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In the figure, < AOC = 2 x < ABC = 60^\circ (angle subtended at the centre) \(\therefore\) Arc AC = \(\frac{60}{360} \times 2 \times 10 \times \frac{22}{7}\) = \(\frac{31.4}{3}\) = 10.466 cm \(\approxeq\) 10.5 cm

x\(^2\) + 4x - 192 x\(^2\) + 16x - 12x - 192 x(x + 16) - 12(x + 16) (x + 16)(x - 12)

2e\(^2\) - 3e + 1 2e\(^2\) - 2e - e + 1 2e(e - 1) - 1(e - 1) (2e - 1)(e - 1)

7y\(^2\) = 3y 7y\(^2\) - 3y = 0 y(7y - 3) = 0 y = 0 or y = \(\frac{3}{7}\)

x\(^2\) + 8 + m To make it a perfect square, we add \((\frac{b}{2})^2\) = \((\frac{8}{2})^2\) = 16

2a\(^2\) - 3a - 27 = 0 2a\(^2\) - 9a + 6a - 27 = 0 a(2a - 9) + 3(2a - 9) = 0 (a + 3)(2a - 9) = 0 a = -3 or a = \(\frac{9}{2}\)

πr2 = 360o 44cm2 = θ; ∴ θ = 44/πr2 x 360o = 102.9o = 103o

Surface area of a sphere = \(4\pi r^2\) = \(4 \times \frac{22}{7} \times 7 \times 7\) = \(616 cm^2\)

\(\theta_{1} = \theta_{2} = 30^\circ\)

TSA of a cylinder = \(2\pi r^2 + 2\pi rh\) = \(2\pi r (r + h)\) = \(2 \times \frac{22}{7} \times 7 (7 + 5)\) = \(44 \times 12\) = \(528 cm^2\)

Volume = \(\pi r^2 h\) = \(\frac{22}{7} \times 7^2 \times 5\) = \(770 cm^3\)

From the figure, < PSR = 32^\circ (base angles of an isos. triangle) \(\therefore\) < PRS = 180^\circ - (32^\circ + 32^\circ) = 116^\circ (sum of angles in a triangle) < QRP = 180^\circ - 116^\circ = 64^\circ (angle on a straight line) < PQR = 64^\circ (base angles of an isos. triangle) < QPR = 180^\circ - (64^\circ + 64^\circ) = 52^\circ

Diagonals bisect at 90^\circ; < YXZ = 90^\circ - 20^\circ = 70^\circ But ZY = XY (sides of a rhombus) \(\therefore\) < XYZ = 70^\circ (base angle of an isos. triangle)

Area of a triangle = \(\frac{1}{2} \times b \times h\) Area of \(\Delta\) SQR = \(\frac{1}{2} \times 6.4 \times 3.2\) = 10.24 cm\(^2\)

< VWZ = < VXZ (angles in the same segment)