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.
Geometry and measures worksheet — GCSE Foundation
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- 1.Write down the name given to a quadrilateral with exactly one pair of parallel sides.
- 2.u is the column vector with top number 5 and bottom number 1. v is the column vector with top number 2 and bottom number 4. Work out u − v, giving your answer as a column vector in the form (top, bottom).
- 3.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 4.A locus consists of all the points that are the same perpendicular distance from two parallel lines 6 cm apart. Describe this locus.
- 5.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.
- 6.A vertical flagpole casts a shadow 6 m long at the same time as a 1.2 m vertical post casts a shadow 1.8 m long. The flagpole and the post are both vertical, and the triangles formed by each object, its shadow, and the sun's ray are similar. Work out the height of the flagpole.
- 7.A tent's sloping side makes an angle of 45° with the ground. The sloping side is 3√2 m long. Using the exact value of sin 45°, work out the exact height of the tent.
- 8.A triangle has vertices A(0, 0), B(10, 0) and C(5, 12). Work out the area of the triangle.
- 9.A triangular bunting flag has two equal sides. The angle between those two equal sides is 80°, and the other two angles of the flag are equal to each other. Work out the size of each of the two equal angles.
- 10.A composite solid is made from a cylinder of radius 3 cm and height 10 cm, with a cone of the same radius and vertical height 4 cm fixed on top, apex upward. Work out the total volume of the solid. Use π = 3.14 and give your answer correct to 1 decimal place.
- 11.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 12.Lines PQ and RS are parallel. A straight line crosses PQ at point X and crosses RS at point Y. Angle PXY = 63°. Angle PXY and angle XYS are co-interior (allied) angles, which add up to 180°. Work out the size of angle XYS.
- 13.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.
- 14.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.
- 15.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
Answer key
- (a) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs, and a kite and rhombus are defined by side lengths, not by having only one pair of parallel sides.
- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (a) a line parallel to both, 3 cm from each — Being equidistant from two parallel lines 6 cm apart means being exactly halfway between them all along their length, tracing out a third line, parallel to both, at 3 cm from each — half of the 6 cm gap. "a line parallel to both, 6 cm from each" repeats the full gap instead of halving it, which puts those points past one of the lines entirely. "a circle of radius 3 cm, centred midway" applies to a locus equidistant from a single fixed POINT, not from two parallel lines running the full length. "the perpendicular bisector of the gap" crosses the gap at right angles and meets each line at only one point — it is not the whole locus, which runs parallel to the lines, not across them.
- (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.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (b) 50° — Method: the three angles of a triangle add up to 180°, and the two angles opposite the equal sides are equal, so subtract the given angle from 180° and halve the remainder. Working: 180° − 80° = 100°, and 100° ÷ 2 = 50°. Answer: 50°. The distractors: 100° comes from subtracting from 180° and forgetting to halve, so it is the two equal angles together; 40° comes from halving the 80° that is given rather than halving what is left of the 180°; 80° comes from assuming that the two base angles must match the angle between the equal sides.
- (b) 320.3 cm³ — Cylinder volume = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. Cone volume = (1/3)πr²h = (1/3) × 3.14 × 9 × 4 = (1/3) × 113.04 = 37.68 cm³. Total = 282.6 + 37.68 = 320.28 cm³, which rounds to 320.3 cm³.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (c) 117° — Co-interior (also called allied) angles between parallel lines add up to 180°. Angle PXY and angle XYS are co-interior angles, so angle PXY + angle XYS = 180°. 180° − 63° = 117°, so angle XYS = 117°. A student who mistakes co-interior angles for alternate angles, which are equal rather than supplementary, answers 63° instead. A student who treats the two angles as complementary, subtracting from 90° instead of 180°, gets 90° − 63° = 27°. A student who assumes the transversal meets the parallel lines at right angles answers 90°.
- (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.
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
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