Area of a Sector

Worked example

1. 1

   $\pi \times 12^2=144\pi$ — Start with the whole circle: $\pi r^2$ with r = 12. Square only the 12: $12^2=144$.

2. 2

   $\frac{110}{360}=\frac{11}{36}$ — Put the angle over 360, the angle of a full turn. This is the fraction of the circle you keep.

3. 3

   $\frac{11}{36}\times 144\pi =44\pi$ — Take that fraction of the circle: 144 ÷ 36 = 4, and 4 × 11 = 44. This is the exact answer.

4. 4

   $44\pi =138{\text{ cm}}^2$ — Only now press π: 44 × 3.14159… = 138.23…, which is 138 cm² to 3 s.f. If the question says “in terms of π”, stop at 44π.

Worked example

1. 1

   $360-45=315$ — The small angle is 45°. Angles round the centre add up to 360°, so the big angle is 315°.

2. 2

   $\frac{315}{360}=\frac{7}{8}$ — Put the big angle over 360. Both divide by 45, so the major sector is seven eighths of the circle.

3. 3

   $\pi \times 8^2=64\pi$ — The whole circle with radius 8: $8^2=64$.

4. 4

   $\frac{7}{8}\times 64\pi =56\pi {\text{ cm}}^2$ — Take seven eighths: 64 ÷ 8 = 8, and 8 × 7 = 56. If the question says “in terms of π”, this is the answer.

5. 5

   $56\pi =176{\text{ cm}}^2$ — For a decimal answer: 56 × 3.14159… = 175.9…, which is 176 cm² to 3 s.f.

Worked example

1. 1

   $R=5+4=9$ — The small radius is 5 and the band is 4 wide. The big radius runs from the centre to the outer arc, so add them.

2. 2

   $\frac{120}{360}\times \pi \times 9^2=27\pi$ — The big sector: 120° is a third of 360°, and a third of 81π is 27π.

3. 3

   $\frac{120}{360}\times \pi \times 5^2=\frac{25\pi }{3}$ — The small sector, same angle: a third of 25π.

4. 4

   $27\pi -\frac{25\pi }{3}=\frac{81\pi -25\pi }{3}$ — Take the small sector from the big one. Write 27π as 81π ÷ 3 so both have the same bottom.

5. 5

   $=\frac{56\pi }{3}{\text{ cm}}^2$ — 81 − 25 = 56. This is the exact answer. As a decimal it is 58.6 cm² to 3 s.f.

Worked example

1. 1

   $\frac{80}{360}\times \pi \times 10^2$ — The sector: radius 10, angle 80°. Write it out first so the method shows.

2. 2

   $=\mathrm{69.81...}$ — Work it out and keep all the digits on your calculator. This is the sector area.

3. 3

   $\frac{1}{2}\times 10\times 10\times \sin 80^{\circ }$ — The triangle: half × side × side × sine of the angle between them. Both sides are the radius, 10.

4. 4

   $=\mathrm{49.24...}$ — Work it out, again keeping the digits. This is the triangle area.

5. 5

   $\mathrm{69.81...}-\mathrm{49.24...}$ — The segment is what is left of the sector once the triangle is cut off, so subtract.

6. 6

   $=20.6{\text{ cm}}^2$ — 69.81… − 49.24… = 20.57…, which is 20.6 cm² to 3 s.f. Round only at the end.

Worked example

1. 1

   $\frac{70}{360}\times \pi r^2=30$ — Put in what you know: angle 70°, area 30 cm². The radius is the unknown.

2. 2

   $70\pi r^2=10800$ — Multiply both sides by 360 to clear the fraction: 30 × 360 = 10800.

3. 3

   $r^2=\frac{10800}{70\pi }=\mathrm{49.11...}$ — Divide both sides by 70π, so $r^2$ is on its own. Keep all the digits.

4. 4

   $r=\sqrt{\mathrm{49.11...}}=7.01\text{ cm}$ — Square root last, because the radius was squared. 7.007… is 7.01 cm to 3 s.f.

Worked example

1. 1

   $\frac{x}{360}\times 100\pi =20\pi$ — Radius 10, so the whole circle is $10^2\pi =100\pi$. The area is 20π.

2. 2

   $\frac{x}{360}=\frac{20\pi }{100\pi }=\frac{1}{5}$ — Divide both sides by 100π. The π cancels, so the sector is a fifth of the circle.

3. 3

   $x=\frac{1}{5}\times 360=72^{\circ }$ — Multiply both sides by 360: a fifth of a full turn is 72°.

Worked example

1. 1

   $\frac{100}{360}\pi r^2-\frac{1}{2}{r}^2\sin 100^{\circ }=40$ — Write sector minus triangle and set it equal to the area, 40 cm². The angle is 100°.

2. 2

   $r^2\left(\frac{100}{360}\pi -\frac{1}{2}\sin 100^{\circ }\right)=40$ — Both terms have r², so factorise it out. The bracket is now just numbers.

3. 3

   $\frac{100}{360}\pi -\frac{1}{2}\sin 100^{\circ }=\mathrm{0.3802...}$ — Work out the bracket: 0.8726… − 0.4924…. Keep all the digits on your calculator.

4. 4

   $r^2=\frac{40}{\mathrm{0.3802...}}=\mathrm{105.19...}$ — Divide both sides by the bracket, so $r^2$ is on its own.

5. 5

   $r=\sqrt{\mathrm{105.19...}}=10.3\text{ cm}$ — Square root, because the radius was squared. 10.256… is 10.3 cm to 3 s.f. Only the positive root counts: a length cannot be negative.

Worked example

1. 1

   $\frac{90}{360}\pi (r+4{)}^2=\pi r^2$ — A quarter sector of radius r + 4 has the same area as a circle of radius r. Write each area and set them equal.

2. 2

   $\frac{1}{4}(r+4{)}^2=r^2$ — 90 out of 360 is a quarter. π is on both sides, so divide both sides by π.

3. 3

   $(r+4{)}^2=4r^2$ — Multiply both sides by 4 to clear the fraction.

4. 4

   $r+4=2r$ — Square root both sides: $\sqrt{4r^2}=2r$. The other root, r + 4 = −2r, gives a negative radius, so reject it.

5. 5

   $r=4$ — Take r from both sides: 4 = r. Check: a quarter of $\pi \times 8^2$ is 16π, and $\pi \times 4^2$ is 16π too.