2022 BTYSS P2 Q8

A piece of wire of length 100 cm is bent into the shape of a trapezium \(PQRS\). Given that \(PQ=SR=x\) cm, \(\angle PQA=\angle SRB=60^\circ\) and \(PABS\) is a rectangle.

1. Show that \(PS=\dfrac{100-3x}2\).[3]
2. Show that the area of the rectangle \(ABSP\), \(A\text{ cm}^2\), is given by \[A=\dfrac{\sqrt3}4(100x-3x^2).\][3]
3. Given that \(x\) can vary, find the value of \(x\) for which \(A\) is maximum.[4]