Let $X$ and $Y$ be the points where the folded portion of the triangle crosses $AB,$ and $Z$ be the location of the original vertex $C$ after folding.

[asy]
draw((0,0)--(12,0)--(9.36,3.3)--(1.32,3.3)--cycle,black+linewidth(1));
draw((1.32,3.3)--(4,-3.4)--(9.36,3.3),black+linewidth(1));
draw((1.32,3.3)--(4,10)--(9.36,3.3),black+linewidth(1)+dashed);
draw((0,-5)--(4,-5),black+linewidth(1));
draw((8,-5)--(12,-5),black+linewidth(1));
draw((0,-4.75)--(0,-5.25),black+linewidth(1));
draw((12,-4.75)--(12,-5.25),black+linewidth(1));
label("12 cm",(6,-5));
label("$A$",(0,0),SW);
label("$D$",(1.32,3.3),NW);
label("$C$",(4,10),N);
label("$E$",(9.36,3.3),NE);
label("$B$",(12,0),SE);
label("$X$",(2.64,0),SW);
label("$Y$",(6.72,0),SE);
label("$Z$",(4,-3.4),W);
[/asy]

We are told that the area of $\triangle XYZ$ is $16\%$ that of the area of $\triangle ABC.$

Now $\triangle ACB$ is similar to $\triangle XZY,$ since $\angle XZY$ is the folded over version of $\angle ACB$ and since $$\angle XYZ=\angle EYB =\angle DEY = \angle CED = \angle CBA$$by parallel lines and folds. Since $\triangle XZY$ is similar to $\triangle ACB$ and its area is $0.16=(0.4)^2$ that of $\triangle ACB,$ the sides of $\triangle XZY$ are $0.4$ times as long as the sides of $\triangle ACB.$

Draw the altitude of $\triangle ACB$ from $C$ down to $P$ on $AB$ (crossing $DE$ at $Q$) and extend it through to $Z.$

[asy]
draw((0,0)--(12,0)--(9.36,3.3)--(1.32,3.3)--cycle,black+linewidth(1));
draw((1.32,3.3)--(4,-3.4)--(9.36,3.3),black+linewidth(1));
draw((1.32,3.3)--(4,10)--(9.36,3.3),black+linewidth(1)+dashed);
draw((0,-5)--(4,-5),black+linewidth(1));
draw((8,-5)--(12,-5),black+linewidth(1));
draw((0,-4.75)--(0,-5.25),black+linewidth(1));
draw((12,-4.75)--(12,-5.25),black+linewidth(1));
label("12 cm",(6,-5));
label("$A$",(0,0),SW);
label("$D$",(1.32,3.3),NW);
label("$C$",(4,10),N);
label("$E$",(9.36,3.3),NE);
label("$B$",(12,0),SE);
label("$X$",(2.64,0),SW);
label("$Y$",(6.72,0),SE);
label("$Z$",(4,-3.4),W);
draw((4,10)--(4,-3.4),black+linewidth(1));
label("$Q$",(4,3.3),NE);
label("$P$",(4,0),NE);
[/asy]

Now $CP=CQ+QP=ZQ+QP=ZP+2PQ.$

Since the sides of $\triangle XZY$ are $0.4$ times as long as the sides of $\triangle ACB,$ then $ZP=0.4CP.$

Since $CP=ZP+2PQ,$ we have $PQ=0.3CP,$ and so $CQ=CP-PQ=0.7CP.$

Since $CQ$ is $0.7$ times the length of $CP,$ then $DE$ is $0.7$ times the length of $AB,$ again by similar triangles, so $DE=0.7(12)=8.4}\text{ cm.$
X$와 $Y$는 삼각형의 접힌 부분이 $AB$와 교차하는 점이고, $Z$는 접힌 후의 원래 꼭짓점 $C$의 위치라고 합니다.

[asy]
draw((0,0)--(12,0)--(9.36,3.3)--(1.32,3.3)--cycle,black+linewidth(1));
draw((1.32,3.3)--(4,-3.4)--(9.36,3.3),black+linewidth(1));
draw((1.32,3.3)--(4,10)--(9.36,3.3),black+linewidth(1)+dashed);
draw((0,-5)--(4,-5),검정+선폭(1));
draw((8,-5)--(12,-5),검정+선폭(1));
draw((0,-4.75)--(0,-5.25),black+linewidth(1));
draw((12,-4.75)--(12,-5.25),black+linewidth(1));
label("12cm",(6,-5));
label("$A$",(0,0),SW);
label("$D$",(1.32,3.3),NW);
label("$C$",(4,10),N);
label("$E$",(9.36,3.3),NE);
label("$B$",(12,0),SE);
label("$X$",(2.64,0),SW);
label("$Y$",(6.72,0),SE);
label("$Z$",(4,-3.4),W);
[/asy]

우리는 $\삼각형 XYZ$의 넓이가 $\삼각형 ABC$의 넓이의 $16\%$라는 것을 알 수 있습니다.

이제 $\삼각형 ACB$는 $\삼각형 XZY$와 유사합니다. $\삼각형 XZY$는 $\삼각형 ACB$의 접힌 버전이고, 평행선과 접힌 선에 의해 $$\각 XYZ=\각 EYB =\각 DEY = \각 CED = \각 CBA$$가 되므로 $\삼각형 XZY$는 $\삼각형 ACB$와 유사합니다. 삼각형 XZY$는 삼각형 ACB$와 유사하고 그 넓이는 $0.16=(0.4)^2$이므로, $\삼각형 XZY$의 변은 $\삼각형 ACB$의 변보다 $0.4$배 길다.

삼각형 ACB$의 고도를 $C$에서 $AB$의 $P$까지 ($Q$에서 $DE$를 교차하여) 그리고 $Z$까지 연장합니다.

[asy]
draw((0,0)--(12,0)--(9.36,3.3)--(1.32,3.3)--cycle,black+linewidth(1));
draw((1.32,3.3)--(4,-3.4)--(9.36,3.3),black+linewidth(1));
draw((1.32,3.3)--(4,10)--(9.36,3.3),black+linewidth(1)+dashed);
draw((0,-5)--(4,-5),검정+선폭(1));
draw((8,-5)--(12,-5),검정+선폭(1));
draw((0,-4.75)--(0,-5.25),black+linewidth(1));
draw((12,-4.75)--(12,-5.25),black+linewidth(1));
label("12cm",(6,-5));
label("$A$",(0,0),SW);
label("$D$",(1.32,3.3),NW);
label("$C$",(4,10),N);
label("$E$",(9.36,3.3),NE);
label("$B$",(12,0),SE);
label("$X$",(2.64,0),SW);
label("$Y$",(6.72,0),SE);
label("$Z$",(4,-3.4),W);
draw((4,10)--(4,-3.4),검정+선폭(1));
label("$Q$",(4,3.3),NE);
label("$P$",(4,0),NE);
[/asy]

이제 $CP=CQ+QP=ZQ+QP=ZP+2PQ.$입니다.

삼각형 $\삼각형 XZY$의 변은 $\삼각형 ACB$의 변의 $0.4$배 길이이므로 $ZP=0.4CP.$가 됩니다.

CP=ZP+2PQ$이므로 $PQ=0.3CP$이므로 $CQ=CP-PQ=0.7CP.$가 됩니다.

CQ$는 $CP$ 길이의 $0.7$배이므로 $DE$는 다시 유사한 삼각형에 의해 $AB$ 길이의 $0.7$배가 되므로 $DE=0.7(12)=8.4}\text{ cm.$입니다.