E5.3 Circles, arcs and sectors

Syllabus
0580–2028–2029
Topic
E5.3
Level
Extended

Calculate with circumference and circle area

A circle is controlled by one length: its radius rr. The diameter crosses the centre from edge to edge, so d=2rd=2r. Circumference measures the boundary and area measures the enclosed surface.

C=2\pi r=\pi d \qquad A=\pi r^2

First decide whether the given length is a radius or diameter. Convert units before substituting, keep π\pi on the calculator until the end, and attach linear units to circumference or square units to area. For an inverse problem, set the formula equal to the known value: divide by 2π2\pi to recover a radius from circumference, or divide by π\pi and then take the positive square root to recover a radius from area.

A circle has diameter 1212 cm. Then r=6r=6 cm, so C=12πC=12\pi cm and A=36π cm2A=36\pi\text{ cm}^2. If instead its area is 150 cm2150\text{ cm}^2, then r=150/π=6.91r=\sqrt{150/\pi}=6.91\ldots cm. Use the requested accuracy; leave the answer in terms of π\pi when asked.

Do not square the diameter in A=πr2A=\pi r^2: halve it first. Circumference cannot be reported in square units, and area cannot be reported in linear units. Compound regions made from several circles belong to a later Topic.

Calculate arc length and sector area

A sector with central angle θ\theta is the fraction θ/360\theta/360 of a full circle. The same fraction scales the full circumference to an arc length and the full area to a sector area.

Quantity Full circle Sector with angle θ\theta Units
boundary along the curve 2πr2\pi r θ360×2πr\dfrac{\theta}{360}\times2\pi r length units
enclosed surface πr2\pi r^2 θ360×πr2\dfrac{\theta}{360}\times\pi r^2 square units

Use the angle at the centre. For a minor sector with angle θ\theta, use θ/360\theta/360; the matching major sector uses (360θ)/360(360-\theta)/360. Choose circumference for an arc or circle area for a sector, then multiply. If total sector perimeter is required, add the two radii to the arc length; those straight edges are not part of the arc.

For r=9r=9 cm and minor angle 140140^\circ, the minor arc length is 140360(2π×9)=7π\frac{140}{360}(2\pi\times9)=7\pi cm. The minor sector area is 140360(π×92)=63π2 cm2\frac{140}{360}(\pi\times9^2)=\frac{63\pi}{2}\text{ cm}^2. The major angle is 220220^\circ, so the major sector area is 220360(81π)=99π2 cm2\frac{220}{360}(81\pi)=\frac{99\pi}{2}\text{ cm}^2.

Do not use an angle on the circumference as the sector angle, and do not scale the radius alone: scale the completed circumference or area formula. A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc and belongs to a later Topic.