ABC is right triagle right angled at C Let BC=a, CA=b,AB=c and let p be the lenght of perpendicular from C on AB. Prove that 1/p square = 1/a square +1/b square

Asked by dhilip n | 26th Aug, 2012, 10:02: AM

Expert Answer:

 
Area of triangle ACB with base AB = (1/2)AB*CD = (1/2)cp
Area of triangle ACB with base AC = (1/2)AC*BC = (1/2)ba
 
Both areas are same. Therefore, on equating, we get,
c = ab/p           ... (1)
 
Now, using Pythagoras theorem in triangle ABC, we have
AB2 = AC2+BC2
c2 = b2+a2
a2b2/ p2 = b2+a2         [Using (1)]
1/ p2= (b2+a2)/ a2b2
Hence, 1/ p2= 1/a2+1/b2

Answered by  | 26th Aug, 2012, 11:11: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.