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In triangle $\triangle LMN$. $LO$ is median. Also $LO$ is the bisector of angle. If $LO=3\text{ cm}$ and $LM=5\text{ cm}$. Then find the area of triangle $\triangle LMN$.

Now here we will use similarity between $\Delta LMO$ and $\Delta LON$ triangles but I could not proceed further.

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    $\begingroup$ Hint: $3^2+4^2=5^2$ $\endgroup$
    – Henry
    Commented Jan 14, 2017 at 10:50
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    $\begingroup$ Make a picture of the problem, that will make it easier! $\endgroup$ Commented Jan 14, 2017 at 11:16
  • $\begingroup$ yes i got it now $\endgroup$
    – Pole_Star
    Commented Jan 14, 2017 at 11:17
  • $\begingroup$ the answer will be 12 cm square.3 will be the height and 8 the base of the triangle $\endgroup$
    – Pole_Star
    Commented Jan 14, 2017 at 11:18
  • $\begingroup$ median=bissector line $\implies$ isosceles triangle. $\endgroup$
    – Jean Marie
    Commented Jan 14, 2017 at 11:33

2 Answers 2

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Since $LO$ is both median and the bisector of $\angle NLM$, $\triangle LMN$ is isosceles and $ LO \perp MN$.

So $MO = NO= \sqrt{5^2-3^2\ }=4$ and area $\triangle LMN = 4\times 3 =12 \text{cm}^2$


To see that $\triangle LMN$ is isosceles, consider that $\angle LON = 180°-\angle LOM$. Flip $\triangle LON$ to superimpose $\angle NLO$ and $\angle MLO$. Then the only way to make $|MO|=|NO|$ is to have $\angle LON = \angle LOM =\perp$

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  • $\begingroup$ hey LO=3cm so,area will be equal to 12cm2 $\endgroup$
    – Pole_Star
    Commented Jan 14, 2017 at 14:41
  • $\begingroup$ @dp1611 thanks, I'm blaming jet-lag for that one. $\endgroup$
    – Joffan
    Commented Jan 14, 2017 at 15:10
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Its' isosceles and acute. Because a acute is smaller than a right angle, and there is two of the sides are the same.

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