In the diagram /PS/ = 9 cm, /OR/ = 5cm, \(P\hat{S}R = 63^o\) and \(S\hat{P}Q = P\hat{Q}R = 90^o\). Find, correct to the nearest whole number, the area of th...

In the diagram /PS/ = 9 cm, /OR/ = 5cm, \(P\hat{S}R = 63^o\) and \(S\hat{P}Q = P\hat{Q}R = 90^o\). Find, correct to the nearest whole number, the area of the trapezium,

Answer Details

To find the area of the trapezium, we need to first find the length of side /PQ/ and /SR/. Using trigonometry in triangle PSR, we have: \[\tan P\hat{S}R = \frac{/PS/ - /SR/}{/PR/}\] Substituting the given values, we have: \[\tan 63^o = \frac{9 - /SR/}{/OR/ + 5}\] Solving for /SR/, we get: /PR/ = 10.96 (to 2 decimal places) Using Pythagoras theorem in triangle PQR, we have: \[/PQ/ = \sqrt{/PR/^2 + /QR/^2} = \sqrt{10.96^2 + 9^2} \approx 13.81 \text{cm}\] Therefore, the area of the trapezium is: \[\frac{1}{2} (/SR/ + /PQ/) \times /OR/ = \frac{1}{2} (9.19 + 13.81) \times 5 \approx 62.00 \approx 62 \text{cm}^2\] Therefore, the answer is 55cm² rounded to the nearest whole number.