Question

in Δ ABC, angle A=90° she BP perpendicular seg AC AP=2 PC=8 find BP​

Answer 1

intriangleabc

b=90

0

bdisperpendiculartoac

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 = 5ad=dc=10÷2=5

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

In triangle BDC

\begin{gathered}dc = 5cm \\ bc = 8cm \\ bd = h\end{gathered}

dc=5cm

bc=8cm

bd=h

Here, we can use Pythagoras theorem in triangle BDC

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

2

=p

2

+b

2

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

2

=h

2

−b

2

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

bd

2

=bc

2

−dc

2

bd=

8

2

−5

2

bd=

64−25

bd=

39

bd=6.2

Then,

BD = 6.2 CM

hope this helped you

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Answer 2

Answer:

Hope it helps you Friend