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Length of an arc = \(\frac{\theta}{360} \times 2\pi r\) 18.33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14\) \(\theta = \frac{18.33 \times 360 \times 17}{2 \times 22 \times 14}\) = 75o (approx.)

Total surface area of cube = 6s\(^2\) 6s\(^2\) = x s\(^2\) = x = \(\frac{x}{6}\).....(1) volume of a cube = s\(^2\) = \(\frac{x}{2}\) s\(^2\) = \(\frac{x}{2}\)......(2) put(1) into (2) s(\(\frac{x}{6}\)) = \(\frac{x}{2}\) s = \(\frac{x}{2} \times \frac{6}{x}\) = 3cm

perimeter of minor segment = Length of arc xy + chord xy where lxy = \(\frac{120}{360} \times 2x \times \frac{22}{7} \times 7\) = 14.67cm perimeter of minor segment = 14.67 + 8 = 22.67cm

p = \(\frac{1}{2}; \frac{1}{p - 1} = \frac{2}{p + x}\) \(\frac{1}{\frac{1}{2} - 1} = \frac{2}{\frac{1}{2} + x}\) \(\frac{1}{\frac{1 - 2}{2}} = \frac{2}{\frac{1 + 2x}{2}}\) \(\frac{1}{-\frac{1}{2}} = \frac{2}{\frac{1 + 2x}{2}}\) -2 = \(\frac{4}{1 + 2x} -2(1 + 2x) = 4\) 1 + 2x = \(\frac{4}{-2}\) 1 + 2x = -2 2x = -2 - 1 2x = -3 x = -\(\frac{3}{2}\) x = -1\(\frac{1}{2}\)

Roots; x and -c sum of roots = c + (-c) = 0 product of roots = c x -c = -c2 Equation; x2 - (sum of roots) x = product of roots = 0 x2 - (0)x + (-c2) = 0 x2 - c2 = 0

1 - 2x -\(\frac{4}{-6}\) = x > \(\frac{2}{3}\)

A = [2, 4, 6, 8} B = {2, 3, 5, 9} Pr = (Prime in B) = \(\frac{3}{4}\)

\(\begin{array}{c|c} x & 2 & 3 & 5 & 9 \\ \hline 2 & 4 & 6 & 10 & 18 \\ \hline 4 & 8 & 12 & 20 & 36 \\ \hline 6 & 12 & 18 & 30 & 54 \\ \hline 8 & 16 & 24 & 40 & 72 \end{array}\) Note: A {vertical} B {horizontal} Pr (Odd Product) = \(\frac{0}{16}\) = 0

\(\begin{array}{c|c} - & 2 & 3 & 5 & 9 \\ \hline 2 & 0 & 1 & 3 & 7 \\ \hline 4 & 2 & 1 & 1 & 5\\ \hline 6 & 4 & 3 & 1 & 3 \\ \hline 8 & 6 &5 & 3 & 1 \end{array}\) Note: A {horizontal} B {vertical} Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)

\(\begin{array}{c|c} x & d = x - \bar{x} & |d| \\ \hline 10 & -4.5 & 4.5 \\ 12 & -2.5 & 2.5\\ 14 & -0.5 & 0.5\\ 15 & 0.5 & 0.5\\ 17 & 2.5 & 2.5\\ 19 & 4.5 & 4.5\\ \hline \sum x = 87 & & \sum|d| = 15\end{array}\) \(\bar{x} = \frac{\sum x}{n}\) = \(\frac{87}{6}\) = 14.5

Given q2 = 25 - r2.....(1) from Pythagoras theorem p2 = q2 + r2....(2) put(1) into (2) p2 = 25 - r2 + r2 p2 = 25 p = \(\sqrt{25}\)

Since |PR| = |RS| = |SP| \(\bigtriangleup\) PRS is equilateral and so o But o (Opposite interior angles of a cyclic quadrilateral) o o But in \(\bigtriangleup\) PQR, PQ = PR, hence o (Angles in a triangle) 2 o From the diagram, o

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS). Thus: \(\bar{PS}^2\) = 52 + 122 \(\bar{PS} = \sqrt{25 + 144}\) = \(\sqrt{169}\) \(\bar{PS}\) = 13cm perimeter = 4 x length of a side = 4 x 13cm = 52cm

o + o - 70o = 78o

From the diagram, < QPS = xo (Corresponding angles) Also, < QPS = < PSR(Alternate angles) x = 42o