Question

In triangle. ABC angle B =. 90° ; BD is perpendicular to AC . BC= 8 cm and AC= 10 cm . Find BD

Answer 1

$in \: triangle \: {abc} \: \\ b = 90 {}^{0} \\ bd \: is \: perpendicular \: to \: {ac} \: \\ bc = 8cm \\ ac = 10cm \\ bd = h$

If BD perpendicular bisect on AC it will be divide in two equal part

Then,

$ad = dc = 10 \div 2 = 5$

Here, this triangle form in two right angled triangle BDC And BDA

In triangle BDC

$dc = 5cm \\ bc = 8cm \\ bd = h$

Here, we can use Pythagoras theorem in triangle BDC

${h} {}^{2} = \: {p} {}^{2} + {b} {}^{2}$

$p {}^{2} = \: h {}^{2} - b {}^{2}$

$bd {}^{2} = bc {}^{2} - dc {}^{2} \\ bd = \sqrt{8{ }^{2} - 5 {}^{2} } \\ bd = \sqrt{64 - 25} \\ bd = \sqrt{39 } \\ bd = 6.2$

Then,

BD = 6.2 CM