#### Question

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

(i) AD = a`sqrt3`

(ii) Area (ΔABC) = `sqrt3` a^{2}

#### Solution

(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

AD^{2} + BD^{2} = AB^{2}

⇒ AD^{2} + (a)^{2} = (2a)^{2}

⇒ AD^{2} + a^{2} = 4a^{2}

⇒ AD^{2} = 4a^{2} − a^{2} = 3a^{2}

⇒ AD = a`sqrt3`

(ii) Area of ΔABC = `1/2xxBCxxAD`

`=1/2xx2axxasqrt3`

`=sqrt3` a^{2}

Is there an error in this question or solution?

#### APPEARS IN

Solution In a Triangle Abc, Ab = Bc = Ca = 2a and Ad ⊥ Bc. Prove that Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.