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प्रश्न
In a ΔABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a`sqrt3`
(ii) Area (ΔABC) = `sqrt3` a2
उत्तर
(i) In ΔABD and ΔACD
∠ADB = ∠ADC [Each 90°]
AB = AC [Given]
AD = AD [Common]
Then, ΔABD ≅ ΔACD [By RHS condition]
∴ BD = CD = a [By c.p.c.t]
In ΔADB, by Pythagoras theorem
AD2 + BD2 = AB2
⇒ AD2 + (a)2 = (2a)2
⇒ AD2 + a2 = 4a2
⇒ AD2 = 4a2 − a2 = 3a2
⇒ AD = a`sqrt3`
(ii) Area of ΔABC = `1/2xxBCxxAD`
`=1/2xx2axxasqrt3`
`=sqrt3` a2
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