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