Properties Of Triangles

The governing relation is Heron's formula, $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, which computes a triangle's area directly from its three sides through the semiperimeter, with no angle measurement required. It can be derived by combining the law of cosines with the area expression $\frac{1}{2}ab\sin C$ and eliminating the angle, which is why it depends only on side lengths. First compute the semiperimeter as half the perimeter: $s=\frac{5+6+7}{2}=9$. Then the three differences are $s-a=9-5=4$, $s-b=9-6=3$, and $s-c=9-7=2$. The product under the radical is $9\cdot 4\cdot 3\cdot 2=216$, so the area equals $\sqrt{216}=\sqrt{36\cdot 6}=6\sqrt{6}$. Option $6\sqrt{5}$ arises from an incorrect semiperimeter difference. Option $15$ guesses a rational value that cannot match an irreducible surd. Option $12\sqrt{2}$ stems from an arithmetic slip in forming the product. This is the standard Heron's-formula application in JEE Advanced triangle questions. As a final plausibility check, $6\sqrt{6}\approx 14.7$ square units is a reasonable area for a triangle with sides near six, and the triangle inequality holds since $5+6>7$, so a valid triangle and a real positive area both exist.