(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).
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°)