$△ A B C$ is right angled at B and the perpendicular drawn to the opposite side bisects it at D. BD = ___ cm if AD = DC = 5cm.

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Solution

Since $△$ ABC is right angled triangle at B and BD is the perpendicular drawn to the opposite side,
${B D}^{2} = A D \times D C$
$⟹ {B D}^{2} = (5 \times 5) {c m}^{2}$
$B D = \sqrt{25} c m = 5 cm$

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△ABC is right angled at B and the perpendicular drawn to the opposite side bisects it at D. BD = ___ cm if AD = DC = 5cm.

Since △ ABC is right angled triangle at B and BD is the perpendicular drawn to the opposite side, BD2=AD×DC ⟹BD2=(5×5)cm2 BD=√25cm=5cm

$△ A B C$ is right angled at B and the perpendicular drawn to the opposite side bisects it at D. BD = ___ cm if AD = DC = 5cm.

Since $△$ ABC is right angled triangle at B and BD is the perpendicular drawn to the opposite side,
${B D}^{2} = A D \times D C$
$⟹ {B D}^{2} = (5 \times 5) {c m}^{2}$
$B D = \sqrt{25} c m = 5 cm$