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Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

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#### Solution

We have,

ΔABC is an equilateral Δ with side 12 cm.

Draw AE ⊥ BC

In ΔABD and ΔACD

∠ADB = ∠ADC [Each 90°]

AB = AC [Each 12 cm]

AD = AD [Common]

Then, ΔABD ≅ ΔACD [By RHS condition]

∴ AD^{2} + BD^{2} = AB^{2}

⇒ AD^{2} + 6^{2} = 12^{2}

⇒ AD^{2} = 144 − 36 = 108

⇒ AD = `sqrt`108 = 10.39 cm

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