The side of an equilateral ΔABC is $3\sqrt{7}$ cm. P is a point on side BC such that BP : PC = 1 : 2. The length (in cm) of AP is:

7

\(\angle\)ABC = \({60}^\circ\) (angle of an equilateral triangle)

Now, according to the question,

BC = BP : PC = 1 : 2

BP = \(\frac{1 \;×\; 3√ 7}{3}\) = \(\sqrt {7 }\)

PC = \(\frac{2 \;×\; 3√ 7}{3}\) = 2\(\sqrt {7 }\)

Now,

Apply cosine rule on triangle APB

= cos 60 = (\( {b }^{2 } \) + \( {c }^{2 } \) - \( {a }^{2 } \))/2bc

= \(\frac{1}{2}\) = (\( {(3√7)}^{2 } \) + \( {√7}^{2 } \) - \( {AP}^{2 } \))/(2 x 3√7 x √7)

= \(\frac{1}{2}\) = [(63 + 7) - \( {AP}^{2 } \)]/42

= \(\frac{1}{2}\) = [70 - \( {AP}^{2 } \)]/42

= \( {AP}^{2 } \) = 70 - 21

= \( {AP}^{2 } \) = 49

= AP = 7

Therefore, AP is 7.