2018 AMC 12A Problem 11

Problem 11 of 25IntermediateGeometry

A paper triangle with sides of lengths

3,

4,

and

5

inches, as shown, is folded so that point

A

falls on point

B.

What is the length in inches of the crease?

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Solution

The crease lies along the perpendicular bisector of

AB,

meeting

AC

E

because

AC \gt BC.

Let

D

be the midpoint of

AB,

AD = \tfrac52

and

\triangle ADE

is right-angled at

D.

Since

\triangle ADE \sim \triangle ACB,

we have

\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34,

DE = \frac52 \cdot \frac34 = \frac{15}{8}.

Thus, the correct answer is D.