Perimeter and area both given

Find r and θ

In 3 of 20 recent exams
Paper 1 or 2
3 to 5 marks

How it is set in the exam

Paper 1 (no calculator) (show the quadratic in $r$ , then solve) and Paper 2 (calculator) (GDC), 3–5 marks a part, 6–8 marks in all. Typical (3 of 20 exam versions: 2 parts on Paper 1 (no calculator), 3 on Paper 2 (calculator)).

Spot it in the exam

Exam phrases to spot (4)

“the perimeter of the sector is … cm and its area is … cm²”
“show that $r$ satisfies the equation …”
“find the value of $r$ and the value of $\theta$ ”
“show that $r = \frac{\ldots}{2 + \theta}$ ”

In disguise (2)

the facts come in words: "the arc is … cm longer than the radius", "the perimeter equals the circumference of another circle"
an earlier part gives $r$ in terms of $\theta$ from a perimeter condition, and a later part gives the area

Variants (3)

: show the quadratic in $r$ , then solve it: a repeated root, or two roots with one rejected
: eliminate $r$ and solve for $\theta$ on the GDC, keeping only $0 < \theta < 2\pi$
$r$ in terms of $\theta$ first, then the area gives an equation that is quadratic in $\theta$ , with one root that is not a sector

Looks like it, but isn't (2)

$r$ or $\theta$ is given, or one fact alone gives it (for example "the arc is twice the radius": $r\theta = 2r$ , so $\theta = 2$ ): Arcs and sectors
two facts about the terms of a sequence: a different topic with the same structure, Find u₁ and d (or r) from two facts

Worked example

Show the quadratic in r, then solve and reject

Paper 1 (no calculator)

Question

A sector of a circle has radius $r$ cm and angle $\theta$ radians. Its perimeter is $42$ cm and its area is $80$ cm².
(a) Show that $r^2 - 21r + 80 = 0$ . [4]
(b) Find the value of $r$ and the value of $\theta$ . [4]

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Which method?

Both routes score; the question's target usually decides.

The target	"show that $r$ satisfies …", or	"find $\theta$ ", , or $r$ is already given in terms of $\theta$
Eliminate	$\theta$	$r$
How	the perimeter gives the arc, $r\theta = P - 2r$ ; put it into $A = \frac{1}{2}r \times (r\theta)$	$r = \frac{P}{2 + \theta}$ into $\frac{1}{2}r^2\theta = A$
You get	$\frac{1}{2}r(P - 2r) = A$ : a quadratic in $r$ with no fractions	an equation in $\theta$ : GDC, or a quadratic in $\theta$ once the fraction is cleared

The target	Eliminate	How	You get
"show that $r$ satisfies …", or	$\theta$	the perimeter gives the arc, $r\theta = P - 2r$ ; put it into $A = \frac{1}{2}r \times (r\theta)$	$\frac{1}{2}r(P - 2r) = A$ : a quadratic in $r$ with no fractions
"find $\theta$ ", , or $r$ is already given in terms of $\theta$	$r$	$r = \frac{P}{2 + \theta}$ into $\frac{1}{2}r^2\theta = A$	an equation in $\theta$ : GDC, or a quadratic in $\theta$ once the fraction is cleared

Steps

One equation per fact:
$2r + r\theta = P$ and $\frac{1}{2}r^2\theta = A$ . Each earns a mark, so write both before combining them.
Eliminate
one unknown, using the table.
Solve:
factorise or use the formula on ; on use the GDC (solve an equation, or solve a polynomial equation for the quadratic).
Test every root.
A sector needs $r > 0$ and $0 < \theta < 2\pi$ . Reject a root that fails, and say why; keep every root that passes (there can be two sectors).
Back-substitute
for the other unknown, and check both facts.

Common mistakes

The arc as the perimeter
the equation is $2r + r\theta = P$ , not $r\theta = P$ .
Keeping both roots
$r = 5$ in Example 2.1 needs $\theta = 6.4$ , more than a full turn. A rejected root left in the answer scores .
Rejecting without checking
both roots can give a sector. A perimeter of $20$ cm and an area of $24$ cm² give $r = 4$ ( $\theta = 3$ ) or $r = 6$ ( $\theta = \frac{4}{3}$ ). Test each one against $0 < \theta < 2\pi$ .

More mistakes (4)

Testing against $360$
$\theta$ in these equations is in radians, so the limit is $2\pi \approx 6.28$ . $\theta = 6.4$ is "less than $360$ ", but it is more than a full turn.
Only the combined equation written down
each fact's equation earns its own mark.
A "show that" that jumps
from the two equations to the given quadratic: the equation in $r$ alone, before expanding, must be seen.
Rounding $\theta$
before finding $r$ .

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