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Volume of the cylinder = \(\frac{\theta}{360} \times \pi r^2h\) \theta = 360\(^o\) - 90\(^o\) = 270\(^o\) ∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)

Volume of the composite solid = Volume of A + Volume of B Volume of a cuboid = length x breadth x height Volume of A = 6 x 26 x 8 = 1248 cm\(^3\) Volume of B = 6 x 10 x 22 = 1320 cm\(^3\) ∴ Volume of the composite solid = 1248 + 1320 = 2568 cm\(^3\)

Volume of a cone = \(\frac{1}{3}\pi r^2 h\) r = 5 cm l = 13 cm Using Pythagoras theorem ⇒ \(13^2 = 5^2 + h^2\) ⇒ \(169 = 25 + h^2\) ⇒ \(169 - 25 = h^2\) ⇒ \(h^2 = 144\) ⇒ \(h = \sqrt144 = 12 cm\) ∴ Volume of the cone = \(\frac{1}{3} \times\pi\times 5^2 x 12 = 100\pi\) cm\(^3\)

Given that radius = 3cm. volume of sphere = \(\frac{4}{3}\times\pi\times r^3\) = \(\frac{4}{3}\times\pi\times 3^3\) = \(\frac{4}{3}\times\pi\times 27\) = \(4\times\pi\times9\) = \(36\pi cm^3\)

given that diameter = 140cm ⇒radius = 70cm, volume of cylinder = 200litres = \(200,000cm^3\) volume of cylinder = base area times height = \(\pi r^2h\) \( 200,000cm^3 = \frac{22}{7} \times 70 \times 70 \times h\) \(200,000 = \frac{107,800h}{7}\) cross multiply 1,400,000 = 107,800h \( h = \frac{1,400,000}{107,800}\) = 13cm (to the nearest cm).

An interior angle of a regular polygon = \(\frac{(2n-4)\times 90}{n}\) An exterior angle of a regular polygon = \(\frac{360}{n}\) \(\frac{(2n-4)\times 90}{n}\) =5 \(\times\) \(\frac{360}{n}\) (Given) = (2n-4) x 90 = 5 x 360 = 180n - 360 = 1800 = 180n = 1800 + 360 = 180n = 2160 = n = \(\frac{2160}{180}\) = 12 The polygon has 12 sides which is dodecagon

∠EHI = ∠DEH = 40^\circ (alternate angles are equal) ∠BEH = 90^\circ (given) ∠BED = 90^\circ - 40^\circ = 50^\circ ∠BEF = 180^\circ - 50^\circ = 130^\circ (sum of angles on a straight line is 180o) ∠BEF = ∠ABE = 130^\circ (alternate angles are equal) ∴ m = 130^\circ

each interior angle of a polygon = \(\frac{(n - 2)\times 180}{n}\) where n = no of side of a polygon each exterior angle of a polygon = \(\frac{360}{n}\) then \(\frac{(n - 2)\times 180}{n}\) = 6\(\times\) \(\frac{360}{n}\) = (n - 2) 180 = 2160 = 180n - 360 = 2160 = 180n = 2160 + 360 = 180n = 2520 therefore, n = \(\frac{2520}{180}\) = 14.

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.

Let the given points be: P(-3, -14) = (x\(_1\), y\(_1\)) Q(t, -5) = (x\(_2\), y\(_2\)) PQ = 9 units (given) Using the distance formula, d = \sqrt [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)] PQ = \sqrt [ \((t - (-3))^2 + (-5 + 14)^2\)] \(\implies\) \sqrt [ \((t + 3)^2 + 81\)] = 9 Squaring on both sides, ⇒ (t + 3)\(^2\) + 81 = 81 ⇒ (t + 3)\(^2\) = 0 ⇒ t + 3 = 0 ∴ t = -3

\(3x + 2y + 4 = 0\) Rearrange: \(2y = -3x - 4\) Divide both sides by 2 y = \(\frac {-3 \times - 4}{2}\) y = \(\frac {-3}{2} \times - 2\) ∴ the gradient of the line 3x + 2y + 4 = 0 is \(\frac {-3}{2}\) If two lines are perpendicular to each other ∴ \(m_1 x m_2\) = -1 Let \(m_1 = \frac {-3}{2} \therefore m_2 = \frac {-1}{m_1} = \frac {-1}{-3/2} = \frac {2}{3}\) From the equation of a line which is given as m = \(\frac {y - y_1}{x - x_1} where (x_1, y_1) = (2,3)\) \(\therefore \frac {2}{3} = \frac {y - 3}{x - 2}\) =3(y - 3) = 2(x - 2) =3y - 9 = 2 x -4 =3y = 2 x -4 + 9 ∴ 3y = 2x + 5

The line: \(3y + 6x\) = 48 Divide through by 3 ⇒ y + 2\(x\) = 16 ⇒ y = -2\(x\) + 16 ∴ The gradient of the line = -2 The points: A(-2, k) and B (4, 8) m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\) ⇒ m =\(\frac[8 - k}{4 + 2} = {8 - k}{6}\) Since the line passes through the points ∴ -2 = \(\frac{8 - k}{6}\) ⇒ \(\frac{-2}[1} = \frac{8 - k]{6}\) ⇒ 8 - k = -12 ⇒ k = 8 + 12 ∴ k = 20

(7, 4) and (-3, 9) = (\( y_1 , x_1)(y_2 , x_2)\) slope\( m_1\) of the points(7, 4) and (-3, 9) = \(\frac{ y_2 - y_1}{ x_2 - x_1}\) \( m_1 = \frac{ 9 - 4}{ -3 - 7} = \frac{5}{-10} = \frac{-1}{2}\) \( m_1 = m_2\) ( condition for parallelism) Since the line L passes through the point (6, -13) and is parallel to the points (7, 4) and (-3, 9) \(\frac{-1}{2} = \frac{ y - (- 13)}{x - 6}\) = \(\frac{-1}{2} = \frac{ y + 13}{x - 6}\) = \(\frac{-1}{2}( x - 6) = y + 13\) \(\frac{-1}{2}x + 3 = y + 13\) = \(\frac{-1}{2}x + 3 - 13 = y\) ∴ y = \(\frac{-1}{2}x - 10\)

Let A = (1, -3), B = (6, 2) and C = (0,4) \(d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}\) |AB| = \(\sqrt{(2 - (-3))^2 + (6 - 1)^2}\) = \(\sqrt{(2 + 3)^2 + (6 - 1)^2}\) = \(\sqrt{5^2 + 5^2}\) = \(\sqrt{25 + 25}\) = \(\sqrt50\) = 5\(\sqrt2\) units |BC| = \(\sqrt{(4 - 2)^2 + (0 - 6)^2}\) = \(\sqrt{2^2 + (-6)^2}\) = \(\sqrt{4 + 36}\) = \(\sqrt40\) = 2\(\sqrt10\) units |AC| = \(\sqrt{(4 - (-3)^2 + (0 - 1)^2}\) = \(\sqrt{(4 + 3)^2 + (0 - 1)^2}\) = \(\sqrt{7^2 + (-1)^2}\) = \(\sqrt{49 + 1}\) = \(\sqrt50\) = 5\(\sqrt{2}\) units Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.

Gradient(slope) m = \(\frac{ y_2 - y_1}{ x_2 - x_1}\) the points are \((\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})\) m = \(\frac{\frac{2}{3} - (\frac{-1}{3})}{3 - \frac{1}{2}}\) = \(\frac{\frac{2}{3} + \frac{1}{3}}{3 - \frac{1}{2}}\) = \(1 \div\frac{5}{2}\) = \(1\times\frac{2}{5}\) Therefore, m = \(\frac{2}{5}\)