Arcs, Sectors and Segments

Use one quarter of the full circumference.

1. The full circumference is 2\pi\times12=24\pi\text{ cm}.
2. The arc length is \frac14\times24\pi=6\pi\text{ cm}.

Use the sector angle as a fraction of a full turn.

1. The sector is \frac{60}{360}=\frac16 of a circle.
2. Its area is \frac16\pi\times6^2=6\pi\text{ cm}^2.

Include the two radii as well as the arc.

1. The arc length is \frac{90}{360}\times2\pi\times4=2\pi\text{ cm}.
2. The two straight radii total 8\text{ cm}.
3. The perimeter is (8+2\pi)\text{ cm}.

Use the 240^\circ angle as a fraction of a full turn.

1. The major arc is \frac{240}{360}=\frac23 of the circumference.
2. Its length is \frac23\times2\pi\times3=4\pi\text{ cm}.

Multiply the full circle area by the fraction of a turn.

1. The area is \frac{72}{360}\pi\times5^2.
2. This is \frac15\times25\pi=5\pi\text{ cm}^2.

Compare the arc length with the full circumference.

1. The full circumference is 18\pi\text{ cm}.
2. The arc is \frac{3\pi}{18\pi}=\frac16 of a circle.
3. The angle is \frac16\times360^\circ=60^\circ.

Write the arc formula with an unknown radius.

1. The arc equation is \frac{120}{360}\times2\pi r=4\pi.
2. This simplifies to \frac23r=4.
3. The radius is r=6\text{ cm}.

Use the sector area equation to find the square of the radius.

1. The area equation is \frac{72}{360}\pi r^2=20\pi.
2. Thus r^2=100.
3. The radius is positive, so r=10\text{ cm}.

Find the fraction of the full circle area occupied by the sector.

1. The full circle area is 36\pi\text{ cm}^2.
2. The sector is \frac{9\pi}{36\pi}=\frac14 of the circle.
3. The angle is \frac14\times360^\circ=90^\circ.

Subtract the two radii before using the arc length.

1. The perimeter of a sector is the arc plus two radii, so the arc length is (24+10\pi)-2\times12=10\pi\text{ cm}.
2. The full circumference is 2\pi\times12=24\pi\text{ cm}, so the arc is \frac{10\pi}{24\pi}=\frac{5}{12} of the circle.
3. The angle is \frac{5}{12}\times360^\circ=150^\circ.

Subtract the matching small sector from the large sector.

1. Area of a sector =\frac{\theta}{360}\times\pi r^2.
2. The large sector has area \frac{90}{360}\times\pi\times8^2=16\pi\text{ cm}^2.
3. The small sector has area \frac{90}{360}\times\pi\times4^2=4\pi\text{ cm}^2.
4. The shaded area is 16\pi-4\pi=12\pi\text{ cm}^2.

The shorter arc occupies one of the six equal parts of the circumference.

1. The circumference is 2\pi\times18=36\pi\text{ m}.
2. There are 5+1=6 ratio parts.
3. The shorter arc length is 36\pi\div6=6\pi\text{ m}.

Use Pythagoras to find the radius squared; a square root is not needed for the area.

1. The radius squared is r^2=4^2+4^2=32.
2. The area is \frac{135}{360}\pi\times32.
3. This simplifies to 12\pi\text{ cm}^2.

Subtract the triangle formed by the radii from the sector.

1. The quarter-sector area is \frac14\pi\times6^2=9\pi\text{ cm}^2.
2. The right-angled triangle area is \frac12\times6\times6=18\text{ cm}^2.
3. The segment area is (9\pi-18)\text{ cm}^2.

Use sector area minus triangle area, keeping full precision until the end.

1. Segment area = sector area - area of the triangle made by the two radii and the chord.
2. The sector area is \frac{120}{360}\times\pi\times8^2=67.020\ldots\text{ cm}^2.
3. The triangle has two sides of 8\text{ cm} with 120^\circ between them, so use area =\frac12ab\sin C: \frac12\times8\times8\times\sin120^\circ=27.712\ldots\text{ cm}^2.
4. The segment area is 67.020\ldots-27.712\ldots=39.307\ldots\text{ cm}^2.
5. To 3 significant figures, this is 39.3\text{ cm}^2.

Split triangle OAB into two right-angled triangles to find angle AOB.

1. Triangle OAB is isosceles because OA=OB=9\text{ cm}. The line from O to the midpoint M of AB is perpendicular to AB, so it makes two right-angled triangles with AM=7\text{ cm}.
2. In triangle OAM, OA=9 is the hypotenuse and AM=7 is opposite angle AOM, so \sin(\angle AOM)=\frac79.
3. \angle AOM=\sin^{-1}\left(\frac79\right)=51.057\ldots^\circ, so \angle AOB=2\times51.057\ldots=102.115\ldots^\circ.
4. Arc length =\frac{\theta}{360}\times2\pi r=\frac{102.115\ldots}{360}\times2\pi\times9=16.040\ldots\text{ cm}.
5. To 3 significant figures, the arc is 16.0\text{ cm}.

The segment boundary consists of its arc and its chord.

1. The arc length is \frac14\times2\pi\times10=5\pi\text{ cm}.
2. The chord is the hypotenuse of a right triangle with two sides of 10\text{ cm}.
3. Its length is \sqrt{10^2+10^2}=10\sqrt2\text{ cm}.
4. The segment perimeter is (5\pi+10\sqrt2)\text{ cm}.

Take away the two radii to find the arc length, then use the arc length to find the angle.

1. The perimeter of a sector is the arc plus two radii, so the arc length is 30-2\times9=12\text{ cm}.
2. Arc length =\frac{\theta}{360}\times2\pi r, so \frac{\theta}{360}\times2\pi\times9=12.
3. So \theta=\frac{12\times360}{18\pi}=76.39\ldots^\circ.
4. Area of the sector =\frac{\theta}{360}\times\pi r^2=\frac{76.39\ldots}{360}\times\pi\times81=54\text{ cm}^2 (keep the full value of \theta in your calculator).
5. To 3 significant figures, the area is 54.0\text{ cm}^2.

The triangle between the two radii is equilateral.

1. Segment area = sector area - triangle area.
2. The sector is \frac{60}{360}=\frac16 of the circle, so its area is \frac16\pi\times6^2=6\pi\text{ cm}^2.
3. The triangle has two sides of 6\text{ cm} (radii) with 60^\circ between them, so its other two angles are also 60^\circ: it is equilateral with side 6\text{ cm}.
4. Its height splits the base in half; by Pythagoras the height is \sqrt{6^2-3^2}=\sqrt{27}=3\sqrt3\text{ cm}, so its area is \frac12\times6\times3\sqrt3=9\sqrt3\text{ cm}^2.
5. The segment area is (6\pi-9\sqrt3)\text{ cm}^2.

Use the area to find what fraction of a full turn the angle is.

1. Let the angle be \theta. The shape's area is the large sector minus the small sector: \frac{\theta}{360}\times\pi\times12^2-\frac{\theta}{360}\times\pi\times8^2=\frac{\theta}{360}\times80\pi.
2. So \frac{\theta}{360}\times80\pi=30\pi, which gives \frac{\theta}{360}=\frac{30}{80}=\frac38 (so \theta=135^\circ).
3. The outer arc is \frac38\times2\pi\times12=9\pi\text{ cm} and the inner arc is \frac38\times2\pi\times8=6\pi\text{ cm}.
4. Each straight edge is 12-8=4\text{ cm}, so the two straight edges total 8\text{ cm}.
5. The perimeter is 9\pi+6\pi+8=(15\pi+8)\text{ cm}.