Additional Mathematics 0606 / Calculus — Differentiation 1 / 2.182.18: Stationary area of a trapezium0606/21/O/N/16 — Question 7 · 9 marksMark as done · Save for later · Show all solutionsThe diagram shows a trapezium PQRSPQRSPQRS in which QRQRQR is parallel to PSPSPS, PQ=3 cmPQ = 3\text{ cm}PQ=3 cm, QR=x cmQR = x\text{ cm}QR=x cm, and PS=(x+14) cmPS = (x + 14)\text{ cm}PS=(x+14) cm.(i)Show that the area, A cm2A\text{ cm}^{2}A cm2, of the trapezium PQRSPQRSPQRS is given by A=(7+x)9−x2A = (7 + x)\sqrt{9 - x^{2}}A=(7+x)9−x2.[2]▸ Answer(ii)Given that xxx can vary, find the stationary value of AAA.[7]▸ Answer▸ Official mark scheme← 2.17: Tangent to a cubic and second intersection2.70: Optimising the volume of a conical tent →
