(−6, 9) by $\binom{5}{−11}$, add 5 to the x-coordinate and −11 to the y-coordinate: (−6 + 5, 9 + (−11)) = (−1, −2). (−1, 9) applies only the x-component and leaves the y-coordinate unchanged. (−6, −2) applies only the y-component and leaves the x-coordinate unchanged. (−11, 20) comes from subtracting the vector instead of adding it.
- (c) 1/2 — sin 45° = √2/2 and cos 45° = √2/2, so sin 45° × cos 45° = √2/2 × √2/2 = 2/4 = 1/2. √2/2 comes from writing down only one of the two factors and forgetting to multiply by the other. √2 comes from adding the two exact values instead of multiplying them: √2/2 + √2/2 = √2. 1 comes from wrongly treating sin 45° × cos 45° as sin(45° + 45°) = sin 90° = 1 — multiplying two ratios is not the same as adding their angles.
- (b) $\binom{8}{3}$ — The column vector is (end − start) in each coordinate: (9 − 1, 4 − 1) = (8, 3), written as $\binom{8}{3}$. $\binom{3}{8}$ swaps the horizontal and vertical components. $\binom{10}{5}$ comes from adding the coordinates instead of subtracting them. $\binom{−8}{−3}$ comes from working out start minus end instead of end minus start.
- (b) 32 cm² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the height. Add the parallel sides: 6 + 10 = 16. Multiply by the height: 16 × 4 = 64. Half of 64 is 32 cm². 64 cm² forgets to halve and just gives (6+10)×4. 8 cm² averages the two parallel sides, (6+10)÷2 = 8, but forgets to multiply by the height. 20 cm² treats it as a triangle using only the longer parallel side as the base: half of 10 × 4.
- (b) SAS — Two pairs of matching sides are equal (AB = DE and BC = EF) and the angle between them is equal (angle ABC = angle DEF), which is exactly the Side-Angle-Side congruence test: SAS. SSS is wrong because only two pairs of sides are given, not three. ASA is wrong because only one angle is given, and it sits between the two sides rather than the side sitting between two angles. RHS is wrong because no right angle is mentioned in either triangle.
- (d) 8 — A cuboid is a 3D shape with 6 rectangular faces, 12 edges and 8 vertices (corners). Counting the corners of a cuboid gives 8. A candidate who mixes up vertices with faces answers 6. A candidate who mixes up vertices with edges answers 12. A candidate who counts only the 4 corners of the top face, forgetting the 4 corners of the bottom face, answers 4. The correct number of vertices is 8.
- (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.
- (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.
- (d) Concentric circles — Method: focus on what the two circles have in common — their centre, not their size. Working: both circles share exactly the same centre point but have different radii, which is the defining feature of this pair of circles. A student who answers congruent circles has confused 'same centre' with 'same size', but congruent circles simply have equal radii and need not share a centre. A student who answers tangential circles has confused circles that touch each other at one point with ones that share a centre. A student who answers similar circles has used the general term for the same shape at different sizes, missing the specific 'same centre' fact. Answer: concentric circles.
- (c) (4, 3) — Method: the midpoint of a segment is the mean of its two end points, so its x-coordinate is the mean of the two x-coordinates and its y-coordinate is the mean of the two y-coordinates. Working: for x, (1 + 7) ÷ 2 = 8 ÷ 2 = 4. For y, (3 + 3) ÷ 2 = 6 ÷ 2 = 3. The midpoint is therefore (4, 3). Answer: (4, 3). The distractors: (3, 3) comes from halving the difference of the x-coordinates, (7 − 1) ÷ 2 = 3, which measures half the distance instead of locating the point; (3.5, 3) comes from halving only the larger x-coordinate and leaving the smaller one out of the working; (4, 0) comes from averaging the x-coordinates correctly but then subtracting the y-coordinates, 3 − 3, rather than averaging them.
- (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (d) 180 — Method: bearings are measured clockwise from north and written using three figures. Working: south is a half turn (180°) clockwise from north. A student who answers 090 has confused south with east. A student who answers 270 has confused south with west. A student who answers 018 has written the correct digits in the wrong order. Answer: 180.
- (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.
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