A trapezium PQRS is such that PQ // RS and the perpendicular P to RS

Explanation

(a) Area of trapezium PQRS = \(\frac{1}{2}(PQ + RS) \times 40\)

= \(\frac{1}{2}(20 + 60) \times 40\)

= \(40 \times 40\)

= \(1600 cm^{2}\)

(b) In \(\Delta\) SPT,

\(|SP|^{2} = |ST|^{2} + |TP|^{2}\) (Pythagoras theorem)

\(|ST|^{2} = |SP|^{2} - |TP|^{2}\)

= \(50^{2} - 40^{2}\)

= \(2500 - 1600\)

\(|ST|^{2} = 900\)

\(\therefore |ST| = 30 cm\)

\(|UR| = 60 cm - (30 + 20) cm\)

= 10 cm

Let < QRS = \(\alpha\),

\(\tan \alpha = \frac{40}{10} = 4\)

\(\alpha = \tan^{-1} (4)\)

= \(75.96° \approxeq 76°\)