Elementary Technical Mathematics

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

Chapter 6 - Section 6.8 - Substituting Data into Formulas - Exercises - Page 258: 15

Answer

$$ \frac{2A-ah}{h}, \ \ 49.0 .$$

Work Step by Step

To solve the given equation for $b$, we multiply both sides by $2$ and then we follow the following steps: $$ A=\frac{(a+b)}{2}h \Longrightarrow 2 A=(a+b) h \Longrightarrow 2 A=ah+bh\\ \Longrightarrow 2A-ah=bh\\ \Longrightarrow b=\frac{2A-ah}{h}.$$ Now, we have $$ b=\frac{2A-ah}{h}=\frac{2\times1160-22.0\times56.5}{22.0}=48.95\approx 49.0 .$$
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