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Area of shaded portion = Area of triangle PQR - Area of inner triangle Area of triangle given 3 sides a, b, c = \(\sqrt{s(s - a)(s - b)(s - c)}\) where \(s = \frac{a + b + c}{2} \) Area of PQR : \(s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7\) Area = \(\sqrt{7(7 - 3)(7 - 5)(7 - 6)}\) = \(\sqrt{7(4)(2)(1)} = \sqrt{56}\) \(\implies K^{2} = \sqrt{56}\) Area of inner triangle : \(s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}\) Area = \(\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}\) = \(\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}\) = \(\sqrt{\frac{896}{81}}\) = \(\sqrt{\frac{16}{81}} \times \sqrt{56}\) = \(\frac{4}{9} K^{2}\) \(\therefore \text{The area of the shaded portion} = K^{2} - \frac{4}{9}K^{2} = \frac{5}{9}K^{2}\)

\(\frac{x + 3}{sin 60^o}\) = \(\frac{x}{sin 306o}\) 3sin 30o = x sin 60o - x sin 30o = x(sin 60o - sin 30o) but sin 30o = \(\frac{1}{2}\) sin 60o = \(\frac{3}{2}\) = \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)

Area of triangle AEF = 6sq. cm area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines) 6 = \(\frac{1}{2}\) x 3 x h 6 = 3h 3h = \(\frac{12}{3}\) h = 4 = DX From D, C x D, CX\(^2\) = 5\(^2\) - 4\(^2\) = 25 - 16 = 9 Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3) BD\(^2\) = 4\(^2\) + 6\(^2\) = 16 + 36 = 52 BD = \(\sqrt{4 \times 13}\) = 2\(\sqrt{13}\)

x\(^2\) + 4 = 0 x\(^2\) = -4 You can't apply a square root to a negative number. So the answer is None of the above.

\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\) Squaring both sides, \(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\) Equating the rational and irrational parts, we have \(a + b = 170 ... (1)\) \(2 \sqrt{ab} = 20 \sqrt{30}\) \(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \) \(ab = 3000 ... (2)\) From (2), \(b = \frac{3000}{a}\) \(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\) \(a^{2} - 170a + 3000 = 0\) \(a^{2} - 20a - 150a + 3000 = 0\) \(a(a - 20) - 150(a - 20) = 0\) \(\text{a = 20 or a = 150}\) \(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\) \(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\) = \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)

\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\) N : b Look for the antilog of .3482 which gives 222.9

Force to the east = 5 units and angle to the east = 0º force to the Northeast = 4 units and angle to northeast = 45º Resolve these forces into x and y components FORCES x - component y - component 5 5cos0 = 5 5sin 0 = 0 4 4cos45 = 4 x \(\frac{\sqrt2}{2}\) 4sin45 = \(\frac{\sqrt2}{2}\) Total force 5 + 2\(\sqrt{2}\) 0 + 2\(\sqrt{2}\) Resultant force = \(\sqrt{(F_x)^2 + (F_y)^2}\) = \(\sqrt{ (5 + 2\sqrt{2})^2 + (2\sqrt{2})^2}\) = \(\sqrt{(25 + 20\sqrt{2} + 8) + 8}\) = \(\sqrt{33 + 20\sqrt{2} + 8}\) = \(\sqrt{41 + 20\sqrt{2}}\)