## Prealgebra (7th Edition)

Published by Pearson

# Chapter 9 - Section 9.3 - Area, Volume, and Surface Area - Practice: 2

#### Answer

$A=48\frac{4}{5}$ sq yd

#### Work Step by Step

$A=\frac{(a+b)}{2}h$ $A=\frac{(11+5)}{2}\times6.1$ $A=\frac{16}{2}\times6\frac{1}{10}$ $A=\frac{8}{1}\times\frac{61}{10}$ $A=\frac{4}{1}\times\frac{61}{5}$ $A=\frac{244}{5}$ $A=48\frac{4}{5}$ sq yd

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