(a) Without using tables, find the value of \(\frac{0.45 \times 0.91}{0.0117}\) (b) Find the number
(a) Without using tables, find the value of \(\frac{0.45 \times 0.91}{0.0117}\)
(b) Find the number which is exactly halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\).
(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?
(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \(π = \frac{22}{7}\)].
Explanation
(a) \(\frac{0.45 \times 0.91}{0.0117} = \frac{45 \times 91}{117}\)
= \(5 \times 7 = 35\)
(b) Halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\)
= \(\frac{1\frac{6}{7} + 2\frac{11}{28}}{2}\)
= \((\frac{13}{7} + \frac{67}{28}) \times \frac{1}{2}\)
= \(\frac{119}{28} \times \frac{1}{2} \)
= \(\frac{119}{56} = 2\frac{7}{56} = 2\frac{1}{8}\)
(c) Let the interior angle = 5x° and the exterior angle = x°.
\(\therefore 5x° + x° = 180°\)
\(6x° = 180° \implies x = 30°\)
Exterior angle = \(\frac{360°}{n} \implies 30° = \frac{360°}{n}\)
\(n = \frac{360°}{30°} = 12\)
The polygon has 12 sides.
(d) Length of pipe = 20 cm
Internal diameter = 6 cm ; radius = 3cm
External diameter = 8 cm ; radius = 4cm
Let the external radius = R and internal radius = r
\(\therefore\) Volume of material used = \(\pi R^{2} l - \pi r^{2} l\)
Where l = length.
\(Volume = \pi l (R^{2} - r^{2}) cm^{3}\)
= \(\frac{22}{7} \times 20 (4^{2} - 3^{2}) = \frac{22}{7} \times 20 \times 7 = 440 cm^{3}\)