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.
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Geometry and measures worksheet — GCSE Higher
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- 1.In triangle ABC, AB = 9 cm, BC = 6 cm and angle BAC = 35°. This description fits two different triangles. Work out the two possible sizes of angle ACB, each to 1 decimal place.
- 2.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = 108°. Work out the size of angle BCD.
- 3.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 4.A ferris wheel's circular frame has centre O. A support strut runs from O to a point A on the rim, and a second strut runs from O to point B on the rim, with angle AOB = 76°. A cabin is mounted at point C on the major arc AB, and D is a separate point on the minor arc AB. Work out the difference between angle ACB and angle ADB.
- 5.A sector of a circle has an angle of 90° and an arc length of 15.7 cm. Using π = 3.14, work out the radius of the circle.
- 6.A sector of a circle has radius 15 cm and angle 216°. Using π = 3.14, work out the arc length of the sector.
- 7.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?
- 8.A solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 9.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 10.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.
- 11.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 12.Two similar triangles have lengths in the ratio 5 : 8. A side of the smaller triangle is 6.5 cm. Work out the length of the corresponding side of the larger triangle.
- 13.A sector of a circle has an angle of 90° and an area of 78.5 cm². Using π = 3.14, work out the radius of the circle.
- 14.A quadrilateral has both pairs of opposite sides parallel and equal in length, but none of its angles are right angles. Which statement about this quadrilateral is correct?
- 15.A regular pentagon has centre O. Each vertex is 8 cm from O, and each of the 5 triangles formed by joining O to the vertices is isosceles with an angle of 72° at O. Work out the area of the pentagon. Give your answer to 1 decimal place.
Answer key
- (d) 59.4° and 120.6° — Method: two sides and an angle that is not between them can describe two triangles, because an acute angle and its supplement have the same sine. Use the sine rule for the angle, then test whether the supplement also fits inside 180°. Working: AB = 9 cm faces angle ACB and BC = 6 cm faces the 35° angle, so sin ACB = 9 × sin 35° ÷ 6 = 5.1622 ÷ 6 = 0.86036. The inverse sine of 0.86036 is 59.358°, and its supplement is 180° − 59.358° = 120.642°. Both survive the angle-sum test, since 35° + 59.4° = 94.4° and 35° + 120.6° = 155.6°, each less than 180°. Answer: angle ACB is 59.4° or 120.6° to 1 decimal place. The distractors: 22.5° and 157.5° come from putting the sides the wrong way up, 6 × sin 35° ÷ 9 = 0.38238; 30.6° and 149.4° come from pressing the inverse cosine key on 0.86036 instead of the inverse sine key; 59.4° and 85.6° are the two unknown angles of the first triangle, 59.358° and 180° − 35° − 59.358°, given by a candidate who has found one triangle and reported its angles rather than the two possible sizes of the same angle.
- (c) 72° — Method: opposite angles in a cyclic quadrilateral always sum to 180°. Working: angle DAB and angle BCD are opposite angles of the cyclic quadrilateral ABCD, so angle BCD = 180 − 108 = 72 degrees. Answer: 72°. Opposite angles in a cyclic quadrilateral are SUPPLEMENTARY, not equal, so do not simply copy the given angle; and do not subtract from 360°, which is the rule for angles round a point, or confuse this with the angle-in-a-semicircle theorem, which gives a fixed 90° that has nothing to do with this quadrilateral.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
- (b) 10 cm — Arc length = (angle ÷ 360) × 2 × π × r, so 15.7 = (90 ÷ 360) × 2 × 3.14 × r = 1.57r, giving r = 15.7 ÷ 1.57 = 10 cm. (2.5 cm comes from forgetting the angle fraction and dividing by 2π alone; 20 cm comes from leaving out the factor of 2 in the arc length formula before dividing; 5 cm comes from dividing the arc length by π alone, ignoring both the angle fraction and the factor of 2.)
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (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.
- (a) 615.44 cm² — Method: the surface area of a sphere is 4πr². Square the radius, multiply by π, then multiply by 4. Working: r² = 7² = 49, then 3.14 × 49 = 153.86, then 4 × 153.86 = 615.44. Answer: 615.44 cm². The distractors: 153.86 cm² comes from stopping at πr², which is the area of a flat circle of radius 7 cm and leaves out the factor of 4 that a curved surface needs; 87.92 cm² comes from 4 × 3.14 × 7, using the radius itself where the formula asks for its square; 1436.03 cm³ comes from working out the volume of the ball with (4 ÷ 3) × π × r³ instead of its surface area, which is a different measure and carries a cubic unit.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (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) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (a) 10.4 cm — The scale factor from the smaller triangle to the larger triangle is 8 ÷ 5 = 1.6, so the larger side is 6.5 × 1.6 = 10.4 cm. The distractor 4.0625 cm comes from using the ratio the wrong way round, 6.5 × 5 ÷ 8 = 4.0625. The distractor 9.5 cm comes from adding the difference between the ratio numbers (8 − 5 = 3) to the given length, 6.5 + 3 = 9.5. The distractor 13 cm comes from doubling the given length, treating the scale factor as 2 instead of 1.6.
- (a) 10 cm — Sector area is (angle ÷ 360) × π × radius². Here 90 ÷ 360 = 1/4, and 1/4 × 3.14 = 0.785, so radius² is 78.5 ÷ 0.785 = 100, and the radius is √100 = 10 cm. Stopping after finding radius² and not taking the square root gives 100 cm. Treating 78.5 as the area of the WHOLE circle, ignoring the 90° fraction, gives radius² = 78.5 ÷ 3.14 = 25, so a radius of 5 cm. Correctly finding a radius of 10 cm but then doubling it, mistaking the question for asking the diameter, gives 20 cm.
- (c) It is a parallelogram but not a rectangle — Method: match the given properties to the definition of each named quadrilateral. Working: having both pairs of opposite sides parallel and equal in length is exactly the definition of a parallelogram; since none of the angles are right angles, it cannot also be a rectangle, which needs four right angles. Options: 'must be a rectangle' wrongly assumes every parallelogram has right angles; 'must be a rhombus' wrongly assumes equal opposite sides means all four sides are equal, but only the opposite pairs are stated as equal here; 'trapezium' is wrong because a trapezium has exactly one pair of parallel sides, while this shape has two pairs, so it is a parallelogram and not a trapezium. Answer: it is a parallelogram but not a rectangle.
- (c) 152.2 cm² — Method: find the area of one of the 5 isosceles triangles with 1/2ab sin C, then multiply by 5 for the whole pentagon. Working: one triangle has area 1/2 × 8 × 8 × sin 72° = 30.4338 cm², so the pentagon's area = 5 × 30.4338 = 152.2 cm² — keep the unrounded triangle area, since 5 × 30.4 would give 152.0. Reporting the area of a single triangle and forgetting to multiply by 5 gives 30.4 cm²; multiplying by 6 instead of 5, as for a hexagon, gives 182.6 cm²; and leaving out the 1/2 from each triangle before multiplying by 5 gives 304.3 cm². A regular pentagon splits into exactly 5 triangles at its centre, each with a 72° angle, since 360° ÷ 5 = 72°.
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