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In a rhombus, the diagonals are perpendicular bisectors of each other, and they bisect the angles of the rhombus. This means that a rhombus is essentially made up of four congruent right-angled triangles. We can use the Pythagorean theorem to find the length of one side of the rhombus (s) \(s^2 = 8^2 + 6^2\) \(s^2 = 64 + 36\) \(s^2 = 100\) s = \(\sqrt{100}\) s =10 cm So, the length of each side of the rhombus is 10 cm.

∆PQR is an equilateral triangle and M is the midpoint of Line QR. Hence, PM is the radius of the circle likewise Line PA and Line PB Using Pythagoras theorem to find Line PM : \( 18^2 = PM^2 + 9^2\) \( PM^2 = 324 - 81 = 243\) PM = \(\sqrt{243} = 9\sqrt{3}\) ∠QPR = \(\frac{180^0}{3} = 60^0\) ( each ∠ of an equilateral △ are equal) Area of △,PQR = \(\frac{1}{2}absin\theta\), but \theta = 60^\circ = \(\frac{1}{2}\times18\times18sin60 = 81\sqrt{3} = 140.30 cm^2\) Area of PAB = \(\frac{\theta}{360} \times\pi r^2\) = \(\frac{60}{360}\times \frac{22}{7} \times(9\sqrt{3})^2\) = \(\frac{1}{6} \times \frac{22}{7} \times243\) = \(127.29cm^2\) therefore, the Area of the shaded portion = area of △ PQR - area of sector PAB = 140. 30 - 127.29 = \(13.01cm^2\) (to 2 d.p)

a. Let the number of sides of polygon Q be x therefore, No. of sides of polygon P = 2x then, \(\frac{360}{x} - \frac{360}{2x}\) = 45, \(\frac{360}{x} - \frac{180}{x}\) = 45 L. C. M = x 360 - 180 = 45x x = \(\frac{180}{4}\) = 4 therefore, the number of P polygon = 2x = 2 x 4 = 8 b. Area of semi-circle = \(\frac{\pi r^2}{2}\) but Area of the semi-circle = 32\pi \(\frac{\pi r^2}{2}\) = 32\pi \(r^2 = 32\times2 = 64\) \(r^2 = \sqrt{64}\) = 8cm. Circumference of semi-circle = \pir + 2r = \pi(8) + 2(8) = 8\pi + 16 = 8(\pi + 2 ) cm

a. Area of sector MON = \(\frac{MON}{360} \times\pi r^2\) = \(\frac{60}{360}\times\frac{22}{7}\times7^2\) = \(\frac{77}{3}cm^2\) \(\frac{NT}{7}\) = sin60^\circ NT = 7sin60^\circ = \(7\times\frac{\sqrt{3}}{2}\) = \(7\times\frac{\sqrt{3}}{2}\)cm. \(\frac{OT}{9} = cos60^0\) OT = 7cos60^\circ = \(7\frac{1}{2}\)cm Area of △ONT = \(\frac{1}{2}\times OT\times NT\) = \(\frac{1}{2}\times\frac{7}{2}\times\frac{7\sqrt{3}}{2} = 49\sqrt{3}{8}cm^2\) Area of shaded portion = Area of sector MON - Area of △ONT = \(\frac{77}{3} - \frac{49\sqrt{3}}{8}\) = 25.67 - 10.61 = \(15.06cm^2\) = 15.1\(cm^2\) b. Given : y = mx + c; hence ; x = \(\frac{ y - c }{ m } = \frac{ y }{ m } - \frac{ c }{ m }\) x-intercept = \(\frac{ - c }{ m } = \frac{ - 3 }{ 4 }\) Therefore, m = \(\frac{4}{3}\)c ....................... (1) But y-intercept, c = \(\frac{2}{7}\) Therefore, m = \(\frac{4}{3}\times\frac{2}{7} = \frac{8}{21}\) Equation of straight: y = mx + c y = \(\frac{8}{21}x + \frac{2}{7}\) OR 21y = 8x + 6 = 8x - 21y = - 6

radius = 8cm , height = 14cm and \(\pi = \frac{22}{7}\) total surface area of a solid cylinder =\( 2\pirh+2\pir^2\) = 2\pir( h + r ) \( 2 \times \frac{22}{7} \times 8( 8 + 14)\) \( 2 \times \frac{22}{7} \times 8 \times 22\) \(\frac{7744}{7}\) = \(1,106.29cm^2\)

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).

∠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.

(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 = \(\frac{2}{5}\)

cos (a + 14)^\circ = sin (4a + 6)^\circ Recall, cos \theta = sin (90 - \theta) ⇒ sin (90 - (a + 14)) = sin (4a + 6) ⇒ sin (90 - a - 14) = sin (4a + 6) ⇒ sin (76 - a) = sin (4a + 6) ⇒ 76 - a = 4a + 6 ⇒ 76 - 6 = 4a + a ⇒ 70 = 5a \(a = \frac{70}{5}\) therefore, a = 14

From the diagram above; Tan\(\theta = \frac{opp}{adj}\) tan50^\circ = \(\frac{124}{d}\) d = \(\frac{124}{tan50}\) therefore, d = 104.05m

Tan\(\theta = \frac{opp}{adj}\) Tan\(\theta = \frac{56}{110}\) \(\theta = tan^1(0.5091)\) therefore, \(\theta\) = 27^\circ ( to the nearest degree)