Elementary Technical Mathematics

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

Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise: 60

Answer

a)$\frac{31}{32}$in. b) 16$\frac{5}{16}$ in.

Work Step by Step

Label diagram as ABCDEF a) Length of missing dimension EF = DC – AB = 3$\frac{9}{32}$ in. - 2$\frac{5}{16}$ in. = $\frac{105}{32}$ - $\frac{37}{16}$ = $\frac{105}{32}$ - $ \frac{37}{16} \times$ $\frac{2}{2}$ =$\frac{105}{32}$ - $\frac{74}{32}$ =$\frac{31}{32}$in. b) Perimeter of the figure = AB+BC+CD+DE+EF+FA = 2$\frac{5}{16}$in + 2$\frac{1}{2}$in + 2$\frac{3}{8}$in + 3$\frac{9}{32}$in + 2$\frac{3}{8}$in + $\frac{31}{32}$in + 2$\frac{1}{2}$in = $\frac{37}{16}$ + $\frac{5}{2}$ + $\frac{1}9{8}$+$\frac{105}{32}$ + $\frac{19}{8}$ + $\frac{31}{32}$i + $\frac{5}{2}$ =$\frac{37}{16}$ + $\frac{10}{2}$ + $\frac{38}{8}$+$\frac{136}{32}$ = $ \frac{37}{16} \times$ $\frac{2}{2}$ + $ \frac{10}{2} \times$ $\frac{16}{16}$ + $ \frac{38}{8} \times$ $\frac{4}{4}$ + $\frac{136}{32}$ =$\frac{74}{32}$ + $\frac{160}{32}$ + $\frac{152}{32}$+$\frac{136}{32}$ =$\frac{522}{32}$ =$\frac{261}{16}$ =16$\frac{5}{16}$ in
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