(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.