from units TRIANGLES
Asked by | 9th Mar, 2009, 11:35: PM
Draw the diagram as mentioned in the question.
Let CD be theperpendicular from C to AB.
Acc to question,
AB=c,BC=a, AB=c,CD=p
CAB is a right triangle and so is DCB.
angle BDC= angle BCA( each is 90 )
angle DBC= angle ABC(common angle to both triangles)
So,
triangle DCB similar to triangle CAB(AA rule)
DC/CA=CB/AB( corr sides of similar trianglesaa re proportional)
p/b=a/c
1/p=c/ab
squaring both sides,
solving one step further we get the answer.
Answered by | 10th Mar, 2009, 12:51: PM
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