geometry question. In \(\triangle ABC\) ,\(\angle B\) is the right angle. \(AC=16\) cm, and \(D\) is the midpoint of \(AC\). find the length of \(BD\). \(\large \color{black}{\normalsize \text{options}\hspace{.33em}\\~\\ a.)10\quad cm\hspace{.33em}\\~\\ b.)\sqrt{15}\quad cm\hspace{.33em}\\~\\ c.)8\quad cm\hspace{.33em}\\~\\ d.)7.5\quad cm\hspace{.33em}\\~\\ }\)
ok
|dw:1427491456182:dw|
what about it
|dw:1427492460840:dw| :D :D
45-45-90 triangles? :D :P
how is \(BD\) perpendicular to \(AC\)
well bec D is midpoint of AC so that's why 16/2
just draw it you don't know the other sides the triangle can take many forms but strangely the BD is always 4
|dw:1427493413670:dw| hmm i just noticed <B is right angle
Nnesha that is not a right angle the only thing the info are saying is that d is the midpoint
i mean not orthogonal :p
which one ?? B or the other two :P
the segment bd is not orthogonal to ac
@mathmath333 would bd be angles bisector?
No the information given does'nt conclude that BD is the angle bisector
okay! i will try and see if i can find anything interesting
tried sin law not good lol
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