In the given figure, ΔABC is a right triangle right angled at B. AD and CE are the two medians drawn from A and C respectivley. If AC =5 cm and AD =3√52 cm, then CE =??
2√5cm
Let AE =BE = a cm
and BD=DC = b cm
Using mid point theorem DE=12AC=52cm
In ΔDBE,DB2+BE2=ED2
⇒a2+b2=254−−−(I)
In ΔABD,AB2+BD2=AD2
(2a)2+b2=(3√52)2
⇒4a2+b2=454−−−(II)
Solving (I) and (II), we get, a2=53 and b2=5512
Now, in ΔBEC,
BE2+BC2=EC2
or,EC2=a2+(2b)2
=a2+4b2
=53+553
=603=20
∴EC=√20 cm=2√5 cm