A) \[\sqrt{3}\]
B) \[\sqrt{3}p\]
C) \[2p\]
D) \[4p\]
Correct Answer: B
Solution :
[b] Given an equilateral triangle ABC in which. |
\[AB=BC=CA=2p\] |
and \[AD\bot BC\] |
In \[\Delta ADB,\] \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] |
(By Pythagoras theorem) |
\[{{(2P)}^{2}}=A{{D}^{2}}+{{p}^{2}}\] |
\[A{{D}^{2}}=\sqrt{3}p\] |
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