+26367 The ratio between the areas of equilateral triangles \(\triangle\) BXY and \(\triangle\) BAC is 1 : 4.
If \(AC=6\), what is XY?
\(\text{Let triangle $ BXY = A_2= \dfrac{XY\cdot h_2}{2} $} \\ \text{Let triangle $ BAC = A = \dfrac{6\cdot h }{2} $} \)
\(\begin{array}{|rcll|} \hline \dfrac{A_2}{A} &=& \dfrac{1}{4} \quad | \quad A_2 = \dfrac{XY\cdot h_2}{2},\ A = \dfrac{6\cdot h }{2} \\\\ \dfrac{\dfrac{XY\cdot h_2}{2}}{\dfrac{6\cdot h }{2}} &=& \dfrac{1}{4} \\\\ \dfrac{XY\cdot h_2} {6\cdot h } &=& \dfrac{1}{4} \\\\ XY \cdot \dfrac{h_2} {h} &=& \dfrac{6}{4} \\\\ XY \cdot \dfrac{h_2} {h} &=& \dfrac{3}{2} \quad | \quad \dfrac{h_2}{h} = \dfrac{XY}{6} \\\\ XY \cdot \dfrac{XY}{6} &=& \dfrac{3}{2} \\\\ XY^2 &=& 3^2 \\\\ \mathbf{XY} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)
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