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.
Non-calculator
Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.Work out the length of the straight line segment joining the points (2, −4) and (−3, −4).
- 2.A cylindrical candle and a cone-shaped candle have the same base radius and the same height. State the cone's volume as a fraction of the cylinder's volume.
- 3.Write down the exact value of tan 60°.
- 4.A roof truss has two horizontal parallel rafters, one above the other. A straight strut crosses both rafters. Where the strut crosses the lower rafter, the angle above the rafter and to the left of the strut is 65°. Work out the size of the angle above the upper rafter and to the right of the strut, where the strut crosses it.
- 5.An angle measures 108°. Write down the name given to an angle of this size.
- 6.Triangle ABC has a right angle at B. Triangle DEF has a right angle at E. The hypotenuse of triangle ABC equals the hypotenuse of triangle DEF, and side AB equals side DE. Which condition proves that the two triangles are congruent?
- 7.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 8.In triangle XYZ, the side opposite vertex X is labelled using the standard lowercase-letter convention. Which label is correct?
- 9.Triangles PQS and RQS share the common side QS. PQ equals RQ, and PS equals RS. Which condition proves that triangle PQS is congruent to triangle RQS?
- 10.A triangle has vertices X, Y and Z, with a right angle at Y. Side XY is twice as long as side YZ. Which of the following correctly describes this triangle?
- 11.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 12.A ramp's sloped surface crosses two horizontal parallel rails. At the top rail, the angle between the ramp and the rail on the right of the ramp is (3x + 10)°. At the bottom rail, the angle between the ramp and the rail on the left of the ramp is (5x − 30)°. Work out the value of x.
- 13.A solid cube has a volume of 8 cm³. Work out the length of one edge of the cube.
- 14.A written description of a diagram states: 'AB is marked with a single arrow, CD is marked with a single arrow, and EF is marked with a double arrow.' Using the standard convention for marking parallel lines, what can be concluded?
- 15.Quadrilateral ABCD has vertices A(1, 1), B(5, 1), C(5, 4) and D(1, 4). The quadrilateral is translated by the vector . Which of these points is NOT a vertex of the image?
Answer key
- (b) 5 — Method: the two points have the same y-coordinate, so the segment joining them runs horizontally and its length is the gap between the two x-coordinates; a length is a distance, so it is never negative. Working: the x-coordinates are 2 and −3, so the gap is 2 − (−3) = 2 + 3 = 5. Both points have y = −4, so there is no vertical part to add on. Answer: 5. The distractors: 1 comes from dropping the minus sign on −3 and working out 3 − 2 instead; 0 comes from subtracting the y-coordinates, which are equal, in place of the x-coordinates; 6 comes from counting the grid lines from −3 across to 2 inclusive, which counts one more than the number of gaps between them.
- (c) One third of the cylinder's volume — Volume of a cylinder = base area × height. Volume of a cone = 1/3 × base area × height. For the same base radius and height, the cone's volume is exactly one third of the cylinder's, so it uses less wax. A student who thinks the cone is half the cylinder's volume has confused it with a different solid's ratio. A student who thinks the two volumes are the same has ignored the 1/3 factor in the cone formula entirely. A student who thinks the cone is two thirds of the cylinder's volume has the right idea that it is a fraction, but the wrong fraction.
- (d) √3 — tan 60° = √3 — one of the exact values you need to know without a calculator. √3/2 is the exact value of sin 60° and of cos 30°, not tan 60°. 1/√3 is the exact value of tan 30°, the reciprocal-angle case. 3 comes from squaring √3 instead of reading off tan 60° itself.
- (c) 115 — Method: use corresponding angles to carry the 65° angle from the lower rafter up to the upper rafter, then use angles on a straight line to move to the other side of the strut. Working: the angle above the upper rafter and to the left of the strut corresponds to the given angle, so it is 65°; the angle above the upper rafter and to the right of the strut lies on a straight line with it, so it is 180 − 65 = 115. Answer: 115°. A candidate who assumes the angle stays 65° without allowing for the move from the left of the strut to the right of it gives 65. A candidate who uses 90° instead of 180°, working out 90 − 65, gets 25. A candidate who adds instead of subtracting, working out 180 + 65, gets 245.
- (d) An obtuse angle — Method: compare the angle with the two markers that separate the angle names, a right angle at 90° and a straight line at 180°. Working: 108° is greater than 90° and smaller than 180°, so it lies between the right angle and the straight line. Answer: an obtuse angle. The distractors: an acute angle is one below 90°, and is chosen by a candidate who checks only that 108° is less than 180°; a reflex angle is one above 180°, and is chosen by a candidate who checks only that 108° is more than 90° and then takes the largest category; a right angle is exactly 90°, and is chosen by reading 108° as near enough to 90° instead of comparing it properly.
- (d) RHS — Both triangles have a right angle, their hypotenuses are equal, and one further pair of sides (AB and DE) are equal, exactly the RHS condition, so RHS is correct. SAS needs two sides and the angle between them; here the right angle at B is not between the hypotenuse and AB, so this is not a valid SAS setup, making SAS wrong. SSS needs three pairs of equal sides, but only two sides are given here, so SSS is wrong. ASA needs two angles and the side between them, but only one angle, the right angle, is given, so ASA is wrong.
- (a) 6 cm — Method: a rectangle is made of two lengths and two widths, so the perimeter is 2 × (length + width). Half the perimeter is therefore one length plus one width, and the width is what is left of that half once the length is taken away. Working: 30 ÷ 2 = 15 cm is one length added to one width, and 15 − 9 = 6. Answer: 6 cm. The distractors: 21 cm comes from taking the length off the whole perimeter, 30 − 9, so three sides are still counted in the figure that is written down; 12 cm comes from removing both lengths correctly, 30 − 2 × 9 = 12, and then recording that remainder as the width instead of halving it, because it is in fact the two widths together; 7.5 cm comes from dividing the perimeter by four, 30 ÷ 4, which is the calculation for the side of a square and throws away the length that was given.
- (c) x — By convention, the side opposite a vertex is labelled with the lowercase version of that vertex's letter, so the side opposite X is labelled x. y is the label for the side opposite Y, not X. z is the label for the side opposite Z, not X. X is the vertex's own uppercase letter — the convention specifically switches to lowercase for the side, so the uppercase letter on its own is not correct.
- (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) A right-angled triangle — The right angle at Y makes this a right-angled triangle, so that description is correct. Since XY is twice YZ, those two sides cannot be equal. The third side XZ is opposite the right angle, so it is the hypotenuse and is longer than either XY or YZ, so it cannot equal either of them. No two sides are equal, which rules out 'an isosceles triangle' and 'a right-angled isosceles triangle', both of which wrongly assume two equal sides. 'An equilateral triangle' would need all three sides equal, which contradicts XY being twice YZ, so it is wrong too.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (d) 20 — Alternate angles between parallel lines are equal, so 3x + 10 = 5x − 30. Rearranging, 10 + 30 = 5x − 3x, so 40 = 2x, and x = 20. −10 comes from a sign error when rearranging, moving a term to the wrong side and getting −20 = 2x instead. 25 comes from wrongly treating the two angles as co-interior and adding them to 180°: (3x + 10) + (5x − 30) = 180 gives 8x − 20 = 180, so x = 25. 47.5 makes the same co-interior mistake but sets the sum equal to 360° instead of 180°, giving 8x − 20 = 360 and x = 47.5.
- (b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (a) (−1, 1) — Translating by $\binom{−2}{3}$ subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. This gives image vertices (−1, 4), (3, 4), (3, 7) and (−1, 7). The point (−1, 1) is not one of these: it has the correct new x-coordinate (1 − 2 = −1) but keeps the original y-coordinate (1) instead of adding 3, as if only the horizontal part of the vector had been applied.
Build your own mix at the worksheet builder.