Sectors with equal area but different arcs

Sectors with equal area but different arcs

IP 2
7 marks

A sector has radius \(r\), angle \(\theta\) radians and arc length \(l\).

  1. Show that the area of the sector is \(\dfrac{1}{2}rl\), and explain why this matches the area of a triangle with base \(l\) and height \(r\).[2]
  2. A second sector has radius \(2r\) and angle \(\dfrac{\theta}{4}\). Compare its arc length and area with those of the first sector.[3]
  3. A third sector has radius \(kr\), where \(k>0\), and the same area as the first. Find its angle in terms of \(\theta\) and \(k\), and its arc length in terms of \(l\) and \(k\).[2]

Solution:

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Answer:(a) \(\dfrac{1}{2}r^2\theta=\dfrac{1}{2}r(r\theta)\) (b) Arc halves, area unchanged (c) Angle \(\dfrac{\theta}{k^2}\), arc \(\dfrac{l}{k}\)

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