Answer:
[tex]\large\boxed{a)\ \dfrac{31}{32}\ in}\\\boxed{b)\ 16\dfrac{5}{16}\ in}[/tex]
Step-by-step explanation:
Look at the picture.
[tex]a)\\3\dfrac{9}{32}\ in-2\dfrac{5}{16}\ in=3\dfrac{9}{32}\ in-2\dfrac{5\cdot2}{16\cdot2}\ in=3\dfrac{9}{32}\ in-2\dfrac{10}{32}\ in\\\\=\bigg((3-2)+\dfrac{9-10}{32}\bigg)\ in=\bigg(1+\dfrac{-1}{32}\bigg)\ in=\bigg(\dfrac{32}{32}-\dfrac{1}{32}\bigg)\ in\\\\=\dfrac{32-1}{32}\ in=\dfrac{31}{32}\ in[/tex]
b)
The perimeter of given figure is equal to the perimeter of the rectangle (look at the second picture).
[tex]P=2\left(3\dfrac{9}{32}\ in+2\dfrac{1}{2}\ in+2\dfrac{3}{8}\ in\right)=2\left(3\dfrac{9}{32}+2\dfrac{1\cdot16}{2\cdot16}+2\dfrac{3\cdot4}{8\cdot4}\right)\ in\\\\=2\bigg(3\dfrac{9}{32}+2\dfrac{16}{32}+2\dfrac{12}{32}\bigg)\ in=2\left((3+2+2)+\dfrac{9+16+12}{32}\right)\ in\\\\=2\left(7+\dfrac{37}{32}\right)\ in=\bigg((2)(7)+(2)\left(\dfrac{37}{32}\right)\bigg)\ in=\left(14+\dfrac{37}{16}\right)\ in\\\\=\bigg(14+2\dfrac{5}{16}\bigg)\ in=16\dfrac{5}{16}\ in[/tex]
