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In equilateral Δ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

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#### Solution 1

In equilateral Δ ABC, AD ⊥ BC.

Therefore, BC = x cm.

Area of equilateral ΔABC = `sqrt3/4 xx "side"^2 = 1/2 xx "base" xx "height"`

= `sqrt3/4 xx x^2 = 1/2 xx x xx "AD"`

AD = `sqrt3/2 x`

#### Solution 2

In △ADC and △ADB,

AD = AD ...(Common)

∠ADB = ∠ADC ...(Each 90°)

AB = AC ....(Given, ABC is an equilateral triangle)

Thus, △ADC ≅ △ADB

BD = DC = `1/2`BC ...(By cpct)

Hence, BD = `1/2`x

In △ADB,

∠D = 90°

△ADB is right angle triangle,

by Pythagoras theorem,

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

`x^2 = (1/2x)^2 + "AD"^2`

`"AD"^2 = x^2 − (x^2)/4`

`"AD"^2 = 3/4x^2`

`"AD" = (sqrt(3))/2x`

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