Area and Perimeter of an Annular Sector

The region between two concentric arcs with the same angle is an annular sector. Its area is the outer sector minus the inner, $\frac12(R^2 - r^2)\theta$. Its perimeter is the two arcs plus the two straight edges.

A shaded region is bounded by two arcs of radii $4$ cm and $7$ cm subtending an angle of $1.2$ radians at the common centre, and the two straight edges joining them. Find (a) the area and (b) the perimeter of the shaded region. [6]

The working

Part (a), area = outer sector − inner sector: $A = \tfrac{1}{2}R^2\theta - \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(R^2 - r^2)\theta \quad \text{(M1)}$ $= \tfrac{1}{2}(7^2 - 4^2)(1.2) = \tfrac{1}{2}(49 - 16)(1.2) = \tfrac{1}{2}(33)(1.2) = 19.8\ \text{cm}^2 \quad \text{(A1, A1)}$

Part (b), perimeter = outer arc + inner arc + two straight edges.

Arc lengths ($s = r\theta$): $\text{outer} = 7(1.2) = 8.4\ \text{cm}, \qquad \text{inner} = 4(1.2) = 4.8\ \text{cm} \quad \text{(M1)}$

Each straight edge is $R - r = 7 - 4 = 3\ \text{cm}$, and there are two: $P = 8.4 + 4.8 + 2(3) = 19.2\ \text{cm} \quad \text{(M1, A1)}$

Where the marks are won and lost

• Use $\frac12(R^2 - r^2)\theta$ for the area, not $\frac12(R - r)^2\theta$. It’s a difference of squares, not a square of a difference.
• The perimeter has four pieces: two arcs and two straight edges of length $R - r$ each. Forgetting the straight edges is the usual slip.
• Keep the calculator in radian mode; $\theta = 1.2$ is already in radians.

Common mistakes

• Squaring $(R - r)$ instead of taking $R^2 - r^2$.
• Including only one straight edge, or none.
• Converting $1.2$ radians to degrees unnecessarily.

Topic home: Circular Measure pillar. More: Worked examples.