In a Triangle Abc, Ab = Bc = Ca = 2a and Ad ⊥ Bc. Prove that - Mathematics


In a ΔABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = a`sqrt3`

(ii) Area (ΔABC) = `sqrt3` a2



(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`


`=sqrt3` a2

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Chapter 7: Triangles - Exercise 7.7 [Page 120]


RD Sharma Class 10 Maths
Chapter 7 Triangles
Exercise 7.7 | Q 13 | Page 120


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