If a $12\times 16$ sheet of paper is folded on its diagonal, what is the area of the region of the overlap? I have tried this problem and keep on getting $96$ as my answer, where the correct answer is $75$. 
Problem: 

If a $12\times 16$ sheet of paper is folded on its diagonal, what is the area of the region of the overlap (the region where paper is on top of paper)?

 A: The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.

A: Referring to the Diagram from Aretino
By the Pythagorean theorem the diagonal [AC] is 20 inches.
Therefore 1/2 the diagonal [AH] is 10 inches.  
By similar triangles:  [AEH] is similar to [ACB].
$$\frac{EH}{AH} = \frac{BC}{AB}  \Rightarrow \frac {h} {10} = \frac {12}{16} $$ 
Solve for height 
$$h=\frac{120}{16}=7.5$$ 
Solve for area of triangle [AEC]
$$Area=\frac 1 2 \cdot Base \cdot Height \Rightarrow \frac 1 2 \cdot AC \cdot EH \Rightarrow \frac 12 \cdot 20 \cdot 7.5 = 75 $$ 
A: Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.
A: The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.
A: The diagonal is $20$ inches long. (Pythagorean Theorem).
The angles well be $A= \arcsin \frac {12}{20}=\arccos \frac {16}{20}$ and $B= \arcsin \frac {16}{20}=\arccos \frac {12}{20}$ with $A < B$ and $A + B = 90^\circ$.  (Basic trig definitions.  Draw a picture to verify.)

The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$.  This cuts in half into two congruent right triangles.
The proportions of one of these right triangles is Hypotenuse = $h$.  Base = $h*\cos A= h*\cos\arccos \frac {16}{20} = h*\frac 45 = 10$.  And height will = $h*\sin A=h\sin \arcsin \frac {12}{20} = h*\frac 35$.
$h* \frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*\frac 35 = 7.5$.
And the area of the triangle is $\frac 12*20*7.5 = 75$.
