Try this beautiful problem from Geometry and solve it by measuring the length in triangle.

## Measuring the length in Triangle- AMC-10B, 2011- Problem 9

The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$. What is $BD$?

- \(8\sqrt 3\)
- \(\frac {4\sqrt3}{3}\)
- \(6\sqrt 3\)

**Key Concepts**

Geometry

Triangle

similarity

## Check the Answer

But try the problem first…

Answer: \(\frac{ 4\sqrt 3}{3}\)

AMC-10B (2011) Problem 9

Pre College Mathematics

## Try with Hints

First hint

We have to find out the length of \(BD\). The given informations are “The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$”

If you notice very carefully about the side lengths of the \(\triangle ABC\) then \(AC=3,BC=4,AB=5\) i.e \((AC)^2+(AB)^2=(3)^2+(4)^2=25=(AB)^2\)……..So from the pythagorean theorm we can say that \(\angle ACB=90^{\circ} \)

Therefore area of \(\triangle ACB=\frac{1}{2} \times 3 \times 4=6\)

so area of the \(\triangle BDE=\frac{1}{3} \times 6=2\)

Now the \(\triangle BDE\) and \(\triangle ABC\) If we can show that two triangles are similar then we will get the value of \(BD\).Can you prove \(\triangle BDE \sim \triangle ABC\) ?

Can you now finish the problem ……….

Second Hint

In \(\triangle BDE\) & \(\triangle ACB\) we have…..

\(\angle B=X\) \(\Rightarrow \angle BED=(90-x)\) and \(\angle CAB=(90-X)\) (AS \(\angle ACB=90\) & sum of the angles of a triangle is 180)

Therefore \(\triangle BDE \sim \triangle BCD\)

can you finish the problem……..

Final Step

**The value of BD:**

Now \(\triangle BDE \sim \triangle BCD\) \(\Rightarrow \frac{(BD)^2}{(BC)^2}=\frac{Area of \triangle BDE}{Area of triangle ACB}\) =\(\frac{(BD)^2}{16}=\frac{2}{6}\)

So \(BD=\frac{ 4\sqrt 3}{3}\)

## Other useful links

- https://www.cheenta.com/ratio-of-lcm-gcf-algebra-amc-8-2013-problem-10/
- https://www.youtube.com/watch?v=PfRqs9W8nPQ

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