Question

∆ABC is an equilateral triangle with each side of length 2p. If AD⊥BC. Then the value of AD is​

Answer 1

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Answer 2

Given:

∆ABC is an equilateral triangle

each side of length 2p

AD⊥BC

Find value of AD=?

In the equilateral ΔABC, we see that

AD⊥BC

AB = BC = CA = 2p    [From the figure shown in the attachment]

$\begin{array}{l}B D=D C=\frac{1}{2} B C \\B D=D C=\frac{2 p}{2} \\B D=D C=p\end{array}$

In ΔADB, using Pythagoras theorem,

$\begin{array}{l}(A B)^{2}=(A D)^{2}+(B D)^{2} \\(2 p)^{2}=(A D)^{2}+p^{2} \\4 p^{2}=(A D)^{2}+p^{2} \\4 p^{2}-p^{2}=A D^{2} \\3 p^{2}=A D^{2} \\A D^{2}=3 p^{2}\end{array}$

$\begin{array}{l}A D=\sqrt{3} p \\A D=p \sqrt{3}\end{array}$

Hence the value of AD is $p \sqrt{3}$