Question

In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a√3
(ii) Area (∆ABC) =√3a²

Answer 1

$<b>$Solution

GIVEN :  

In ∆ABC , AB = BC = CA = 2a and AD ⟂BC.

(i) In ∆ABD and ∆ACD

∠ADB = ∠ADC     [Each 90°]

AB = AC                  [Given]

AD = AD                 [Common]

∆ABD ≅ ∆ACD     [By RHS condition]

∴ BD = CD            [By c.p.c.t]

BC = BD + CD

BC = BD + BD

BC = 2BD

2a = 2BD

BD = a ……………(1)

In ∆ADB, by Pythagoras theorem

AB² = AD² + BD²

(2a)² = AD² + a²

[From eq 1 , BD = a]

4a² = AD² +a²

4a² - a² = AD²

3a² = AD²

AD = √(3a²)

AD = √3a

Hence, AD = √3a

(ii) Area of ∆ABC = ½ × BC × AD

= ½ × 2a × √3a = √3a²

Area of ∆ABC = √3a²

Answer 2

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