Fixed-Perimeter Sectors | CAIE 9709 Pure Mathematics 1 Notes

Chapters

Pure Mathematics 1

Mathematics9709

Use the wire length first

Worked example

A 24 cm wire forms a sector: express its area using r

The boundary contains an arc and two radii:

$r\theta+2r=24$ , so $r\theta=24-2r$ .

Swipe left or right to see the whole diagram →

The sector has two straight radii and one curved arc.

Write the area so that it contains this known combination:

$A=\frac12r(r\theta)=\frac12r(24-2r)$ .

Answer

A=12r-r^2

Both radius and arc must be positive. Also check $0<\theta<2\pi$ for an ordinary sector without a full turn.

Find the greatest possible area

Complete the square, as in Quadratics. Since $(r-6)^2=r^2-12r+36$ , we have:

$A=12r-r^2=36-(r-6)^2$ .

A square cannot be negative. The largest area is therefore 36 cm², when the subtracted square is zero: $r=6$ cm.

Recover the angle from the perimeter equation: $6\theta+12=24$ gives $\theta=2$ rad. This is between 0 and $2\pi$ , so the maximum describes a valid sector.

Change the wire length

Your turn

A 20 cm wire forms the boundary of a sector. Find its greatest possible area and the corresponding radius and angle.

Show worked answer

$r\theta=20-2r$ , hence $A=10r-r^2=25-(r-5)^2$ . The largest area is 25 cm² at $r=5$ cm. Substitution gives $\theta=2$ rad, a valid sector angle.