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16th KVJ Solutions10

Probability Introduction :

Here you find the solutions of  16th KVJ Solutions10 in detail explanation .

10Q) Let PQRS be a rectangle such that PQ= a and QR =b. Suppose \(r_{1}\) is the radius of the circle passing through P and Q and touching RS and \(r_{2}\) is the radius of the circle passing through Q and R and touching PS. Show that :

\( 5(a+b) \leq 8(r_{1}+r_{2} )\).

Solution:

Please see the figure below for this Question.

      PQRS be a rectangle .

      PQ= a , QR =b .

      Let B be the center , \(r_{1}\) be the radius , draw a circle passing through P , Q and touches RS.

      Let E be the center , \(r_{2}\) be the radius , draw a circle passing through Q, R and touches PS.

        BC \(\perp\) RS and C is the midpoint of RS, extend CB to meet PQ at D.

  PD=DQ=\(\frac{a}{2}\)

  CD =b, BD+BC =b

   BD =b - \(r_{1}\) ---------(1)

    From triangle BDQ ,

    \(BD^{2}= BQ^{2} - DQ^{2}\)

   BD = \(\sqrt{(r_{1})^{2} - \frac{a^{2}}{4}}\) -----(2)

 From (1) and (2)

   b - \(r_{1}\) = \(\sqrt{(r_{1})^{2} - \frac{a^{2}}{4}}\)

  Squaring on both sides, we get

   \( b^{2} -2br_{1} +(r_{1})^{2} = (r_{1})^{2} - \frac{a^{2}}{4} \)

   \( 2br_{1} = b^{2} + \frac{a^{2}}{4} \)

   \( r_{1} = \frac{b}{2} + \frac{a^{2}}{8b}\) -------(3)

Similarly we get,

   \( r_{2} = \frac{a}{2} + \frac{b^{2}}{8a} \) -------(4)

 From (3) and (4) , we add

  \( r_{1} + r_{2} = \frac{a+b}{2} + \frac {a^{3}+b^{3}}{8ab}\)

 \( r_{1} +  r_{2} = \frac{a+b}{2} + \frac{(a+b)(a^{2}-ab+b^{2}}{8ab}\)

 \( r_{1} +r_{2} = (a+b)(\frac{1}{2} + \frac{a^{2}-ab+b^{2}}{8ab})\)

 \(r_{1} + r_{2} =(a+b)(\frac{1}{2} + \frac{(a-b)^{2}+ab}{8ab})\)

 \( r_{1} +r_{2} = (a+b)(\frac{1}{2} + \frac{(a-b)^{2}}{8ab} +\frac{1}{8})\)

 \( r_{1} + r_{2} = (a+b)(\frac{5}{8} + \frac{(a-b)^{2}}{8ab})\)

  \( r_{1} +r_{2} \geq \frac{5(a+b)}{8}\)

  \( 8(r_{1}+r_{2}) \geq 5(a+b) \)

Therefore \( 5(a+b) \leq 8(r_{1}+r_{2}) \).

Hence proved.

 This  is the  explanation for  16th KVJ Solutions10 . Please feel free to share you opinion on this .

 
                     Figure 10

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