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Let the slant height = L ; height = 24cm ; r = 7cm. \(\therefore L^{2} = 24^{2} + 7^{2} = 576 + 49 = 625\) \(L = \sqrt{625} = 25cm\) Surface area of the structure = Area of cone + Area of hemisphere = \(\pi r L + 2 \pi r^{2} = \pi r (L + 2r)\) = \(\frac{22}{7} \times 7 (25 + 2(7))\) = \(22 \times 39 = 858 cm^{2}\)

(a) (i) Area of minor segment POQ = \(\frac{84}{360} \times \frac{22}{7} \times (3.2)^{2}\) = \(\frac{11 \times 10.24}{15}\) = \(7.5093 cm^{2} \approxeq 7.51 cm^{2}\) (2 decimal places) (ii) Area of \(\Delta\) in POQ = \(\frac{1}{2} \times 3.2 \times 3.2 \times \sin 84^\circ\) = \(\frac{127.296}{25} = 5.0918 cm^{2}\) Area of segment PQ (shaded part) = Area of sector POQ - Area of triangle = \(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\) \(\approxeq 2.42 cm^{2}\) (b) Area of sector POQ = Curved surface area of the cone \(7.5093 cm^{2} = \pi rl\) where r = base radius, L = slant height = 3.2cm \(7.5093 = \frac{22}{7} \times r \times 3.2 = \frac{70.4r}{7}\) \(70.4r = 7.5093 \times 7 \implies r = \frac{7.5093 \times 7}{70.4} = 0.7467 cm\) \(\approxeq 0.7cm\)

(a)(i) Given a circle APB with centre O. To prove that < AOB = 2 < APB. Construction: Draw PO and produce it to Q. Proof : Referring to the lettering in the figure, |OA| = |OB| (radius) \(\therefore x_{1} = x_{2}\) (base angles of isosceles triangle) \(< AOQ = x_{1} + x_{2}\) (exterior angles of triangle AOP) \(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\) Similarly, \(< BOQ = 2y_{2}\) \(< AOB = < AOQ + < BOQ \) Hence, \(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\) = \(2 \times < APB\) (ii) From the figure, \(|OQ| = |QT|\) (radius) \(< OQT = 22^\circ = < QTO\) (base angles of isosceles triangle) \(\therefore < POQ = 44^\circ \) (exterior angle of \(\Delta\) QOT) \(< QOR = 180^\circ - (44^\circ + 98^\circ) = 180^\circ - 142^\circ = 38^\circ\) (Angles on a straight line) \(< QOR = 2 \times < QSR = 38^\circ\) \(\therefore < QSR = \frac{38^\circ}{2} = 19^\circ\) (b) \(< BHC = (180^\circ - 70^\circ) = 110^\circ\) (angle on a straight line) \(< DAC = 41^\circ = < DBC\) (angles on the same segment) But \(< DBC = < HBC\) \(< AHB = < HBC + < HCB \) (exterior angles of \(\Delta\) BHC) \(70^\circ = 41^\circ + < HCB \implies < HCB = 70^\circ - 41^\circ = 29^\circ\) \(< ACB = < HCB = 29^\circ\)

(101.5)2 - (100.5)2;(101.5 + 100.5) 202 x 1 = 202

36\(^\frac{1}{2}\) x 64\(^{-\frac{1}{3}}\) x 5\(^0\) = \(6 \times \frac{1}{4} \times 1\) = \(1\frac{1}{2}\)

From the diagram: PQ = 3 ; PR = 3+2 → 5 PQ: PR = \(\frac{3}{5}\)

Let \(\theta\) be the angular difference between P (50^\circN. 50^\circW), and Q (50^\circN, 150^\circW), \(\theta\) = 150^\circ - 50^\circ =100^\circ

No expanded working has been attached for this question yet. The verified answer is Option C.

PQ\(^2\) = 24\(^2\) + 7\(^2\) PQ\(^2\) = 625 PQ = 25

(271/3)2 = (33 x 1/3)2 = 32 = 9

Error = 1.25 - 1.15 = 0.10 0.10/1.25* 100 = 8.0%

The shape is a prism so volume of the tank = base area x height = \(\frac{1}{2}\)( a + b) x height = \(\frac{1}{2}\)(4 + 7) x 4 x 6 = 132cm\(^3\) But 1000cm\(^3\) = 1litres 132cm\(^3\) =

log106 + log1045 - log1027 log10 \(\frac{6 \times 45}{27}\) = log102 x 5 log1010 = 1

5y 2 + 5ay - 3ay - 3a 2 5y(y + a) - 3a(y + a) (y + a) (5y - 3a)