Chords and Segment Perimeters | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Split the radius triangle in half

Join A and B to centre O. Since $OA=OB=r$ , triangle AOB is : it has two equal sides. Draw OM to meet AB at 90°. Symmetry halves both the chord and the central angle.

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Equal radii make two matching right triangles: AM = MB.

In right triangle OMA, OA is the hypotenuse, opposite the right angle. AM is opposite angle $\theta/2$ . The sine ratio is opposite divided by hypotenuse:

$\sin(\theta/2)=\frac{AM}{r}$ .

Multiply by $r$ , then double the half-chord: $AB=2r\sin(\theta/2)$ .

Worked example

Radius 7 cm and central angle 2 radians: find AB

The half-angle is 1 rad. In RAD mode, $AB=2(7)\sin1$ .

Answer

AB\approx11.8\text{ cm}

The diameter is 14 cm. A chord cannot be longer than the diameter, so this length is possible.

Work back from a chord

Divide by $2r$ to get $\sin(\theta/2)=AB/(2r)$ . The calculator's $\sin^{-1}$ function finds an angle from its sine; it does not mean $1/\sin$ .

For a chord of 6 cm in a circle of radius 5 cm:

$\theta=2\sin^{-1}(6/10)\approx1.29\text{ rad}$ .

This gives the smaller central angle because $0<\theta/2\le\pi/2$ . For the major angle, subtract it from $2\pi$ . A diameter gives $\theta=\pi$ .

A segment has a different boundary

A is the region between a chord and an arc. Its perimeter is arc + chord, not arc + two radii. The next lesson finds its area.

For radius 16 cm and minor angle 0.6 rad, the arc is 9.6 cm and the chord is $32\sin0.3$ cm. Their sum is approximately 19.1 cm. (9709/13 May/Jun 2023 Q6(b))

Your turn

A minor segment has radius 8 cm and central angle 1.4 rad. Find its perimeter. State what changes if the major segment is requested instead.

Show worked answer

Arc $=8(1.4)=11.2$ cm; chord $=16\sin0.7$ cm. The perimeter is approximately 21.5 cm. For the major segment, use arc $8(2\pi-1.4)$ ; the chord stays the same.