Statement-1: In a `!ABC`, if
`2a^(2)+4b^(2)+c^(2)=4ab+2ac`, then `cosA=1/4`
Statement-2: In a `DeltaA BC` if `cosA=1/4`, then
`(a+b+c)(b+c-a)=5/2bc`

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement- 2 is True.

Explanation

We have,
`2a^(2)+4b^(2)+c^(2)=4ab+2ac` …(i)
`rArr(a^(2)-2ac+c^(2))+(a^(2)-4ab+4b^(2))=0`
`rArr(a-c)^(2)+(a-2b)^(2)=0rArra=c` and a=2b
CASE I When a = c
Putting a = c in (i), we get c = 2b
`thereforea=c=2b`
`rArrcosA=(b^(2)+c^(2)-a^(2))/(2bc)=(b^(2))/(2ab)=(b^(2))/(4b^(2))=1/4`
CASE II When a = 2b
Putting a = 2b in (i), we get c = 2b.
`thereforea=c=2brArrcosA=1/4`
So, statement-I is true.
In a `DeltaABC`, if cosA`=1/4`, then
`(b^(2)+c^(2)-a^(2))/(2bc)=1/4rArrb^(2)+c^(2)-a^(2)=1/2bc`
`rArr(b+c)^(2)-a^(2)=5/2bcrArr(b+c-a)(b+c-a)=5/2bc`
So, statement-2 is also true.