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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.

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

Pre College Mathematics

AMC-10A, 2018 Problem-13

$\frac{15}{8}$

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?

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?

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 $\frac{15}{8}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

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