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In a $∆ABC,\ AB\ =\ BC\ =\ CA\ =\ 2a$ and $AD\ ⊥\ BC$. Prove that:
(i) $AD\ =\ a\sqrt{3}$
(ii) Area $(∆ABC)\ =\ \sqrt{3}a^2$
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Given:


In $∆ABC,\ AB\ =\ BC\ =\ CA\ =\ 2a$ and $AD\ ⊥\ BC$.


To do:


We have to prove that:


(i) $AD\ =\ a\sqrt{3}$ 


(ii) Area $(∆ABC)\ =\ \sqrt{3}a^2$


Solution:


(i) In $∆ABD$ and $∆ACD$,


$\angle ADB = \angle ADC = 90^o$


$AB = AC$  (given)


$AD = AD$  (Common)


Therefore,


$∆ABD ≅ ∆ACD$   (By RHS congruency)


This implies,


$BD = CD = a$  (Corresponding parts of congruent triangles are equal)


In $∆ABD$,


By Pythagoras theorem,


$AD^2 + BD^2 = AB^2$


$AD^2 + a^2 = (2a)^2$


$AD^2 = 4a^2 – a^2 = 3a^2$


$AD = \sqrt{3a^2}$


$AD = \sqrt{3}a$


(ii) Area $(∆ABC) = \frac{1}{2} \times BC \times AD$


                               $= \frac{1}{2} \times (2a) \times (\sqrt{3}a)$


                                $= \sqrt{3}a^2$


Hence proved.

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Updated on: 10-Oct-2022

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