The base of a triangular piece of paper $A B C$ is $12 \mathrm{~cm}$ long. The paper is folded down over the base, with the crease $D E$ parallel to the base of the paper. The area of the triangle that projects below the base is $16 \%$ that of the area of the triangle $A B C$. The length of $D E$, in $\mathrm{cm}$, is
(A) 9.6
(B) 8.4
(C) 7.2
(D) 4.8
(E) 6.96

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-113.jpg?height=452&width=466&top_left_y=1411&top_left_x=1293)
Solution 1

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

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-208.jpg?height=445&width=466&top_left_y=1423&top_left_x=884)

We are told that the area of $\triangle X Y Z$ is $16 \%$ that of the area of $\triangle A B C$.

Now $\triangle A C B$ is similar to $\triangle X Z Y$, since $\angle X Z Y$ is the folded over version of $\angle A C B$ and since $\angle X Y Z=\angle E Y B=\angle D E Y=\angle C E D=\angle C B A$ by parallel lines and folds.

Since $\triangle X Z Y$ is similar to $\triangle A C B$ and its area is $0.16=(0.4)^{2}$ that of $\triangle A C B$, then the sides of $\triangle X Z Y$ are 0.4 times as long as the sides of $\triangle A C B$.

Draw the altitude of $\triangle A C B$ from $C$ down to $P$ on $A B$ (crossing $D E$ at $Q$ ) and extend it through to $Z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-209.jpg?height=455&width=469&top_left_y=149&top_left_x=882)

Now $C P=C Q+Q P=Z Q+Q P=Z P+2 P Q$.

Since the sides of $\triangle X Z Y$ are 0.4 times as long as the sides of $\triangle A C B$, then $Z P=0.4 C P$.

Since $C P=Z P+2 P Q$, then $P Q=0.3 C P$, and so $C Q=C P-P Q=0.7 C P$.

Since $C Q$ is 0.7 times the length of $C P$, then $D E$ is 0.7 times the length of $A B$, again by similar triangles, so $D E=0.7(12)=8.4$.

Solution 2

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

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-209.jpg?height=444&width=466&top_left_y=1163&top_left_x=884)

We are told that the area of $\triangle X Y Z$ is $16 \%$ that of the area of $\triangle A B C$.

Now $\triangle A C B$ is similar to $\triangle X Z Y$, since $\angle X Z Y$ is the folded over version of $\angle A C B$ and since $\angle X Y Z=\angle E Y B=\angle D E Y=\angle C E D=\angle C B A$ by parallel lines and folds.

Since $\triangle X Z Y$ is similar to $\triangle A C B$ and its area is $0.16=(0.4)^{2}$ that of $\triangle A C B$, then the sides of $\triangle X Z Y$ are 0.4 times as long as the sides of $\triangle A C B$.

Draw perpendiculars to $A B$ at $X$ and $Y$, intersecting $A C$ and $B C$ and $P$ and $Q$, respectively, and $D E$ at $R$ and $S$, respectively.

By symmetry, $P Q$ and $R S$ are parallel to $X Y$ and the same length, so let $P Q=R S=X Y=s$. Since the sides of $\triangle X Z Y$ are 0.4 times as long as the sides of $\triangle A C B$, then $s=0.4 \times 12=4.8$. Since $\triangle C D E$ is congruent to $\triangle Z D E$ (since one is the folded over version of the other), then by symmetry, $P R=R X$ and $Q S=S Y$.

Let $D R=x$ and $E S=y$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-210.jpg?height=455&width=469&top_left_y=149&top_left_x=882)

Then $A X=2 x$, since $\triangle P X A$ is similar to $\triangle P R D$ and has sides twice as long (since $P X=2 P R$. Similarly, $B Y=2 y$.

Now looking at $A B$ as a whole, we have $A B=2 x+s+2 y=12$, so $x+y=\frac{1}{2}(12-s)=3.6$. Looking at $D E$, we have $D E=s+x+y=4.8+3.6=8.4$.