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In ΔLMN, LO is the median. Also LO is the bisector of ∠MLN. If LO = 3 cm, and LM = 5 cm, then find the area of ΔLMN.

(a) 12 cm2 

(b) 10 cm2 

(c) 4 cm2 

(d) 6 cm2 

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(a) 12 cm2.

LO being the median on MN, 

MO = ON 

Also, LO being the internal bisector of ∠MLN,

\(\frac{LM}{LN}=\frac{MO}{ON}=1\)              (By bisector theorem)

⇒ LM = LN 

⇒ ΔLMN is isosceles triangle. 

Now ΔLOM ≅ ΔLON    (By SSS) 

⇒ ∠LOM = ∠LON = 90º (cpct)

∴ In ΔLOM, MO = \(\sqrt{LM^2-LO^2}\)

\(\sqrt{25-9}=\sqrt{16}\) = 4 cm.

Area of ΔLMN = \(\frac{1}{2}\times{MN}\times{LO}\)

\(\frac{1}{2}\times{8}\times{3}\) cm2 = 12 cm2.

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