Paper Strip

Label the rectangle and its fold as follows:

Since $\angle DEA'$ and $\angle BEC$ are vertical angles, $\angle DEA' \cong \angle BEC$ . Since $\angle DA'E$ and $\angle BCE$ are right angles, $\angle DA'E \cong \angle BCE$ . Also, $A'D \cong AD \cong BC$ . Therefore, $\triangle BCE \cong \triangle DA'E$ by AAS congruence. This means that $BE = DE = x$ .

Then by Pythagorean's Theorem on $\triangle BCE$ , $12^2 + (24 - x)^2 = x^2$ , which solves to $x = 15$ .

Therefore, the area of the overlapping triangle is $A = \frac{1}{2}\cdot 15 \cdot 12 = \boxed{90}$ .

Let the paper strip be $ABCD$ , $M$ be the point of intersection of $AB$ and $CD$ after the fold along the diagonal and $MN$ be perpendicular to $BD$ .

We note that $\triangle BCD$ and $\triangle MND$ are similar. Therefore, $\dfrac {MN}{DN} = \dfrac {BC}{CD} = \dfrac {12}{24} = \dfrac 12$ .

Since $BD$ is the diagonal, by Pythagorean theorem , $BD = \sqrt{12^2 + 24^2} = 12 \sqrt 5 \text{ cm}$ , and $DN = \dfrac 12 BD = 6 \sqrt 5 \text{ cm}$ , and $MN = \dfrac 12 DN = 3\sqrt 5 \text{ cm}$ .

Now, the area of the overlapped part is $A_\triangle = \dfrac 12 \times BD \times MN = \dfrac 12 \times 12 \sqrt 5 \times 3 \sqrt 5 = \boxed{90} \text{ cm}^2$ .

Label the rectangular strip and its coordinates by O(0,0), A(0,12),B(24,12), D(24,0). The equation of line )B is y = x/2. Let P be the point on )B such that AP is perpendicular to OB. The equation of line AP is y = 12 - 2x. Solving the equations simultaneously gives the coordinates of P +(4.8, 2.4). When the fold is made, point A is reflected to point A' with coordinates (9.6, -7.2). Let point C be the intersection of BA' and the x axis. C has coordinates (15,0). The overlap is the triangle OCB. Its area is the area of OBD - area of CBD = 144 - 54 = 90. Ed Gray