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ABC is a right angle triangle. AD is drawn perpendicular to BC. If BC = 9cm, BD = 4 cm then find AB?

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11
Jan 7, 2017

Answer:

#AB=6#

Explanation:

As #AD# is drawn perpendicular to #BC# in right angled #DeltaABC#, it is apparent that #DeltaABC# is right angled at #/_A# as shown below (not drawn to scale).
enter image source here
As can be seen #/_B# is common in #Delta ABC# as well as #DeltaDBA# (here we have written two triangles this way as #/_A=/_D#, #/_B=/_B# and #/_C=/_BAD#) - as both are right angled (obviously third angles too would be equal) and therefore we have

#Delta ABC~~DeltaDBA# and hence

#(BC)/(AB)=(AB)/(BD)=(AC)/(AD)#..............(1)

therefore, we have #(BC)/(AB)=(AB)/(BD)# or

#AB^2=BCxxBD=9xx4=36#

Hence #AB=6#

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dk_ch Share
Jan 7, 2017

drawn

Given that #DeltaABC# is right angled and #AD# is drawn perpendicular to #BC# . So #/_BAC# is right angle.

Given #BC = 9cm and BD =4cm-> CD = BC-BD=5cm#

In #Delta ABC and Delta ABD#

  • #/_ABD=/_ABC#

  • #/_ADB=/_BAC= "right angle"#

  • #/_BAD=/_ACD ("remaining")#

So #Delta ABC and Delta ABD# are similar

Hence

#(AB)/(BC)=(BD)/(AB)#

#=>AB^2=BDxxBC=4xx9=36#

#=>AB=sqrt36cm=6cm#

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