Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Geometry and measures worksheet — GCSE Higher
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- 1.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 2.A ramp rises at 30° to the horizontal. Its sloping surface is 4.8 m long. Safety rules say the vertical rise of a ramp must be no more than 2.5 m. Work out how far below that limit the rise of this ramp is.
- 3.A sector of a circle has radius 6 cm and an arc length of 15.7 cm. Using π = 3.14, work out the angle of the sector.
- 4.In a right-angled triangle, the side opposite angle θ is 6 cm and the hypotenuse is 10 cm. Work out the size of angle θ. Give your answer correct to 1 decimal place.
- 5.A path goes from A(1, 1) to B(1, 5), then from B to C(6, 5). Work out the total length of the path from A to C.
- 6.Triangle T has a vertex A at (3, 1). It is enlarged by a scale factor of −2, centre (1, 1). Work out the coordinates of the image of point A.
- 7.Triangle XYZ has angle X = 90°. Angle Y = 34°. Work out angle Z.
- 8.Triangle T has a vertex at (8, 4). It is enlarged by a scale factor of 1/2, centre (2, 4). Work out the coordinates of the image of this vertex.
- 9.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 10.From an external point P, two tangents PA and PB touch a circle with centre O at A and B. Angle AOB = 112°. Work out the size of angle APB.
- 11.A student is asked to rearrange the cosine rule a² = b² + c² − 2bc cos A to make cos A the subject. They write: cos A = (a² − b² − c²) / (2bc). Is the student's rearrangement correct?
- 12.A quadrilateral has all four sides equal in length and all four interior angles equal to 90°. Its diagonals are equal in length and bisect each other at right angles. Which quadrilateral matches this description exactly?
- 13.Three of these are rational numbers and one is irrational. Write down the one that is irrational.
- 14.A patio plan states: 'Side AB of the patio is parallel to side DC, and AB is longer than DC. Sides BC and AD are not parallel to each other and are not equal in length.' The four sides measure AB = 5.1 m, BC = 3.2 m, DC = 2.8 m and AD = 4.5 m. The tiler needs to order edging strip only for the sides that are not parallel to any other side. What total length of edging strip, in metres, should the tiler order?
- 15.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
Answer key
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
- (a) 150° — The arc length is the same fraction of the circumference as the angle is of 360°. The full circumference is 2 × 3.14 × 6 = 37.68 cm, so the angle is 15.7 ÷ 37.68 × 360 = 150°. Treating the 6 cm as a diameter instead of a radius gives a circumference of 18.84 cm and an angle of 300°. Using π = 3 instead of the given 3.14 gives a circumference of 36 cm and an angle of 157°. Treating the fraction 15.7 ÷ 37.68 as being out of 100 rather than 360 gives about 42°.
- (d) 36.9° — sin θ = opposite/hypotenuse = 6/10 = 0.6, so θ = sin⁻¹(0.6) = 36.86...° ≈ 36.9°. "53.1°" finds the OTHER acute angle in the triangle, 90° − 36.9°, instead of θ itself, as if the two acute angles had been swapped. "31.0°" comes from using the tangent ratio instead of sine, working out tan⁻¹(6/10) = 31.0° with the wrong ratio for the two sides given. "36.8°" rounds sin⁻¹(0.6) = 36.86...° down to 36.8° instead of correctly rounding it up to 36.9°.
- (a) 9 — AB is a vertical segment, since A and B share the x-coordinate 1, and its length is the difference in y-coordinates: 5 − 1 = 4. BC is a horizontal segment, since B and C share the y-coordinate 5, and its length is the difference in x-coordinates: 6 − 1 = 5. The total path length is 4 + 5 = 9. 20 comes from multiplying the two lengths, 4 × 5, instead of adding them. 5 is only the length of BC, forgetting to include AB. 4 is only the length of AB, forgetting to include BC.
- (c) (−3, 1) — Method: for an enlargement about a centre, first find the vector from the centre to the point, multiply it by the scale factor, INCLUDING its sign, then add the result back onto the centre. Working: the vector from the centre (1, 1) to A(3, 1) is (3 − 1, 1 − 1) = (2, 0). Multiplying by the scale factor −2 gives −2 × 2 = −4 and −2 × 0 = 0, so the scaled vector is (−4, 0). Adding this to the centre gives 1 + (−4) = −3 and 1 + 0 = 1, so the image is (−3, 1). Answer: (−3, 1). A NEGATIVE scale factor keeps its sign all the way through the calculation: do not treat −2 as +2, and do not treat it as a fraction like 1/2, which is the rule for a scale factor between 0 and 1, not a negative one. Always measure the vector from the CENTRE of enlargement, never from the origin, unless the two happen to coincide.
- (a) 56° — The angles of a triangle add up to 180°. Add the two known angles: 90° + 34° = 124°. Subtract from 180°: 180° − 124° = 56°.
- (a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (a) 68° — Method: a tangent meets a radius at 90°, so quadrilateral OAPB has two right angles at A and B; its four angles sum to 360°, so angle APB = 360° − 90° − 90° − angle AOB. Working: angle APB = 360° − 90° − 90° − 112° = 68°. Halving angle AOB, confusing this with the angle-at-centre theorem for a chord, gives 56°; assuming the angle between two tangents is always 90° (true only for the angles at A and B, not at P) gives 90°; and assuming angle APB simply equals angle AOB by symmetry gives 112°. The two right angles at A and B, from the tangent-radius property, are what link angle AOB and angle APB.
- (b) No — should divide by −2bc, not +2bc; sign is wrong. — Method: rearrange a² = b² + c² − 2bc cos A step by step and compare with the student's version. Working: subtracting b² + c² from both sides gives a² − b² − c² = −2bc cos A, then dividing both sides by −2bc gives cos A = (a² − b² − c²) / (−2bc), which is the same as cos A = (b² + c² − a²) / (2bc). The student divided by +2bc instead of −2bc, so their expression is the negative of the correct one — the verdict is No. Saying the algebra is fine because 'either sign order gives a valid result' ignores that only one of the two signed expressions matches the original equation. Saying the denominator should be bc rather than 2bc is a different, unrelated error — the coefficient 2bc in the original formula is correct and must stay.
- (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.
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (a) 7.7 m — AB is parallel to DC, so those two sides are each parallel to another side. BC and AD are stated to be not parallel to each other, so neither one is parallel to any other side — these are the two sides that need edging. Adding these: 3.2 + 4.5 = 7.7 m, so 7.7 m is correct. 7.6 m comes from an arithmetic slip when adding 3.2 and 4.5. 15.4 m comes from doubling the correct total, mistakenly assuming edging strip is needed along both faces of each side. 4.5 m comes from using only the longer of the two non-parallel sides and forgetting to add the shorter one.
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
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