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Mathematics 10 Online
OpenStudy (mathmath333):

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 \(BD\). \(\large \color{black}{\normalsize \text{options}\hspace{.33em}\\~\\ a.)10\quad cm\hspace{.33em}\\~\\ b.)\sqrt{15}\quad cm\hspace{.33em}\\~\\ c.)8\quad cm\hspace{.33em}\\~\\ d.)7.5\quad cm\hspace{.33em}\\~\\ }\)

OpenStudy (anonymous):

ok

OpenStudy (mathmath333):

|dw:1427491456182:dw|

OpenStudy (anonymous):

what about it

Nnesha (nnesha):

|dw:1427492460840:dw| :D :D

Nnesha (nnesha):

45-45-90 triangles? :D :P

OpenStudy (mathmath333):

how is \(BD\) perpendicular to \(AC\)

Nnesha (nnesha):

well bec D is midpoint of AC so that's why 16/2

OpenStudy (anonymous):

just draw it you don't know the other sides the triangle can take many forms but strangely the BD is always 4

Nnesha (nnesha):

|dw:1427493413670:dw| hmm i just noticed <B is right angle

OpenStudy (xapproachesinfinity):

Nnesha that is not a right angle the only thing the info are saying is that d is the midpoint

OpenStudy (xapproachesinfinity):

i mean not orthogonal :p

Nnesha (nnesha):

which one ?? B or the other two :P

OpenStudy (xapproachesinfinity):

the segment bd is not orthogonal to ac

OpenStudy (xapproachesinfinity):

@mathmath333 would bd be angles bisector?

OpenStudy (mathmath333):

No the information given does'nt conclude that BD is the angle bisector

OpenStudy (xapproachesinfinity):

okay! i will try and see if i can find anything interesting

OpenStudy (xapproachesinfinity):

tried sin law not good lol

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