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Let `D` be the middle point of the side `B C` of a triangle `A B Cdot` If the triangle `A D C` is equilateral, then `a^2: b^2: c^2` is equal to `1:4:3` (b) `4:1:3` (c) `4:3:1` (d) `3:4:1`

A

`1 : 4 : 3`

B

`4 : 1 : 3`

C

`4 : 3 : 1`

D

`3 : 4 : 1`

Text Solution

Verified by Experts

The correct Answer is:
B

In `Delta ABD, cos 120^(@) = (x^(2) + x^(2) - AB^(2))/(2x^(2))`
`rArr (2x^(2) - AB^(2))/(2x^(2)) = (-1)/(2)`
`rArr 3x^(2) = AB^(2)`
`rArr AB = x sqrt3`
`rArr a^(2) : b^(2) : c^(2) = (2x)^(2) : x^(2) : (x sqrt3)^(2)`
`= 4x^(2) : x^(2) : 3x^(2) = 4 : 1 : 3`
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