Question

∆ABC is a right angled at B. BD is perpendicular upon AC. If AD = a, CD= b, then (AB)2=

Answer 1

Step-by-step explanation:

AB*2=a*2+BD*2...

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Answer 2

Answer:

$In \:\ triangle \: ABC \: \angle B = 90 \degree \: \\ and \: BD \perp \: AC \\ \therefore \: by \: geometric \: mean \\ \: property \colon \\ BD^{2} = AD \times CD \\ \implies \: BD^{2} \: = \: ab \\ in \: \triangle \: ADB \: \angle D = 90 \degree \\ \therefore\: by \: Pythagoras\: Theorem \\ AB^{2} = AD^{2} + BD^{2} \\ \: \: \:= {a}^{2} + ab \\ \: \: \:= a(a + b) \\ AB^{2} =a(a + b) \:$