1995 - WAEC Mathematics Past Questions and Answers - page 3

Area of rhombus = \(\frac{pq}{2}\)

where p and q are the diagonals of the rhombus.

\(\therefore A = \frac{16 \times 12}{2}\)

= 96 cm\(^2\)

Volume of water taken by the tank = \(\frac{3}{2} \times \frac{3}{2} \times \frac{1}{2}\)

= \(\frac{9}{8} m^3\)

= \(\frac{9}{8} \times 1000\)

= 1125 litres

\(\therefore\) Water left = (1500 - 1125) litres

= 375 litres.

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° - 15° 

= 50°

The length of the arc subtended by the sector of angle 120° = 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 = \(\frac{\theta}{360°} \times 2\pi r + 2r\)

= \(\frac{108}{360} \times 2 \times \frac{22}{7} \times \frac{7}{2} + 2(\frac{7}{2})\)

= \(6 \frac{3}{5} + 7\)

= \(13 \frac{3}{5} cm\)