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# Length of the crease | AMC 10A, 2018 | Problem No 13

Try this beautiful Problem on Geometry based on Length of the crease from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Length of the crease- AMC-10A, 2018- Problem 13

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?

,

• $1+\frac{1}{2} \sqrt{2}$
• $\sqrt 3$
• $\frac{7}{4}$
• $\frac{15}{8}$
• $2$

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-13

#### Check the answer here, but try the problem first

$\frac{15}{8}$

## Try with Hints

#### First Hint

Given that ABC is a right-angle triangle shape paper. Now by the problem the point $A$ move on point $B$ . Therefore a crease will be create i.e $DE$ . noe we have to find out the length of $DE$?

If you notice very carefully then $DE$ is the perpendicular bisector of the line $AB$. Therefore the $\triangle ADE$ is Right-angle triangle. Now the side lengths of $AC$,$AB$,$BC$ are given. so if we can so that the $\triangle ADE$ $\sim$ $\triangle ABC$ then we can find out the side length of $DE$?

Now can you finish the problem?

#### Second Hint

In $\triangle ABC$ and $\triangle ADE$ we have ...

$\angle A=\angle A$( common angle)

$\angle C=\angle ADE$ (Right angle)

Therefore the remain angle will be equal ....

Therefore we can say that $\triangle ADE$ $\sim$ $\triangle ABC$

Now Can you finish the Problem?

#### Third Hint

As $\triangle ADE$ $\sim$ $\triangle ABC$ therefore we can write

$\frac{B C}{A C}=\frac{D E}{A D} \Rightarrow \frac{3}{4}=\frac{D E}{\frac{5}{2}} \Rightarrow D E=\frac{15}{8}$

Therefore the length in inches of the crease is $\frac{15}{8}$

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