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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

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

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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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

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

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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