(a)
In the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\(cm^{2}\), calculate the value of r. [Take \(\pi = \frac{22}{7}\)].
(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \(< PQR = 66°\) and \(< PSQ = 109°\), calculate the value of \(< RQS\).
In the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\(cm^{2}\), calculate the value of r. [Take \(\pi = \frac{22}{7}\)].
(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \(< PQR = 66°\) and \(< PSQ = 109°\), calculate the value of \(< RQS\).
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Correct Answer: Option n
Explanation:
(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.
\(\implies 504 = \frac{\theta}{360°} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\)
\(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\)
\(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\)
\(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\)
\(r = \sqrt{1764} = 42 cm\).
(b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle)
\(\therefore a_{1} + a_{2} = 2a_{1}\)
\(\therefore 2a_{1} + 109° = 180° \implies 2a_{1} = 180° - 109° = 71°\)
\(a_{1} = a_{2} = \frac{71°}{2} = 35.5°\)
Hence, \(< RQS = 66° + 35.5°\)
= \(101.5°\)
(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.
\(\implies 504 = \frac{\theta}{360°} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\)
\(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\)
\(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\)
\(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\)
\(r = \sqrt{1764} = 42 cm\).
(b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle)
\(\therefore a_{1} + a_{2} = 2a_{1}\)
\(\therefore 2a_{1} + 109° = 180° \implies 2a_{1} = 180° - 109° = 71°\)
\(a_{1} = a_{2} = \frac{71°}{2} = 35.5°\)
Hence, \(< RQS = 66° + 35.5°\)
= \(101.5°\)