Using Hero's formula

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)

1)\(y^2\)-14y+48=0

On simplifying this equation

we get \(y^2 - 6y - 8y + 48 = 0\) -> \(y(y-6) - 8(y-6) = 0\) -> \((y-6)(y-8) = 0\)

Therefore, y could take the values 6 or 8.

For both the values of y for triangle ABC, the Area is 12.

(Sufficient)

2) \(y^2\)-16y+60=0

On simplifying this equation,

we get \(y^2 - 10y - 6y + 60 = 0\) -> \(y(y-10) - 6(y-10) = 0\) -> \((y-6)(y-10) = 0\)

Therefore, y could take the values 6 or 10

A triangle with y = 10 is not possible.

The Area of Triangle ABC with sides a=b=5 and c=6 is 12.

(Sufficient - Option D)
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