$\triangle \mathrm{ABC}$ is an equilateral triangle. $\mathrm{D}$ is a point on side $\mathrm{BC}$ such that $\mathrm{BD}: \mathrm{BC}=1: 3$. If $\mathrm{AD}=5 \sqrt{7} \mathrm{~cm}$, then the side of the triangle is:
Answer & explanation
Correct answer: option 3
Using cosine rule
= cos 60 = [\( {(1x) }^{2 } \) + \( {(3x) }^{2 } \) - \( {(5√7) }^{2 } \)]/(2 × 1x × 3x)
= \(\frac{1}{2}\) = (\( {1x }^{2 } \) + \( {9x }^{2 } \) - 175)/(2 × \( {3x }^{2 } \))
= \( {3x }^{2 } \) = \( {10x }^{2 } \) - 175
= (\( {10x }^{2 } \) - \( {3x }^{2 } \)) = 175
= \( {7x }^{2 } \) = 175
= \( {x }^{2 } \) = \(\frac{175}{7}\)
= \( {x }^{2 } \) = 25
= x = 5 cm
Now,
The side of an equilateral triangle = 3x = (3 x 5) = 15 cm.
Therefore, answer is 15 cm.