If you know Heron's formula, this is easy.
[tex]\text{area}=\sqrt{s(s-6)(s-10)(s-14)}=15\sqrt3[/tex]
where [tex]s=\dfrac{6+10+14}2=15[/tex] is the semiperimeter of the triangle.
The area is also equal to half the product of the base and height of the triangle, which happen to be
[tex]15\sqrt3=\dfrac{14x}2\implies x=\dfrac{15\sqrt3}7\approx3.711[/tex]
so the answer is B.
