Question

In BAC =90 AD perpendicular to BC, BD =4cm , DC=5cm , find AB and AD

Answer 1

$by \: theorem \: o f \: geomentric \: \: mean \: \\ ad { }^{2} = bd \times dc \\ = 4 \times 5 \\ = 20 \\ taking \: squer \: root \: \\ \\ad = \sqrt{20 } \: \\ ad = \sqrt{4 \times 5 } \: \\ ad = 2 \sqrt{5}$