A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\). (a) Calculate, correct to 1 decimal place, the perimeter of the sector. (b) If the sect...
Question 1 Report
A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\).
(a) Calculate, correct to 1 decimal place, the perimeter of the sector.
(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \(\pi = \frac{22}{7}\)].
Answer Details
(a) Perimeter of the sector. First find the angle \(\theta\) from the area:
\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]
\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]
Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)
\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]
(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:
\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]
Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).
\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]
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