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

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 $BD$?

Measuring the length in Triangle- Problem

$8\sqrt 3$
$\frac {4\sqrt3}{3}$
$6\sqrt 3$

Key Concepts

Geometry

Triangle

similarity

Check the Answer

Answer: $\frac{ 4\sqrt 3}{3}$

AMC-10B (2011) Problem 9

Pre College Mathematics

Try with Hints

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 ..........

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........

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 $BD=\frac{ 4\sqrt 3}{3}$

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