Prealgebra (7th Edition)

$T=45$ square in.
The area of this figure can be calculated as the sum of the area of the square and the area of the parallelogram. $T=T_1+T_2$ Square: $T_1=a^2$ $T_1=5^2=25$ square in. Parallelogram: $T_2=h\times a=4\times 5=20$ square in $T=T_1+T_2=25+20=45$ square in.