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Volume of a cone = \(\frac{1}{3} \pi r^2 h\) r = 4\(\frac{1}{2}\) cm; h = 14 cm Volume of cone = \(\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 14\) = 297 cm\(^3\) When half- filled, the volume of the water = \(\frac{297}{2} = 148.5 cm^3\)

Area of rhombus = \(\frac{pq}{2}\) where p and q are the two diagonals of the rhombus. \(\therefore 60 = \frac{12 \times q}{2}\) 6q = 60 \(\implies\) q = 10 cm

Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm

< STQ = < SQR = 68^\circ (alternate segment) \(\therefore\) < STQ = 68^\circ < TQS = \(\frac{180^\circ - 68^\circ}{2}\) = \(\frac{112}{2} = 56^\circ\) \(\therefore\) < PQT = 180^\circ - (68^\circ + 56^\circ) = 180^\circ - 124^\circ = 56^\circ

Sum of interior angles in a polygon = \((2n - 4) \times 90^\circ\) \(\therefore (2n - 4) \times 90^\circ = 30 \times 90^\circ\) \(\implies 2n - 4 = 30 \) \(2n = 34 \implies n = 17\) The polygon has 17 sides.

\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles) \(\therefore \frac{PS + 10}{12} = \frac{12}{10}\) \(10 (PS + 10) = 12 \times 12\) \(10 PS + 100 = 144 \implies 10 PS = 44\) \(PS = 4.4 cm\)

Join PR QRP = QPR = 180 - 80 = \(\frac{100}{20}\) = 50º SRP = SPR = 180 - 30 = \(\frac{150}{2}\) = 75º ∴ SPQ = SPR - QPR = 75 - 50 = 25º

QPS - QRK = 31o QRK + RKQ + KQR = 180 31 + 42 + KQR = 180o KQR = 180 - 73 = 107o

< PQR = 180^\circ - (84^\circ + 56^\circ) = 40^\circ \(\therefore\) x = 40^\circ (corresponding angles QP// RT)

\(\frac{2.25}{0.015}\) = \(\frac{2250}{15}\) = \(150\) = \(1.5 \times 10^{2}\)

\(-2, -3, -4\frac{1}{2}, ...\) This is a G.P with r = 1\(\frac{1}{2}\). \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (-2)(\frac{3}{2})^{5 - 1}\) = \(-2 \times \frac{81}{16}\) = \(-\frac{81}{8}\) \(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\) = \(-2 \times \frac{729}{64}\) = \(-\frac{729}{32}\)

complete x as in the diagram. x/90 = sin30o x = 90 x sin30 = 45o ∴Dist = 150 + 45 = 195km

No expanded working has been attached for this question yet. The verified answer is Option A.

x2 = 242 + 72 = 576 + 49 x2 = 625 x = √625 = 25

\(\tan x = \frac{opp}{adj} = \frac{5}{12}\) \(Hyp^2 = opp^2 + adj^2\) \(Hyp^2 = 5^2 + 12^2\) = \(25 + 144 = 169\) \(Hyp = \sqrt{169} = 13\) \(\sin x = \frac{5}{13}; \cos x = \frac{12}{13}\) \(\sin x + \cos x = \frac{5}{13} + \frac{12}{13}\) = \(\frac{17}{13}\)