\(\sqrt{108}\) cm
9 cm
\(\sqrt{92}\) cm
10 cm
Correct answer is C
Using the cosine rule, we have
\(p^{2} = q^{2} + r^{2} - 2qr \cos P\)
\(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\)
= \(64 + 36 - 8\)
\(p^{2} = 92 \therefore p = \sqrt{92} cm\)