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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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (b) 2 — A rhombus is a special parallelogram, so opposite sides are parallel — that is two pairs of parallel sides. A quadrilateral with only one pair of parallel sides is a trapezium, not a rhombus, so 1 is wrong. A kite has 0 pairs of parallel sides, not a rhombus. There are only two pairs of opposite sides in a quadrilateral altogether, so 4 is not possible. The correct answer is 2.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (d) Alternate angles are equal — The two 65° angles are on opposite sides of the line that crosses the parallel lines, in the shape of a Z, so they are alternate angles, and alternate angles between parallel lines are always equal. Corresponding angles are equal too, but they sit in matching positions at each crossing point, in the shape of an F — a different pair from the one shown here. Co-interior angles add up to 180°, not to each other's value, and they lie between the parallel lines on the same side, in the shape of a C. Angles on a straight line add up to 180°, but that rule is about two angles at a single point on one line, not about a pair of angles formed where a line crosses two parallel lines.
- (d) Minor arc — Method: compare the lengths of the two arcs formed by the two points, and recall the term for the shorter one. Working: the two points split the circumference into two arcs; the shorter one is the minor arc and the longer one is the major arc. A student who answers major arc has picked the longer arc by mistake. A student who answers minor segment has confused the curved boundary with the enclosed two-dimensional region. A student who answers chord has named the straight line joining the two points instead of the curved arc. Answer: minor arc.
- (c) 1/2 — Method: cos 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2, base 1 and height √3. Working: cosine is the adjacent side over the hypotenuse, the side alongside the 60° angle is the base of length 1, and the hypotenuse is 2, so cos 60° = 1 ÷ 2. Answer: cos 60° = 1/2. A candidate who assumes the cosine falls steadily from cos 0° = 1 to cos 90° = 0 puts 60° two thirds of the way along and writes 1/3; the cosine does not change at a constant rate. Reading the start of the cosine row gives cos 0° = 1, and reading its end gives cos 90° = 0.
- (b) AB = DE — RHS needs a right angle, the hypotenuse, and one OTHER side to be equal; the right angles and hypotenuses are already equal, so a matching pair of the remaining sides, AB = DE, completes RHS. Angle A = angle D is an extra ANGLE fact, not the extra SIDE fact that RHS specifically requires. AC being parallel to DF says nothing about either triangle's side lengths, so it cannot complete a congruence condition. Being drawn the same way up is about orientation on the page, not about any measurement, so it proves nothing about congruence.
- (b) √2/2 — cos 45° comes from a right-angled isosceles triangle with both shorter sides 1 and hypotenuse √2, giving cos 45° = 1/√2 = √2/2. 1/2 is the value of cos 60° (mixing up the two angles). √2 is the hypotenuse length itself, not divided by it (forgetting to divide by the hypotenuse). √3/2 is the value of cos 30° (using the wrong special triangle).
- (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) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (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) second quadrant — The point (−3, 5) has a negative x-coordinate and a positive y-coordinate, and this combination lies in the second quadrant. The first quadrant needs both coordinates positive. The third quadrant needs both coordinates negative. The fourth quadrant needs a positive x-coordinate and a negative y-coordinate.
- (a) 49 cm² — The area of a square is side × side = 7 × 7 = 49 cm². 28 cm² is the perimeter of the square (4 × 7), not its area. 14 cm² adds just two of the sides together. 21 cm² multiplies the side length by 3 instead of squaring it.
- (b) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (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.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
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