WAEC Mathematics Past Questions & Answers - Page 63

Using the two - point from

\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)

\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)

\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)

\(\frac{-12(y -2)}{-4}\) = x - 4

3(y -2) = x -4

3y - 6 = x - 4

3y = x - 4 + 6

3y = x + 2...

By comparing the equations;

3y = px + , p = 1

|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)

= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\)

= \(\sqrt{(-8)^2 + (8 + 7)^2}\)

= \(\sqrt{64 + (15)^2}\)

= \(\sqrt{64 + 225}\)

= \(\sqrt{289}\)

= 17 units

x +  2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)

8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\)

8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\)

8x = 720\(^o\) - 40\(^o\) = 680\(^o\)

x = \(\frac{680^o}{8}\)

= 85\(^o\)

Surface area of a sphere = \(4 \pi r^2\) \(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number)

Volume of a cylinder = \( \pi r^2\)h

385 = \(\frac{22}{7}\) x \(r^2\) x 10

385 x 7 = 22 x \(r^2\) x 10

\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)

= 12.25

r = \(\sqrt{12.25}\)

= 3.5m

Hence, diameter of tank = 2r

= 2 x 3.5 = 7m