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.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 2.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 3.A wooden block is a cuboid measuring 2 cm by 10 cm by 15 cm. Work out the volume of the block.
- 4.A tangent touches a circle with centre O at the point P. Q is a point on the tangent. Write down the circle fact that tells you the size of angle OPQ.
- 5.A square-based pyramid is cut by a flat plane parallel to its base, partway up between the base and the apex. What shape is the cross-section?
- 6.A sector of a circle is bounded by two straight edges and one curved edge. Write down the mathematical term for each of the two straight edges.
- 7.A line segment joins A(1, 1) and B(4, 7). Write down the sign of the gradient of this line.
- 8.Lines AB and CD are parallel. A straight line EF crosses AB at point P and crosses CD at point Q. At P, angle APE = 65°. Angle APE and angle BPQ are vertically opposite. Angle BPQ and angle PQD are co-interior (allied) angles. Work out angle PQD.
- 9.A straight line passes through points A, B and C, with B between A and C. A fourth point D is not on the line. Angle ABD = 132°. Work out the size of angle DBC.
- 10.A scale drawing of a bridge uses a scale of 1 : 25. A support beam on the drawing measures 4.4 cm. What is the real length of the beam, in metres?
- 11.In triangle ABC and triangle DEF, the angle at A and the angle at D are both 40°, and the angle at B and the angle at E are both 70°. No side lengths are given. Write down whether the two triangles must be similar, with the reason.
- 12.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 13.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?
- 14.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 15.Write down the mathematical name for a quadrilateral that has exactly one pair of parallel sides.
Answer key
- (a) 65 — There are 10 millimetres in every centimetre, so to convert from cm to mm, multiply by 10: 6.5 × 10 = 65 mm. A candidate who forgets to convert at all writes down the original number, 6.5. A candidate who multiplies by 100 instead of 10, confusing cm-to-mm with m-to-cm, gets 650. A candidate who divides by 10 instead of multiplying gets 0.65. The correct length in millimetres is 65.
- (a) 143° — Angles on a straight line add up to 180°. Set up 37° + x = 180°. Subtract: x = 180° − 37° = 143°. 53° comes from using 90° as the total, as if the two angles made a right angle, instead of the 180° of a straight line.
- (d) 300 cm³ — Method: the volume of a cuboid is length × width × height. Working: 2 × 10 = 20, then 20 × 15 = 300. Answer: 300 cm³. The distractors: 27 cm³ comes from adding the three edges, 2 + 10 + 15, instead of multiplying them; 400 cm² comes from working out the surface area, 2 × (2 × 10 + 2 × 15 + 10 × 15) = 400, which answers a different question and carries a different unit; 150 cm³ comes from multiplying 10 × 15 and leaving the 2 cm edge out of the calculation altogether.
- (d) The angle between a tangent and a radius is 90° — OP is the radius drawn to the point of contact P, and a circle theorem states that a tangent always meets that radius at a right angle, so angle OPQ = 90°. A tangent is not parallel to the radius it touches — at the point of contact it is perpendicular to that radius, not parallel to it. A tangent does not pass through the centre — a straight line through the centre that also touches the circle at one point would have to be a diameter, which is a different line entirely. The angle-in-a-semicircle theorem needs a triangle drawn inside the circle with a diameter as its longest side; there is no such triangle here, just a tangent and a radius.
- (c) Square — Because the cut is parallel to the square base, it produces a shape that is similar to the base — the same shape, scaled down — so it is also a square, with four equal sides, smaller than the base. A candidate who looks at the sloped triangular faces instead of at the horizontal cut answers triangle. A candidate who pictures a slanted cut, which would shorten one pair of sides, answers trapezium, but a cut parallel to the base scales all four sides by the same amount. A candidate who counts an extra side, perhaps including an edge of the base, answers pentagon. The cross-section is a square.
- (c) Radius — Method: recall that a sector is formed using two straight lines drawn from the centre out to the circle's edge. Working: each straight edge of a sector runs from the centre of the circle to a point on the circumference. A student who answers chord has confused a straight edge from the centre with one joining two points on the circumference. A student who answers diameter has wrongly assumed the two straight edges must form a single full diameter. A student who answers arc has named the curved edge instead of the straight edges. Answer: radius (radii).
- (d) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (b) 48° — Angles on a straight line add up to 180°, so angle ABD + angle DBC = 180°. 180° − 132° = 48°, so angle DBC = 48°. A student who uses angles around a point (360°) instead of a straight line (180°) finds 360° − 132° = 228°. A student who halves angle ABD by mistake, thinking DBC must be half of ABD, gets 132° ÷ 2 = 66°.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (d) Yes, by the AA condition — Method: similarity is decided by the angles, and because the three angles of a triangle add up to 180°, two matching pairs force the third pair to match as well. Working: the angles at A and D are both 40° and the angles at B and E are both 70°, so the third angles are both 180° − 40° − 70° = 70° and all three pairs are equal. Two pairs were enough, and that is the AA condition. Answer: yes, by the AA condition. The distractors: 'Yes, by the SSS condition' names a condition about three pairs of sides in proportion, and the question gives no side lengths at all; 'No, because no side lengths are given' treats side information as necessary, which it is for congruence but not for similarity; 'No, because the triangles may be different sizes' turns the definition of similarity into an objection, since similar figures are allowed to differ in size and only their shape must match.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (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.
- (c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (b) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs of parallel sides, not one, so it does not fit the definition. A kite has no parallel sides at all. A rhombus, like a parallelogram, has two pairs of parallel sides. The correct name is trapezium.
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