prove this ?

Asked by Jagmeet Singh | 16th Oct, 2013, 09:54: AM

Expert Answer:

PQRS is a rectangle, PQ = a, QR = b

Let B be the center, r1 be the radius, draw a circle passing through P, Q and touching RS. Let E be the center, r2 be the radius, draw a circle passing through Q, R and touching PS.

BC ? RS and C is the midpoint of RS, extend CB to meet PQ at D.


  CD =b, BD+BC =b

   BD =b - r1 ---------(1)

    From triangle BDQ,    


   BD = root [(r1) 2? a2/ 4] -----(2)

 From (1) and (2)

   b - r1 = root [(r1) 2? a2/ 4]

  Squaring both sides, we get

   b2?2br1+(r1)2=(r1) 2? a2/ 4

   2br1=b2+ a2/ 4

   r1=b/ 2+a2 / 8b-------(3)

Similarly we get,

   r2=a/ 2+b2 / 8a -------(4)

 From (3) and (4), we have

  r1+r2= a+b/ 2+a2 + b2/ 8ab

 r1+ r2=a+b/ 2+(a+b)(a2?ab+b2/ 8ab)

 r1+r2=(a+b)(1/2+ a2?ab+b2/ 8ab)

 r1+r2=(a+b)(1/2+(a?b)2+ab / 8ab)

 r1+r2=(a+b)(1/2+(a?b)2 /8ab+1/8)

 r1+r2=(a+b)(5/8+(a?b)2 /8ab)



Hence, 5(a+b)?8(r1+r2)

Answered by  | 16th Oct, 2013, 03:18: PM

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