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I am stumped on the following question:

In the figure below $AD=4$ , $AB=3$ , and $CD=9$. What is the area of Triangle AEC? enter image description here

I need to solve this using trigonometric ratios however if trig. ratios makes this problem a lot easier to solve I would be interested in looking how that would work. Currently I am attempting to solve this in the following way

$\bigtriangleup ACD_{Area}=\bigtriangleup ECD_{Area} + \bigtriangleup ACE_{Area}$

The only problem with this approach is finding the length of AE or ED. How would I do that ? Any suggestion or is there a better method

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  • $\begingroup$ Consider using similar triangles. $\endgroup$ – Joe Z. Dec 15 '12 at 2:42
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$AE+ED=AD=4$, and $ECD$ and $EAB$ are similar (right?), so $AE$ is to $AB$ as $ED$ is to $CD$. Can you take it from there?

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  • $\begingroup$ Thanks I believe I can take it from there .. though . How did you know that ECD and EAB are similar ? $\endgroup$ – Rajeshwar Jul 24 '12 at 3:40
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    $\begingroup$ They are right-angled, and share an angle at $E$ (so all corresponding pairs of angles are equal, so triangles are similar). $\endgroup$ – Gerry Myerson Jul 24 '12 at 4:52
  • $\begingroup$ Angle-Angle. Thanks This helped $\endgroup$ – Rajeshwar Jul 24 '12 at 9:50
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Hint: $\bigtriangleup DCE \sim \bigtriangleup ABE$ as $BC$ is crossing parallel lines.

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