## Basic College Mathematics (10th Edition)

$\frac{49}{32}in^{2}$
Area = Length $\times$ width = $1\frac{3}{4}in \times \frac{7}{8}in$ = $\frac{7}{4} \times \frac{7}{8}in^{2}$ = $\frac{7\times7}{4\times8}in^{2}$ (multiply nominator and denominator) = $\frac{49}{32}in^{2}$