What's The Area?

In Rectangle $ABCD$ let $BM = y$ and $AN = x \implies MC = 16 - y$ and $ND = 16 - x$ .

Folding Rectangle $ABCD$ about line $MN$ so that point $C$ goes to point $A$ we obtain

where $\angle{AD'N}$ is a right angle so that $\triangle{AD'N}$ and $\triangle{ABM}$ are right triangles.

From the diagram above the Area of pentagon $ABMND' = Area(\triangle{ABM}) + Area(\triangle{AMN}) + Area(\triangle{AD'N})$

For $\triangle{ABM}$ :

$(16 - y)^2 = 16 + y^2 \implies 256 - 32y + y^2 = 16 + y^2 \implies 240 = 32y$

$\implies y = \dfrac{15}{2} = BM \implies \boxed{Area(\triangle{ABM}) = \dfrac{1}{2} * 4 * \dfrac{15}{2} = 15}$

For $\triangle{AD'N}$ :

$(16 - x)^2 + 16 = x^2 \implies 256 - 32x + x^2 + 16 = x^2 \implies 272 = 32x \implies x = \dfrac{17}{2}$ $\implies D'N = \dfrac{15}{2} \implies\boxed{Area(\triangle{AD'N}) = \dfrac{1}{2} * 4 * \dfrac{15}{2} = 15}$

For $\triangle{AMN}$ :

$\boxed{Area(\triangle{AMN}) = \dfrac{1}{2} * 4 * \dfrac{17}{2} = 17}$

$\implies$ Area of pentagon $ABMND' = Area(\triangle{ABM}) + Area(\triangle{AMN}) + Area(\triangle{AD'N}) = \boxed{47}$ .