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I have tried this problem and keep on getting $96$ as my answer, where the correct answer is $75$.

Problem:

If a $12times 16$ 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)?

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 $EH=15/2$.

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].
$$fracEHAH = fracBCAB Rightarrow frac h 10 = frac 1216 $$
Solve for height
$$h=frac12016=7.5$$
Solve for area of triangle [AEC]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$

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.

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.

The diagonal is $20$ inches long. (Pythagorean Theorem).

The angles well be $A= arcsin frac 1220=arccos frac 1620$ and $B= arcsin frac 1620=arccos frac 1220$ 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 = $h*cos A= h*cosarccos frac 1620 = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac 1220 = h*frac 35$.

$h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.

And the area of the triangle is $frac 12*20*7.5 = 75$.