Question

In the given figure, triangle ABC is right angled at B. If AD
and CE are the two medians drawn from A and C, such
that AC = 5 cm, and AD = 3/2√5 cm, find the length of CE.​

Answer 1

Answer:

We apply Pythagoras theorem.

ΔABC: AC² = AB² + BC²

5² = (2 BE)² + BC²

25 = 4 BE² + BC² -- (1)

Also, BE² + BC² = CE² ----(2)

ΔABD, AD² = AB² + BD²

= 4 BE² + BC²/4

=> 4 (3√5/2)² = 45 = 16 BE² + BC² -- (3)

Solve the equations (1) and (2).

55 = 3 BC² => BC = √(55/3)

=> BE² = (AC² - BC²)/4 = (5² - 55/3)/4 = 5/3

By (2), CE² = 5/3 + 55/3 = 20/3

=> CE = 2 √(5/3)

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Since △ABD is a right-angled triangle at B. Therefore,

      AD2=AB2+BD2

⇒ AD2=AB2+(2BC)2             [∵BD=DC]

⇒ AD2=AB2+41BC2.......(i)

Again, △BCE is right-angled triangle at B

∴  CE2=BC2+BE2

⇒ CE2=BC2+(2AB)2         [∵BE=EA]

⇒ CE2=BC2+41AB2.........(ii)

Adding (i) and (ii), we get

     AD2+CE2=AB2+41BC2

I hope this helps you :)