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) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (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) 10 cm — For a triangle to exist, any two sides must add up to more than the third side. 9 + 10 = 19 > 15, and 15 − 9 = 6 < 10, so 10 cm satisfies the triangle inequality. The other lengths fail: 6 cm gives 9 + 6 = 15, which is not more than 15; 24 cm and 26 cm are each at least as large as 9 + 15 = 24.
- (d) 60 — Method: write the two supplementary angles as x and 2x, since one is twice the other, and solve x + 2x = 180. Working: 3x = 180, so x = 60. Answer: the smaller angle is 60°. A candidate who gives the larger angle, 2x, instead of the smaller angle gets 120. A candidate who uses a complementary sum of 90° instead of a supplementary sum of 180° gets 30. A candidate who divides 180 by 2 instead of by 3 gets 90.
- (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.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (b) 4 — A triangular prism has 5 faces in total: 2 triangular ends and 3 rectangular faces. One rectangular face is the groundsheet, lying on the ground, so the other 5 − 1 = 4 faces are above the ground. A candidate who forgets one of the triangular ends when counting the remaining faces answers 3. A candidate who forgets to subtract the groundsheet at all answers 5. A candidate who only counts the two sloped rectangular faces, forgetting the two triangular ends, answers 2. The number of faces above the ground is 4.
- (c) 1/4 — A full turn at the centre of a circle is 360°, so a sector's fraction of the circle is its angle divided by 360°: 90 ÷ 360 = 1/4. 1/2 would be the fraction for a sector with an angle of 180°, not 90°. 3/4 is the fraction of the rest of the circle, the major sector left over from the 270° that is not part of this sector. 1/8 would be the fraction for a sector with an angle of 45°, half of 90°.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (c) 32 cm — Method: a perimeter is lengths added together, so it scales by the length scale factor itself, which is 4 ÷ 3 going from the smaller triangle to the larger one — not by its square. Working: 24 ÷ 3 = 8, and 8 × 4 = 32. Answer: 32 cm. The distractors: 18 cm comes from multiplying by 3 ÷ 4, scaling from the larger triangle down to the smaller one; 25 cm comes from adding the difference between the parts of the ratio, 4 − 3 = 1, to the perimeter; 8 cm comes from dividing by 3 and stopping there, before multiplying by 4.
- (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.
- (c) 52° — When two straight lines cross, the angles opposite each other at the crossing point are vertically opposite and are equal. Ray OQ is opposite ray OP and ray OS is opposite ray OR, so angle QOS is vertically opposite angle POR. Therefore angle QOS = 52°.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (a) segment — A chord cuts a circle into two regions, and each of these regions is called a segment. A sector is a different region, bounded by two radii and an arc, not by a chord. An arc is a curved part of the circumference, a length, not a region. A radius is a straight line from the centre to the edge, also a length, not a region.
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