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Compute the area via Heron’s formula. Semiperimeter s=(13+14+15)/2=21s = (13+14+15)/2 = 21. Then

A=s(s−a)(s−b)(s−c)=21⋅8⋅7⋅6=7056=84.A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84.

The altitude hh to side b=14b=14 satisfies A=12bhA = \frac{1}{2} b h. Thus

84=12⋅14⋅h⇒84=7h⇒h=12.84 = \tfrac{1}{2}\cdot 14 \cdot h \Rightarrow 84 = 7h \Rightarrow h = 12.

Choice B. (Note that option C simplifies to 6, which is not correct.) A quick check: the area can also be found using base 15 and its altitude; the result must be consistent. Recognizing the classic 13–14–15 triangle is helpful: it has integer area 84, and the altitude to the middle side (14) is an integer (12), which is a common GRE-friendly feature. Avoid arithmetic slips when taking the square root: 7056=8427056=84^2, not ±84 \pm 84; area is positive.