2002 - WAEC Mathematics Past Questions and Answers - page 1

Given, Length of minor arc = 5.5m

Angle subtended by minor arc = 360° - 210° = 150°

\(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \)

\(\frac{55r}{21} = 5.5\)

\(r = \frac{5.5 \times 21}{55}\)

r = 2.1m

Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\)

= \(7.7m \approxeq 8m\) (to the nearest metre)

\(BD^2 = 4^2 + 4^2\)

\(BD = \sqrt{16 + 16} = \sqrt{32}\)

\(BD = 4\sqrt{2} cm\)

\((2\sqrt{3})^2 = (2\sqrt{2})^2 + h^2\)

\(h^2 = 12 - 8 = 4\)

\(h = \sqrt{4} = 2 cm\)

Volume of pyramid = \(\frac{a^2 h}{3}\)

= \(\frac{4^2 \times 2}{3}\)

= \(\frac{32}{3} = 10\frac{2}{3} cm^3\)

Curved surface area or a cone \(=\pi rl\)
from the information \(l^2 = 4^2 + 3^2 = 16+9\
l = \sqrt{25} = 5; ∴ C.S.A\hspace{1mm} = \frac{22}{7}\times 3 \times 5\
Since \frac{22}{7}=\pi ∴ C.S.A\hspace{1mm} =\hspace{1mm}15\pi\)

< UQR = 29° (alternate angles)

< PQR = < PQU + < UQR

= 36° + 29°

= 65°

\(5\frac{1}{4}\div \left(1\frac{2}{3}- \frac{1}{2}\right)\
\frac{21}{4}\div \left(1\frac{4-3}{6}\right)\
\frac{21}{4}\div \left(1\frac{1}{6}\right)\
\frac{21}{4} \times \frac{6}{7}= 4\frac{1}{2}\)