Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 3 - Section 3.4 - Volume and Area - Exercise - Page 147: 52

Answer

$$12.5 \ ha$$

Work Step by Step

The area of the field is $$\frac{1}{4} \times \frac{1}{2} \ km^2=\frac{1}{8} \ km^2=\frac{10^6}{8} \ m^2 .$$ Now change it to $ha$ by dividing by $10^4$; that is $$\frac{10^6}{8} \ m^2= \frac{10^6}{8\times 10^4} \ ha =\frac{100}{8} \ ha=12.5 \ ha.$$
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