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.A cone is cut by a flat plane parallel to its circular base, partway up between the base and the apex. What shape is the cross-section?
- 2.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 3.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
- 4.A sector of a circle has an angle of 150°. Work out what fraction of the circle's area the sector is.
- 5.A trapezium has an area of 45 cm², a perpendicular height of 5 cm, and one parallel side of length 10 cm. Work out the length of the other parallel side.
- 6.A square-based pyramid has a square base and four identical triangular sloping faces, with its apex directly above the centre of the base. How many planes of symmetry does it have?
- 7.A straight line is drawn from the centre of a circle to a point on the circle. Write down the name of this line.
- 8.ABCD is an isosceles trapezium. The sides AB and DC are parallel, and the two sloping sides AD and BC are equal in length. Angle D is 112°. Work out the size of angle B.
- 9.Work out the exact value of (sin 30°)² + (cos 30°)².
- 10.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 11.Two straight lines cross at a point. Which statement about a pair of vertically opposite angles is always true?
- 12.Priya says: 'To find the arc length of a sector, you divide the circumference by the angle in degrees.' Which statement about Priya's method is correct?
- 13.A perpendicular bisector is drawn through the midpoint of a line segment AB. Which set of points does this construction show?
- 14.Triangle ABC has AB = 10 cm, BC = 7 cm and angle A = 40°. Triangle DEF has DE = 10 cm, EF = 7 cm and angle D = 40°. Sam says the two triangles are congruent by SAS. Which statement about Sam's claim is correct?
- 15.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder. Give your answer to the nearest whole number of cubic centimetres.
Answer key
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (a) 5/12 — A sector's area is always the same fraction of the circle as its angle is of the full 360° turn. Here that fraction is 150 out of 360, which simplifies (divide both numbers by 30) to 5/12. Treating a full turn as 180° instead of 360° gives 5/6. Confusing the 150° given with the 180° of a half turn gives 1/2. Working out the fraction OUTSIDE the sector instead of inside it gives 7/12.
- (c) 8 cm — Area of a trapezium = (sum of parallel sides) ÷ 2 × height, so 45 = (a + 10) ÷ 2 × 5. Dividing 45 by 5 gives 9, so (a + 10) ÷ 2 = 9, meaning a + 10 = 18, so a = 18 − 10 = 8 cm. A pupil who correctly finds that the two parallel sides add up to 18 but forgets to subtract the known side of 10 cm gives the sum of both parallel sides, 18 cm, as the answer. A pupil who then adds 10 again by mistake instead of subtracting gets 18 + 10 = 28 cm. A pupil who halves the correct answer by mistake gets 8 ÷ 2 = 4 cm. The correct length of the other parallel side is 8 cm.
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (b) The radius — Method: each of the four names describes a line in a fixed position relative to the circle, so match the description in the question against those positions. Working: the line described has one end at the centre and its other end on the circle. A chord joins two points that both lie on the circle, so it is ruled out. A diameter joins two points on the circle by passing through the centre, so it is twice as long as the line described. A tangent touches the circle at exactly one point and never reaches the centre. The line with one end at the centre and one end on the circle is a radius. Answer: The radius. The distractors: The diameter comes from remembering only that the line involves the centre, and not noticing that it stops there instead of carrying on to the far side; The chord comes from seeing one end on the circle and assuming both ends are; The tangent comes from matching the phrase 'a point on the circle' to the line that touches at exactly one point, ignoring that a tangent never reaches the centre.
- (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.
- (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (d) They are always equal to each other — Method: label the four angles a, b, c, d in order around the crossing point, then use the fact that neighbouring angles lie on a straight line. Working: a and b lie on a straight line, so a + b = 180°; b and c also lie on a straight line, so b + c = 180°. Both a and c are therefore 180° minus b, which forces a = c. Answer: a pair of vertically opposite angles is always equal to each other. The distractors: the claim that each is 90° holds only when the two lines happen to be perpendicular, so it is not always true; the claim that they add to 180° confuses the opposite pair with the neighbouring pair that lies along a straight line, and again holds only in the perpendicular case; the claim that they add to 360° uses the total of all four angles at the point rather than of the opposite pair.
- (a) Wrong: multiply by angle ÷ 360, not divide by it. — The arc length is the fraction angle ÷ 360 of the circumference, and that fraction is MULTIPLIED by the circumference — it is never divided by the angle itself. So Priya's method, dividing by the angle, is wrong, and the correct statement names both the fraction and the multiplication. Saying she is correct accepts her wrong division. Saying the arc length is the full circumference multiplied by the angle in degrees, with no division at all, drops the 360 that turns the angle into a fraction. Saying the arc length is the radius multiplied by the angle in degrees swaps the circumference for the radius, which is a different and, in degrees, incorrect formula.
- (d) points the same distance from A as from B — The perpendicular bisector of AB is, by definition, the locus of every point equidistant from A and B — any point on it forms two congruent right-angled triangles with A and B, which is exactly the property the construction guarantees. "points as far from A as the length AB" describes a circle centred at A with radius AB, not a bisector. "the single point exactly halfway along AB" names only the midpoint, one point, not the whole locus the construction produces. "points twice as far from A as from B" describes a different curve entirely, not a straight line construction.
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (d) 283 cm³ — Volume of a cylinder = πr²h. Using π = 3.14: V = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6, which rounds to 283 cm³. A pupil who uses the diameter, 6 cm, instead of the radius gets 3.14 × 6² × 10 = 1130.4 ≈ 1130 cm³. A pupil who forgets to square the radius gets 3.14 × 3 × 10 = 94.2 ≈ 94 cm³. A pupil who uses the circumference formula 2πr instead of πr² gets 2 × 3.14 × 3 × 10 = 188.4 ≈ 188 cm³. The correct volume is 283 cm³.
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