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Area of the triangle XYZ = \(\frac{1}{2}\)bh = \(\frac{1}{2}\)ZY x XY = \(\frac{1}{2}\) x 3 x 4 = 6cm2.

From similar triangle, \(\frac{x}{6}\) = \(\frac{8}{12}\) 12x = 48 x = \(\frac{48}{12}\) = 4

Rearranging in increasing order 4, 4, 7, 9, 10, 13, 14, 17, the median = \(\frac{(9 + 10)}{2}\) = \(\frac{19}{2}\)

Constant Angle Rule: In geometry, the locus of points from which a line segment subtends a constant angle is an arc of a circle. Inscribed Angle Theorem: This theorem states that angles subtended by the same arc at the circumference of a circle are equal. Consequently, as the point moves along the arc, the subtended angle remains constant.

10% of 400 = N40, Remainder = N360 40% of 360 = N144 Altogether they 144 + 40 = N184

(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......) nthterm is (-1)n\(\frac{2}{n + 1}\) Since (-1)n will always be negative for n = odd

\(\frac{26}{130}\) = \(\frac{x}{360}\) 130x = 26 x 360 = 9360 x = \(\frac{9360}{130}\) = 72o

The sum of interior angle of pentagon (5 sides) = (2n - 4) x 90o n = 5 (2 x 5 - 4) x 90o = 6 x 90o = 540o 540o - 60o = 480o \(\div\) 4 = 120o

\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a a2 = \(\frac{1 - cosx}{1 + cosx}\) \(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\) = \(\frac{(1 - cosx)^2}{cos^2 x}\) a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\) a = \(\frac{1 - cos x}{sin x}\)

\(\frac{5}{x + 1} - \frac{3}{1 - x} - \frac{7x - 1}{x^{2} - 1} = \frac{?}{x + 1}\) \(\frac{5(x - 1) - 3(- (x + 1)) - (7x - 1)}{x^{2} - 1}\) = \(\frac{5x - 5 + 3x + 3 - 7x + 1}{x^{2} - 1}\) = \(\frac{x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{(x - 1)(x + 1)}\) = \(\frac{1}{x + 1}\). The numerator = 1.

By Pythagoras QN = 4 Since, QR = 9, and QN = 4, then, NR = 5 Tan x = \(\frac{3}{5}\)

from the diagram above, angle b = 180º - 55º - x ( 55º not 56º) sum of angle at Q = y + 90º + 180º - 55º - x = 180º( sum of angles on a straight line) y - x = -35 ..................................... eq1 sum of angle in triangle TPQ = 180º - x + 2y = 180º ( sum of angles in a triangle ) 2y - x = 0 ............................................eq2 solving eq1 and eq 2 simultaneously x = 70º

Perimeter of lunar portion = 3\(\pi\) + \(\frac{3\pi \sqrt{2}}{2}\) = 3 \(\pi\)(1 + \(\frac{2}{2}\))

Sum of external angles = 360o 3x + 5x + x + 2x + x = 360o 12x = 360o x = 30o

Source: Myschool Past Questions Archive (Record #32750)