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about to only mathematics

A

16

B

18

C

24

D

22

Text Solution

Verified by Experts

The correct Answer is:
B, D


Let `s-a = 2k -2, s-b = 2k, s -c = 2k + 2, k in I, k gt 1`
Adding we get,
`s = 6k`. So, `a = 4k +2, b = 4k, c = 4k -2`
Now, `cos P = (1)/(3) rArr (b^(2) + c^(2) -a^(2))/(2bc) = (1)/(3)`
`rArr 3[(4k)^(2) + (4k -2)^(2) - (4k + 2)^(2)] = 2 xx 4k (4k -2)`
or `3[16 k^(2) - 4 (4k) xx 2] = 8k (4k -2)`
or `48k^(2) - 96k = 32 k^(2) - 16k or 16k^(2) = 80k or k = 5`
So, sides are 22, 20, 18
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