please give me the answer as follows
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Step-by-step explanation:
∠ADC = ∠ADB = 90°
BD = DC { perpendicular bisects the side BC in two equal halves }
➼ BD = DC = BC/2
➼ BD = DC = a/2
Now , in right angled ∆ADC by using Pythagoras Theorem -
AD (perpendicular)
DC (base)
AC (hypotenuse)
★ H² = Perpendicular² + Base² ★
\implies{\rm }⟹ AC² = AD² + DC²
\implies{\rm }⟹ a² = AD² + (a/2)²
\implies{\rm }⟹ a² = AD² + a²/4
\implies{\rm }⟹ a² – a²/4 = AD²
\implies{\rm }⟹ 4a² – a²/4 = AD²
\implies{\rm }⟹ √3a²/4 = AD
\implies{\rm }⟹ √3a/2= AD
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