Question

In an equilateral triangle ABC , AD perpendicular to BC and AD^2 = P.BC^2, then P is equal to ------------

Answer 1

Given:

In an equilateral triangle ABC , AD perpendicular to BC and $\[A{D^2} = p \times B{C^2}\]$

Solution:

Assume that the side of the equilateral triangle is $a$.

$\[AB = BC = AC = a\]$

Refer the following figure and apply Pythagoras theorem,

$\[ \Rightarrow A{D^2} = {a^2} - \frac{{{a^2}}}{4}\]$

$\[ \Rightarrow A{D^2} = \frac{{4{a^2} - {a^2}}}{4}\]$

$\[ \Rightarrow A{D^2} = \frac{{3{a^2}}}{4}\]$

Put, $\[a = BC\]$

$\[ \Rightarrow A{D^2} = \frac{3}{4}B{C^2}\]$

Now compare with $\[A{D^2} = p \times B{C^2}\]$

$\[\therefore p = \frac{3}{4}\]$

Hence the value of p is $\[\frac{3}{4}\]$.