Question

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?

$1+\dfrac{1}{2} \sqrt{2}$

$\sqrt{3}$

$\dfrac{7}{4}$

$\dfrac{15}{8}$

$2$

Answer: D

Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1420

Solution:

The crease is the perpendicular bisector of $\overline{AB}.$ Let $\overline{DE}$ be the crease.

By $AA$ similarity, $\triangle ADE\sim\triangle ACB.$ Therefore, $\dfrac{BC}{AC}=\dfrac{DE}{AD}.$ Plugging in the side lengths gives $\dfrac34=\dfrac{DE}{5/2},$ so $DE=\dfrac{15}{8}.$

Thus, D is the correct answer.

2018 AMC 10A Problem 13

Answer: D

Difficulty rating: 1420

Solution:

Problem 13 in Other Years