A trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50 cm and |SR| = 60 cm. Calculate, correct to 2 signi...
Question 1 Report
A trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50 cm and |SR| = 60 cm. Calculate, correct to 2 significant figures, the
(a) Area of the trapezium ; (b) < QRS.
Answer Details
(a) Area. The perpendicular distance between the parallel sides is 40 cm.
\[A = \tfrac{1}{2}(PQ + SR) \times h = \tfrac{1}{2}(20 + 60) \times 40 = 1600 \text{ cm}^2\](b) Angle QRS. Place S at \((0, 0)\) and R at \((60, 0)\). P is 40 cm above RS; since \(SP = 50\), its horizontal offset is \(\sqrt{50^2 - 40^2} = 30\), so P is at \((30, 40)\). As \(PQ \parallel SR\) with \(PQ = 20\), Q is at \((50, 40)\).
At R, \(\vec{RQ} = (-10, 40)\) and \(\vec{RS} = (-60, 0)\):
\[\cos(\angle QRS) = \frac{(-10)(-60) + 0}{\sqrt{10^2 + 40^2} \times 60} = \frac{600}{2474} = 0.2425\] \[\angle QRS = 76^{\circ} \text{ (2 s.f.)}\]Download The App On Google Playstore
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