1990 - JAMB Mathematics Past Questions and Answers - page 5

yzx + 105^o + 30^o = 180^o

yzx = 180 - 155 = 45^o

\(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\)

yz = \(\frac{8 \sin 30}{sin 45}\)

= 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\)

= 4 \(\div\) \(\frac{1}{\sqrt{2}}\)

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

\(\frac{x}{4} = \frac{6 + x}{6}\)

6x = 4(6 + x) = 24 + 4x

x = 12 = c = \(\pi RL - \pi L\)

= \(\pi (6) \sqrt{18^2} + 6^2 - \pi \times 4 \times \sqrt{12^2} + 4^2\)

= 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\)

= 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\)

= 20\(\pi \sqrt{10}\)cm²