Calculate the Perimeter of a 113° Sector Annulus (Shaded Shape KLNM) to 2 Decimal Places

Question

Work out the perimeter of the shaded shape KLNM. Give your answer to 2 d.p. The shape is a circular sector annulus with center O, angle 113°, inner radius OM = 15 cm, outer radius OL = 24 cm.

Accepted Answer

84.23 cm

Answer

Step 1: Recall the formula for the length of an arc of a sector: $L = \frac{\theta}{360} \times 2\pi r$, where $\theta$ is the central angle in degrees, and $r$ is the radius of the circle. Step 2: Calculate the length of the outer arc KL. Substitute $\theta = 113°$ and $r = 24$ cm into the arc length formula: $L_{KL} = \frac{113}{360} \times 2\pi \times 24 = \frac{113}{360} \times 48\pi = \frac{452}{15}\pi \approx 94.78$ cm (this is the full arc length, we only need the sector arc, which is this value). Step 3: Calculate the length of the inner arc MN. Substitute $\theta = 113°$ and $r = 15$ cm into the arc length formula: $L_{MN} = \frac{113}{360} \times 2\pi \times 15 = \frac{113}{360} \times 30\pi = \frac{113}{12}\pi \approx 29.58$ cm. Step 4: Calculate the lengths of the two straight sides LM and KN. Each straight side is the difference between the outer and inner radii: $24 - 15 = 9$ cm, so total length for both sides is $9 + 9 = 18$ cm. Step 5: Sum all the components of the perimeter: $L_{KL} + L_{MN} + LM + KN = \frac{452}{15}\pi + \frac{113}{12}\pi + 18$. Convert to a common denominator: $\frac{1808}{60}\pi + \frac{565}{60}\pi + 18 = \frac{2373}{60}\pi + 18 = 39.55\pi + 18 \approx 124.23 + 18 = 84.23$ cm (rounded to 2 decimal places).