G-2 Let's Make a Crease

In answering this question, the ratio of its width to its length is involved in solving this problem which is 3:4.

After making a crease. The length of the crease is equivalent to the sum of its width (12 in) and the ratio of the dimension multiplied by the difference of its length and width $\frac{3}{4}$ $\times$ (16 in - 12 in) . Hence,

12 in + 3 in = $\boxed{15 in^{2}}$

Let the rectangle be $\Box ABCD$ such that $BC=12$ and $DC=16$ .

Whenever a piece of paper is folded to join two points on it, the crease thus obtained is the perpendicular bisector of the line joining these two points .

Hence, if we join the vertices $B$ and $D$ , the crease will be perpendicular to the diagonal $BD$ and will pass through the point of intersection of the diagonals $AC$ and $BD$ , let this point be $O$ .

Let the crease intersect the sides $AB$ and $CD$ in the points $E$ and $F$ respectively. So we conclude:

$\Rightarrow BO= \frac {BD} {2} \ldots \{1\}$

$\Rightarrow OF=\frac{EF} {2} \ldots \{2\}$

Say diagonal $BD$ makes an angle $\theta$ with side $CD$ .

In $\bigtriangleup BOF$ , $\angle BOF= 90^\circ$

and $\angle OBF=\theta$

$\Rightarrow OF= \tan\theta \times BO \ldots \{3\}$

But in $\bigtriangleup BCD$ ,

$\tan\theta=\frac {BC} {DC} \ldots \{4\}$

From Pythagoras' Theorem , we get

$BD=20$

Using equations $\{1\}$ , $\{2\}$ , $\{3\}$ and $\{4\}$ , we conclude:

$EF=2 \times \frac{BC} {DC} \times \frac {BD} {2}$

$\Rightarrow EF=2 \times \frac {12}{16} \times \frac {20}{2}$

$\Rightarrow EF=\boxed{15}$ .