2013 - WAEC Mathematics Past Questions and Answers - page 4

Volume of a cone = (\frac{1}{3} \pi r^2h)

h² = 5² = 3²

= 25 - 9 = 16

h = (\sqrt{16})

h = 4cm

v = (\frac{1}{3} \times \frac{22}{7} \times 3^2 \times 4)

(\frac{1}{3} \times \frac{22}{7} \times 9 \times 4)

= (\frac{22 \times 3 \times 4}{7})

= 37.7cm³

= 38cm³

1km = 100,000cm

on the map 1 cm represent every 10 km which is equal to (10 x 100,000cm)

= 1,000,000cm

the scale is 1:1,000,000

(x = 2)(x² - 3x + 2) + 2(x + 2)(x - 1) = (x + 2)m

(m + 2)[(x² - 3x + 2) + 2(x - 1)] = (x + 2)M

divide both side by (x + 2)

(x² - 3x + 2) + 2(x - 1) = M

x² - 3x + 2 + 2x - 2 = M

x² - 3x + 2 + 2x - 2 = M

x² - 3x + 2x = M

x² - x = M

M = x² - x

L² = 96² + 28²

= 9216 + 784

= 10000

L = (\sqrt{10000})

= 100cm

curved surface area = (\pi r l)

= (\frac{22}{7} \times 28 \times 100)

= 8800cm²

area of cone = area of sector

area of sector = 8800cm²

by Pythagoras x² = 6² + 8²

36 + 64 = 100

x = (\sqrt{100})

x = 10

The diagonals of rhombus bisects its angles

(\bigtriangleup) ABC, A + 153 = 180 (angles on straight line)

< A = 180 - 153 = 27

< B = 2m(vertically opposite angles)

< C = 7m (corresponding ngles)

< A + < B + < C = 180 (sum of int. angles of (\bigtriangleup))

i.e. 27 + 2m + 7m = 180^o

9m = 180 - 27

9m = 153

m = (\frac{153}{9})

= 17