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

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

T.S.A of a cone = \(\pi r^2 + \pi rl\) = \(\frac{22}{7} \times 3^2 + \frac{22}{7} \times 3 \times 4\) = \(\frac{198}{7} + \frac{264}{7}\) = \(\frac{462}{7}\) = 66 cm\(^2\)

\(Volume = \frac{4}{3} \pi r^3 = k\) ...(i) \(S.A = 4\pi r^2 = k\) ... (ii) Divide (i) by (ii), \(\frac{4}{3} \pi r^3 \div 4\pi r^2 = \frac{k}{k}\) \(\frac{r}{3} = 1 \implies r = 3cm\) Diameter = 2 x 3cm = 6cm

Considering the smaller and larger triangle, these two are similar triangles. Hence, If the height of the smaller triangle = h, \(\therefore \frac{h}{6} = \frac{h + 12}{12}\) \(12h = 6h + 72 \implies 6h = 72\) \(h = 12 cm\) \(\therefore\) The height of the cone = 12 + 12 = 24 cm

Latitudinal difference = 65^\circ - 15^\circ = 50^\circ

The length of the arc subtended by the sector of angle 120^\circ = circumference of the base of the cone. \(\frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = 2\pi r\) \(44 = 2\pi r\) \(r = 22 \div \pi\) \(r = 22 \times \frac{7}{22}\) r = 7 cm

Perimeter of sector = \(\frac{\theta}{360^\circ} \times 2\pi r + 2r\) = \(\frac{108}{360} \times 2 \times \frac{22}{7} \times \frac{7}{2} + 2(\frac{7}{2})\) = \(6 \frac{3}{5} + 7\) = \(13 \frac{3}{5} cm\)

< DBC = 35^\circ (base angles of an isosceles triangle) < CDB = 180^\circ - (35^\circ + 35^\circ) = 110^\circ < ADB = 70^\circ; < ABD = 90^\circ(< in a semi-circle) \(\therefore\) < BAO = 180^\circ - (70^\circ + 90^\circ)( sum of

Source: Myschool Past Questions Archive (Record #23748)