Points A, B and C are on the circumference of a circle, centre O and radius \(r\) . A trapezium OABC is formed such that AB is parallel to OC, and the angle \({\rm{A}}\hat {\text{O}}{\text{C}}\) is \(\theta\) , \(\frac{\pi }{2} \leqslant \theta  \leqslant \pi \) .

(a)     Show that angle \({\rm{B\hat OC}}\) is \(\pi - \theta \).

(b)     Show that the area, T, of the trapezium can be expressed as

\[T = \frac{1}{2}{r^2}\sin \theta - \frac{1}{2}{r^2}\sin 2\theta .\]

(c)     (i)     Show that when the area is maximum, the value of \(\theta \) satisfies

\[\cos \theta = 2\cos 2\theta .\]

  (ii)     Hence determine the maximum area of the trapezium when r = 1.

    (Note: It is not required to prove that it is a maximum.)

(a)     \({\rm{O\hat AB}} = \pi - \theta \,\,\,\,\,\)(allied)     A1

recognizing OAB as an isosceles triangle     M1

so \({\rm{A\hat BO}} = \pi - \theta \)

\({\rm{B\hat OC}} = \pi - \theta \,\,\,\,\,\)(alternate)     AG

Note: This can be done in many ways, including a clear diagram.

[3 marks]

(b)     area of trapezium is \(T = {\text{are}}{{\text{a}}_{\Delta {\text{BOC}}}} + {\text{are}}{{\text{a}}_{\Delta {\text{AOB}}}}\)     (M1)

\( = \frac{1}{2}{r^2}\sin (\pi - \theta ) + \frac{1}{2}{r^2}\sin (2\theta - \pi )\)     M1A1

\( = \frac{1}{2}{r^2}\sin \theta - \frac{1}{2}{r^2}\sin 2\theta \)     AG

[3 marks]

(c)     (i)     \(\frac{{{\text{d}}T}}{{{\text{d}}\theta }} = \frac{1}{2}{r^2}\cos \theta - {r^2}\cos 2\theta \)     M1A1

for maximum area \(\frac{1}{2}{r^2}\cos \theta - {r^2}\cos 2\theta = 0\)     M1

\(\cos \theta = 2\cos 2\theta \)     AG

(ii)     \({\theta _{\max }} = 2.205 \ldots \)     (A1)

\(\frac{1}{2}\sin {\theta _{\max }} - \frac{1}{2}\sin 2{\theta _{\max }} = 0.880\)     A1

[5 marks]

Total [11 marks]

In part (a) students had difficulties supporting their statements and were consequently unable to gain all the marks here. There were some good attempts at parts (b) and (c) although many students failed to recognise r as a constant and hence differentiated it, often incorrectly.