Try this beautiful problem from Geometry based on Side Length of Rectangle.

## Side Length of Rectangle – AMC-10A, 2009- Problem 17

Rectangle $A B C D$ has $A B=4$ and $B C=3 .$ Segment $E F$ is constructed through $B$ so that $E F$ isperpendicular to $D B$, and $A$ and $C$ lie on $D E$ and $D F$, respectively. What is $E F$ ?

- $9$
- $10$
- $\frac{125}{12}$
- \(\frac{103}{9}\)
- \(12\)

**Key Concepts**

Triangle

Rectangle

Geometry

## Check the Answer

But try the problem first…

Answer: $\frac{125}{12}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

First hint

We have to find out the length of \(EF\)

Now $BD$ is the altitude from $B$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$. ( as \(\triangle BDE \sim \triangle BDF\)).so we have to find out \(BE\) and \(BF\)

Can you now finish the problem ……….

Second Hint

Now Clearly, $\triangle BDE \sim \triangle DCB$. Because of this, $\frac{A B}{C B}=\frac{E B}{D B}$. From the given information and the Pythagorean theorem, $A B=4, C B=3$, and $D B=5 .$ Solving gives $E B=20 / 3$

We can use the above formula to solve for $B F . B D^{2}=20 / 3 \cdot B F$. Solve to obtain $B F=15 / 4$

can you finish the problem……..

Final Step

Therefore $E F=E B+B F=\frac{20}{3}+\frac{15}{4}=\frac{80+45}{12}$

## Other useful links

- https://www.cheenta.com/problem-from-probability-amc-8-2004problem-no-21/
- https://www.youtube.com/watch?v=VLyrlx2DWdA&t=15s

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