Geometry – Finding Length of AR in Triangle ABC

euclidean-geometrygeometrytriangles

I need your help with a question on triangles. It is from the entrance exam of an institute in India for students in 10th grade.

In the given triangle ABC, if
DP, MQ AR and ES are perpendiculars to BC such that DP = 6 cm, MQ = 4 cm and ES = 8 cm
Then find AR.

The image of the figure.

The image of the figure👆

The image of the original question with options 👇
The original question with options

MY ATTEMPT:
I tried to use Pythagoras theorem and tried substituting Pythagorean triplets to find the base.
Then I also tried similarity of triangles â–³DPC & â–³MQC and â–³ESB &â–³MQB to find the base BC.
I even tried to join AM and then extend it to intersect BC at F so as to use the Menelaus theorem and Ceva theorem. However, that didn't work for me too.

Any help would mean a lot to me.

Best Answer

Like you thought, this is a question about similar triangles, so you should write down whatever ratios you know. You've implied a few of them already, and those are the important ones. You could be really close to a solution, but are just missing one important observation.
Unforuntately, this isn't a question about nice Pythagorean triples like you thought. Pythagorean theorem might be helpful, but it won't be my first choice.

The hint I gave was

Let $AM$ intersect $BC$ at $F$. What is $\frac{ FM}{FA} $?

If we knew the answer to the hint, then using similar triangles, we can obtain $ \frac{FM}{FA} = \frac{ MQ}{AR} = \frac{4}{AR}$. So it remains to find $ \frac{ FM}{FA}$.

Recall the well-known fact

$$\frac{FM}{FA} + \frac{ EM}{EB} + \frac{DM}{DC} = 1.$$

Hence, conclude that $ \frac{ FM}{FA} = \frac{1}{6} $ and so $AR = 24$.

Note

  • The well-known fact can be proved using areas.
  • Given that OP quoted Menelaus and Ceva's Theorem, I believe that they have seen this fact before.
Related Question