GCSE Maths: Segments, algebraic forms and exact answers (Higher)
Part of the GCSE Maths perimeter, area and circles chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Segments, algebraic forms and exact answers
A segment is the region between a chord and an arc. To find its area, start with the sector that has the same two radii, then take away the triangle formed by the two radii and the chord:
Area of segment = area of sector − area of triangle
The triangle has two sides equal to the radius r and the angle θ between them, so its area is ½ab sin C = ½r2 sin θ (this area formula is covered fully in Chapter 20). For a calculator answer, use your calculator in degree mode. For an exact answer you need the exact values of sin: sin 30° = ½, sin 60° = √3/2, sin 90° = 1, sin 120° = √3/2, sin 150° = ½.
Exact answers with π and surds
An exact answer keeps π and square roots (surds, Chapter 6) as symbols instead of rounding: for example 6π − 9√3 cm². Terms with π, terms with √3 and plain numbers cannot be merged, so leave them separate. A regular hexagon splits into 6 equilateral triangles, so its area contains √3, and a square inside a circle makes the diagonal equal to the diameter.
Algebraic forms
Sometimes lengths are written with letters. Treat them like numbers: write the area or perimeter as an expression, simplify (collect like terms, remember (3x)2 = 9x2), and, if you are told the area, solve for x. When asked to “show that”, finish on the exact expression you were given.
Key ideas
- Segment area = sector area − triangle area (½r2 sin θ).
- Exact values: sin 30° = ½, sin 60° = √3 ÷ 2, sin 90° = 1.
- Keep π and √ in the answer for an exact value; do not merge unlike terms.
- A regular hexagon = 6 equilateral triangles; an equilateral triangle of side a has area (√3/4)a2.
- Algebraic shapes: write the area or perimeter as an expression, simplify, then solve.
- Perimeter of a segment = arc length + chord length.
Watch out
Do not subtract the triangle by using the sector area minus the radius or similar. The triangle area needs ½r2 sin θ. And never round the sector or triangle area before you subtract.Method. A segment is the part of a sector left when you cut off the triangle formed by the two radii and the chord. Find the sector area, find the triangle area using ½ab sin C, then subtract.
- Area of the sector. Use (θ ÷ 360) × πr2.16 × π × 82 = 33.5103… cm²
- Area of the triangle. The two sides of length 8 cm are radii and the angle between them is 60°. Use area = ½ × a × b × sin C.½ × 8 × 8 × sin 60°
= ½ × 64 × 0.8660… = 27.7128… cm² - Subtract. The segment is the sector without the triangle.33.5103… − 27.7128… = 5.79751… cm²
- Round to 2 decimal places. Round only at the end.segment area = 5.80 cm²
Check. The segment must be smaller than the sector (33.51) and positive. Rough check with π ≈ 3 and sin 60° ≈ 0.9: 32 − 28.8 ≈ 3.2. ✓
Watch out. Make sure your calculator is in degree mode before using sin. And do not round the sector or the triangle before subtracting.
Method. Find the sector area (keeping π), then the triangle area (keeping the exact value of sin), and subtract.
- Area of the sector. Use (θ ÷ 360) × πr2 and leave π as a symbol.16 × π × 62 = 16 × 36π = 6π cm²
- Area of the triangle. Two sides are radii (6 cm) and the angle between them is 60°, so use ½ab sin C. The exact value is sin 60° = √32.½ × 6 × 6 × √32 = ½ × 36 × √32 = 9√3 cm²
- Subtract. Segment = sector − triangle. The two parts cannot be merged, so leave the answer as a difference.(6π − 9√3) cm²
Check. As a decimal this is about 3.3 cm², which is positive and smaller than the sector (18.8 cm²). ✓
Watch out. Do not subtract the numbers inside the π term: π and a plain number (or a surd) are different kinds of term.
Method. The shaded area is the circle minus the square. The square can be split into four right-angled triangles whose short sides are radii.
- Area of the circle. A = πr2.π × 52 = 25π cm²
- Area of the square. The four triangles each have two sides of 5 cm at a right angle at the centre.one triangle: ½ × 5 × 5 = 12.5 cm²
four triangles: 4 × 12.5 = 50 cm² - Subtract. Shaded area = circle − square. π and a plain number cannot be merged.(25π − 50) cm²
Check. As a decimal: 78.54 − 50 = 28.54 cm², which is positive and smaller than the circle. ✓
Watch out. The diagonal of the square is the diameter, so the square is NOT 5 by 5.
Method. Split the hexagon into 6 identical triangles meeting at the centre, find the area of one with ½ab sin C, then multiply by 6.
- Find the angle at the centre. The 6 triangles fill a full turn.360° ÷ 6 = 60°. Each triangle has two equal sides and a 60° angle, so it is equilateral with side 6 cm.
- Area of one triangle. Use ½ × 6 × 6 × sin 60° and sin 60° = √3 ÷ 2.½ × 6 × 6 × (√3 ÷ 2) = 18 × (√3 ÷ 2) = 9√3 cm²
- Multiply by 6. There are six triangles.6 × 9√3 = 54√3 cm²
Check. As a decimal 54 × 1.732 = 93.5 cm², about 2.6 times the 36 cm² square on one side. ✓
Watch out. Keep √3 as a symbol: do not turn it into a decimal in an exact question.
Method. Use sector area = (θ/360) × πr2 with r = 3x and θ = 120.
- Write the fraction of the circle. 120/360 simplifies.120 ÷ 360 = ⅓
- Substitute the radius. Square the whole radius, including the number: (3x)2 = 9x2.Area = ⅓ × π × (3x)2 = ⅓ × π × 9x2
- Simplify. ⅓ of 9 is 3.3πx2 cm²
Check. With x = 1: radius 3, area = ⅓ × 9π = 3π. ✓
Watch out. (3x)2 is 9x2, not 3x2.
Exercise 18I Segments, algebraic forms and exact answers
Calculator allowedMethod. A segment is the part of a sector left when you cut off the triangle formed by the two radii and the chord. Find the sector area, find the triangle area using ½ab sin C, then subtract.
- Area of the sector. Use (θ ÷ 360) × πr2.16 × π × 82 = 33.5103… cm²
- Area of the triangle. The two sides of length 8 cm are radii and the angle between them is 60°. Use area = ½ × a × b × sin C.½ × 8 × 8 × sin 60°
= ½ × 64 × 0.8660… = 27.7128… cm² - Subtract. The segment is the sector without the triangle.33.5103… − 27.7128… = 5.79751… cm²
- Round to 2 decimal places. Round only at the end.segment area = 5.80 cm²
Check. The segment must be smaller than the sector (33.51) and positive. Rough check with π ≈ 3 and sin 60° ≈ 0.9: 32 − 28.8 ≈ 3.2. ✓
Watch out. Make sure your calculator is in degree mode before using sin. And do not round the sector or the triangle before subtracting.
Method. A segment is the part of a sector left when you cut off the triangle formed by the two radii and the chord. Find the sector area, find the triangle area using ½ab sin C, then subtract.
- Area of the sector. Use (θ ÷ 360) × πr2.14 × π × 102 = 78.5398… cm²
- Area of the triangle. The two sides of length 10 cm are radii and the angle between them is 90°. Use area = ½ × a × b × sin C.½ × 10 × 10 × sin 90°
= ½ × 100 × 1.0000… = 50.0000… cm² - Subtract. The segment is the sector without the triangle.78.5398… − 50.0000… = 28.5398… cm²
- Round to 1 decimal place. Round only at the end.segment area = 28.5 cm²
Check. The segment must be smaller than the sector (78.54) and positive. Rough check with π ≈ 3 and sin 90° ≈ 1.0: 75 − 50.0 ≈ 25.0. ✓
Watch out. Make sure your calculator is in degree mode before using sin. And do not round the sector or the triangle before subtracting.
Method. A segment is the part of a sector left when you cut off the triangle formed by the two radii and the chord. Find the sector area, find the triangle area using ½ab sin C, then subtract.
- Area of the sector. Use (θ ÷ 360) × πr2.13 × π × 62 = 37.6991… cm²
- Area of the triangle. The two sides of length 6 cm are radii and the angle between them is 120°. Use area = ½ × a × b × sin C.½ × 6 × 6 × sin 120°
= ½ × 36 × 0.8660… = 15.5885… cm² - Subtract. The segment is the sector without the triangle.37.6991… − 15.5885… = 22.11065… cm²
- Round to 2 decimal places. Round only at the end.segment area = 22.11 cm²
Check. The segment must be smaller than the sector (37.70) and positive. Rough check with π ≈ 3 and sin 120° ≈ 0.9: 36 − 16.2 ≈ 19.8. ✓
Watch out. Make sure your calculator is in degree mode before using sin. And do not round the sector or the triangle before subtracting.
Method. Segment = sector − triangle. Find each with your calculator in degree mode, keep full values, and subtract.
- Area of the sector. (θ/360) × πr2.75360 × π × 122 = 94.2478… cm²
- Area of the triangle. ½ × 12 × 12 × sin 75°.72 × 0.9659… = 69.5467… cm²
- Subtract. Sector minus triangle.94.2478… − 69.5467… = 24.7011…
- Round to 3 s.f. Look at the 4th significant figure.24.7 cm²
Check. The segment must be smaller than the sector (94.2 cm²) and positive. ✓
Watch out. Check your calculator is in degree mode, and do not round the sector or triangle early.
Method. Find the sector area (keeping π), then the triangle area (keeping the exact value of sin), and subtract.
- Area of the sector. Use (θ ÷ 360) × πr2 and leave π as a symbol.16 × π × 62 = 16 × 36π = 6π cm²
- Area of the triangle. Two sides are radii (6 cm) and the angle between them is 60°, so use ½ab sin C. The exact value is sin 60° = √32.½ × 6 × 6 × √32 = ½ × 36 × √32 = 9√3 cm²
- Subtract. Segment = sector − triangle. The two parts cannot be merged, so leave the answer as a difference.(6π − 9√3) cm²
Check. As a decimal this is about 3.3 cm², which is positive and smaller than the sector (18.8 cm²). ✓
Watch out. Do not subtract the numbers inside the π term: π and a plain number (or a surd) are different kinds of term.
Method. Find the sector area (keeping π), then the triangle area (keeping the exact value of sin), and subtract.
- Area of the sector. Use (θ ÷ 360) × πr2 and leave π as a symbol.14 × π × 102 = 14 × 100π = 25π cm²
- Area of the triangle. Two sides are radii (10 cm) and the angle between them is 90°, so use ½ab sin C. The exact value is sin 90° = 1.½ × 10 × 10 × 1 = ½ × 100 × 1 = 50 cm²
- Subtract. Segment = sector − triangle. The two parts cannot be merged, so leave the answer as a difference.(25π − 50) cm²
Check. As a decimal this is about 28.5 cm², which is positive and smaller than the sector (78.5 cm²). ✓
Watch out. Do not subtract the numbers inside the π term: π and a plain number (or a surd) are different kinds of term.
Method. Find the sector area (keeping π), then the triangle area (keeping the exact value of sin), and subtract.
- Area of the sector. Use (θ ÷ 360) × πr2 and leave π as a symbol.13 × π × 62 = 13 × 36π = 12π cm²
- Area of the triangle. Two sides are radii (6 cm) and the angle between them is 120°, so use ½ab sin C. The exact value is sin 120° = √32.½ × 6 × 6 × √32 = ½ × 36 × √32 = 9√3 cm²
- Subtract. Segment = sector − triangle. The two parts cannot be merged, so leave the answer as a difference.(12π − 9√3) cm²
Check. As a decimal this is about 22.1 cm², which is positive and smaller than the sector (37.7 cm²). ✓
Watch out. Do not subtract the numbers inside the π term: π and a plain number (or a surd) are different kinds of term.
Method. Find the sector area (keeping π), then the triangle area (keeping the exact value of sin), and subtract.
- Area of the sector. Use (θ ÷ 360) × πr2 and leave π as a symbol.512 × π × 122 = 512 × 144π = 60π cm²
- Area of the triangle. Two sides are radii (12 cm) and the angle between them is 150°, so use ½ab sin C. The exact value is sin 150° = 12.½ × 12 × 12 × 12 = ½ × 144 × 12 = 36 cm²
- Subtract. Segment = sector − triangle. The two parts cannot be merged, so leave the answer as a difference.(60π − 36) cm²
Check. As a decimal this is about 152.5 cm², which is positive and smaller than the sector (188.5 cm²). ✓
Watch out. Do not subtract the numbers inside the π term: π and a plain number (or a surd) are different kinds of term.
Method. Circle minus square. Split the square into four right-angled triangles with two radii as the short sides.
- Area of the circle. A = πr2.25π cm²
- Area of the square. Each of the four triangles has area ½ × 5 × 5.4 × 12.5 = 50 cm²
- Subtract. Circle minus square.(25π − 50) cm²
Check. 28.54 cm² as a decimal, which is positive. ✓
Watch out. The square’s diagonal is the diameter. Its side is not 5 cm.
Method. The shaded corners are the square with the circle taken away. The circle touches all four sides, so its diameter equals the side of the square.
- Find the radius of the circle. The circle touches opposite sides, so its diameter is the side of the square.radius = 12 ÷ 2 = 6 cm
- Area of the square. side × side.12 × 12 = 144 cm²
- Area of the circle. The diameter is 12 cm, so the radius is half of it.π × 62 = π × 36 = 36π cm²
- Combine the pieces. Subtract the parts that are taken away. Whole numbers go with whole numbers and the π terms go with π terms. A number and a π term cannot be merged into one, so leave them as a sum.Area = 144 − 36π = (144 − 36π) cm²
Check. Estimate with π ≈ 3: the answer should be about 36 cm², which is close to the value found. ✓
Watch out. Radius is half the side of the square, not the whole side.
Method. Six equilateral triangles of side 6 cm make the hexagon. Find one triangle with ½ab sin C, then multiply by 6.
- Triangle shape. Each has a 60° angle at the centre and two equal sides, so it is equilateral.side 6 cm, angle 60°
- Area of one triangle. ½ × 6 × 6 × sin 60° with sin 60° = √3 ÷ 2.18 × (√3 ÷ 2) = 9√3
- Multiply by 6. Six triangles.6 × 9√3 = 54√3 cm²
Check. 54 × 1.732 = 93.5 cm². ✓
Watch out. Keep √3 exact.
Method. All three sides are equal.
- Write the sides. An equilateral triangle has three equal sides.8 cm, 8 cm, 8 cm
- Add the sides. Perimeter is the distance round the outside.8 + 8 + 8 = 24 cm
Check. 3 × 8 = 24. ✓
Method. Use area = ½ab sin C with two sides of 8 cm and the angle 60° between them.
- Write the formula. Equilateral triangles have all angles 60°.½ × 8 × 8 × sin 60°
- Use the exact value. sin 60° = √3 ÷ 2.= 32 × (√3 ÷ 2)
- Simplify. 32 ÷ 2 = 16.= 16√3 cm²
Check. 16 × 1.732 = 27.7 cm², about half of 8 × 6.9. ✓
Watch out. Keep the surd.
Method. Use (θ/360) × πr2 with r = 3x and θ = 120.
- Fraction of the circle. Simplify 120/360.⅓
- Substitute and square. (3x)2 = 9x2.⅓ × π × 9x2
- Simplify. ⅓ of 9 is 3.3πx2 cm²
Check. With x = 1: ⅓ × 9π = 3π. ✓
Watch out. Square the 3 as well: (3x)2 = 9x2.
Method. Find the area of the rectangle and of the semicircle in terms of x and add.
- Area of the rectangle. length × width.4x × 2x = 8x2
- Radius of the semicircle. The diameter is 2x.radius = x
- Area of the semicircle. ½πr2.½ × π × x2 = πx2 ÷ 2
- Add over a common denominator. 8x2 = 16x2 ÷ 2.16x2 ÷ 2 + πx2 ÷ 2 = (16 + π)x2 ÷ 2
Check. With x = 1: 8 + 1.57 = 9.57 and (16 + 3.1416) ÷ 2 = 9.57. ✓
Watch out. The radius is half the diameter, so it is x, not 2x.
Method. Add the arc length and the two radii, then combine over a common denominator.
- Arc length. (60/360) = ⅙ of 2πr.⅙ × 2πr = πr ÷ 3
- Two radii. Each is r.2r = 6r ÷ 3
- Add. Same denominator.(πr + 6r) ÷ 3 = r(π + 6) ÷ 3
Check. With r = 6: arc = 2π = 6.28, radii 12, total 18.28; 6 × 9.1416 ÷ 3 = 18.28. ✓
Watch out. Include both radii.
Method. The perimeter of a segment is the arc plus the chord. Find each.
- Arc length. (60/360) × 2π × 6.⅙ × 12π = 2π cm
- Chord length. The triangle has two sides of 6 cm and a 60° angle at the top, so its other two angles are equal and add to 120°: each is 60°. All three angles are 60°, so it is equilateral.chord = 6 cm
- Add. Perimeter = arc + chord.2π + 6 cm
Check. As a decimal: 6.28 + 6 = 12.28 cm. ✓
Watch out. The chord is the straight edge of the segment and counts in its perimeter.
Method. Sector minus triangle, both in terms of r.
- Area of the sector. (90/360) = ¼ of πr2.πr2 ÷ 4
- Area of the triangle. Two radii at a right angle: ½ × r × r (sin 90° = 1).r2 ÷ 2 = 2r2 ÷ 4
- Subtract over a common denominator. Same denominator 4.(πr2 − 2r2) ÷ 4 = r2(π − 2) ÷ 4
Check. With r = 10: 100 × 1.1416 ÷ 4 = 28.54, matching the segment worked out earlier. ✓
Watch out. Write both parts over the same denominator before subtracting.