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Let 'l' is the length of median from the...

Let 'l' is the length of median from the vertex A to the side BC of a `Delta ABC`. Then

A

`4l^(2)=2b^(2)+2c^(2)-a^(2)`

B

`4l^(2)=b^(2)+c^(2)+2bc cos A`

C

`4l^(2)=a^(2)+4bc cos A`

D

`4l^(2)=(2s-a)^(2)"sin"^(2)(A)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
D is the midpoint of BC

Using Apollonius theorem, we have
`AB^(2)+AC^(2)=2(AD^(2)+BD^(2))`
`rArr c^(2)+b^(2)=2[l^(2)+((a)/(2))^(2)]` (where AD = l)
`rArr 4l^(2)=2b^(2)+2c^(2)-a^(2)`
`=b^(2)+c^(2)+(b^(2)+c^(2)-a^(2))`
`=b^(2)+c^(2)+2bc cos A`
`=(b^(2)+c^(2)-a^(2))+a^(2)+2bc cos A`
`=2bc cos A + a^(2)+2bc cos A`
`=4bc cos A + a^(2)`
Also `4l^(2)=b^(2)+c^(2)+2bc cos A`
`= b^(2)+c^(2)+2bc(1-2"sin"^(2)(A)/(2))`
`=(b+c)^(2)-4bc "sin"^(2)(A)/(2)`
`= (2s-a)^(2)-4bc "sin"^(2)(A)/(2)`
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