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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (a) where the angle bisector meets the posts' perpendicular bisector — Being equidistant from the two walls means lying on the angle bisector of the corner; being equidistant from the two posts means lying on the perpendicular bisector of the 4 m segment joining them. A single point satisfying both conditions is wherever those two loci cross. "where the angle bisector meets the line joining the posts" uses the straight line between the posts instead of its perpendicular bisector — a point on that line is not generally equidistant from both posts. "the perpendicular bisector of the posts, alone" satisfies only the posts condition, ignoring the walls entirely. "the angle bisector of the corner, alone" satisfies only the walls condition, ignoring the posts entirely.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (a) 6 cm — Method: a diameter runs right across a circle through its centre, so it is made of two radii laid end to end, which gives diameter = 2 × radius. Working: the radius is 3 cm, so the diameter is 2 × 3 = 6 cm. Answer: 6 cm. The distractors: 1.5 cm comes from dividing by 2 instead of multiplying by it, which is the relationship applied in the wrong direction; 3 cm comes from copying the radius straight down, treating the two words as names for the same measurement; 5 cm comes from adding 2 to the radius instead of multiplying the radius by 2.
- (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.
- (a) 3 squares — Method: the plan view shows only the floor positions that have at least one cube standing on them; height does not add extra squares to the plan. Working: the base row occupies three floor positions in a line. The two extra cubes stand on top of two of those same three positions, so they do not create any new floor position. Answer: 3 squares. The distractors: 5 squares comes from adding the total number of cubes used (3 + 2 = 5) instead of counting distinct floor positions. 2 squares comes from counting only the raised two-cube section and ignoring the single cube at the other end of the row. 4 squares comes from counting one of the shared positions twice.
- (d) 9√3 cm — DE is opposite the 60° angle at F, and EF is adjacent to it, so DE = EF × tan 60° = 9 × √3 = 9√3 cm. 9√3/2 cm comes from using sin 60° = √3/2 instead of tan 60°. 3√3 cm comes from using tan 30° = 1/√3 instead of tan 60° (9 × 1/√3 = 9/√3 = 3√3). 18 cm is the hypotenuse DF, not DE: it comes from using cos 60° = 1/2 and working out 9 ÷ 1/2 = 18, which finds the wrong side of the triangle.
- (d) 71° — Alternate angles between parallel lines are equal. Since angle HMN and angle MNK are alternate angles, angle MNK = angle HMN = 71°.
- (a) Yes - SSS, the three side lengths all match — Bracket P's sides (12 cm, 16 cm, 20 cm) can each be matched to one of Bracket Q's sides (16 cm, 20 cm, 12 cm) — the same three lengths, just listed differently — so the brackets are congruent by SSS. SAS is wrong here because no angle is stated for either bracket, only three sides. The order the sides are listed in does not matter for SSS — only whether the SET of three lengths matches, and it does, so 'listed in a different order' is not a reason to say no. Nothing extra is needed: SSS proves congruence from side lengths alone, without any angles, so it can be determined.
- (b) A triangular prism standing on its triangular end — Method: a rectangular elevation with no sloping sides means the solid keeps the same cross-section all the way from the bottom to the top; work out which solid, standing the right way up, has a triangular cross-section that stays that shape as you go higher. Working: a triangular prism standing upright on its triangular end has a triangle as its plan view, and because the cross-section is constant all the way up, both the front and side elevations are plain rectangles. Answer: a triangular prism standing on its triangular end. The distractors: a triangle-based pyramid standing on its triangular base does give a triangle as its plan view, but its cross-section shrinks towards the apex, so its front and side elevations come to a point and are triangles, not rectangles. A cuboid standing on a rectangular face is wrong because its plan view is a rectangle, not a triangle. A triangular prism lying on one of its rectangular faces is wrong because it is then the triangular end that faces the side, so its plan view is a rectangle and one of its elevations is a triangle.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (b) Square — A square has all four sides equal, all four angles equal to 90°, and diagonals that are equal in length and bisect each other at right angles — every part of the description matches, so Square is correct. A rhombus has all four sides equal and diagonals bisecting at right angles, but its interior angles are not generally 90° (only a square, a special rhombus, has that), so it does not fully match. A rectangle has four 90° angles and equal diagonals, but its sides are not all equal in general, so it fails the equal-sides condition. A kite has two pairs of adjacent equal sides rather than all four sides equal, and its diagonals are not generally equal in length, so it fails both conditions.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (c) 5 hours — The distance from the origin to (7, 24) is √(7² + 24²) = √(49 + 576) = √625 = 25 km. Travelling at 5 km per hour, the time taken is 25 ÷ 5 = 5 hours. 25 hours mistakes the distance itself for the time, forgetting to divide by the speed. 125 hours comes from multiplying the distance by the speed, 25 × 5, instead of dividing. 0.2 hours comes from dividing the speed by the distance, 5 ÷ 25, the wrong way round.
- (d) The centre — Method: sort the four names by what kind of object each one is, because only one of them names a single point rather than a line or a curve. Working: a chord is a straight line joining two points on the circle; a diameter is the special chord that runs right across the circle through the middle, so it too is a line; an arc is a piece of the circle's own curve. That leaves one name, and it belongs to the point from which every radius is drawn, which is why every point on the circle is the same distance from it. Answer: The centre. The distractors: The chord is picked by a candidate who reads 'the same distance' as a line of fixed length rather than as a point; The diameter is picked by a candidate who names the line that passes through the point instead of the point itself; The arc is picked by a candidate who hunts for a part of the circle's own curve rather than for a point inside it.
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