In an equilateral triangle \(ABC\), \(AD\) is an altitude. Then \(4AD^{2}\) is equal to（ ）

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In an equilateral triangle \(ABC\), \(AD\) is an altitude. Then \(4AD^{2}\) is equal to（ ）

C

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In an equilateral triangle, all sides are equal.

In \(\Delta ADC\), applying Pythagoras theorem, we get

\(AD^2=\sqrt{AC^2-DC^2}\)

As, \(DC=\frac{BC}{2}\)

\(\Rightarrow\) \(AD=\sqrt{AC^2-(\frac{BC}{2})^2}\)

\(\Rightarrow\) \(AD=\sqrt{AC^2-\frac{BC^2}{4}}\)

\(\Rightarrow\) \(AD=\sqrt{\frac{4\ AC^2-BC^2}{4}}\)

As all sides are equal i.e. \(AB=BC=CA\)

\(\Rightarrow\) \(AD=\sqrt{\frac{4\ AB^2-AB^2}{4}}\)

\(\Rightarrow\) \(AD^2=\frac{4\ AB^2-AB^2}{4}\)

\(\Rightarrow\) \(4AD^2=3\ AB^2\)

In \(\Delta ADC\), applying Pythagoras theorem, we get

\(AD^2=\sqrt{AC^2-DC^2}\)

As, \(DC=\frac{BC}{2}\)

\(\Rightarrow\) \(AD=\sqrt{AC^2-(\frac{BC}{2})^2}\)

\(\Rightarrow\) \(AD=\sqrt{AC^2-\frac{BC^2}{4}}\)

\(\Rightarrow\) \(AD=\sqrt{\frac{4\ AC^2-BC^2}{4}}\)

As all sides are equal i.e. \(AB=BC=CA\)

\(\Rightarrow\) \(AD=\sqrt{\frac{4\ AB^2-AB^2}{4}}\)

\(\Rightarrow\) \(AD^2=\frac{4\ AB^2-AB^2}{4}\)

\(\Rightarrow\) \(4AD^2=3\ AB^2\)

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