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April 14, 2020

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




Check the Answer

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

AMC-10B (2011) Problem 9

Pre College Mathematics

Try with Hints

Find BD

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 ..........

Finding the measurement

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

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