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16th KVJ Solutions4

Here solution for 16th KVJ Solutions4 part a in detail . Please feel free to send your opinion

4Q) a) Let a, b,c be the length of the sides of a triangle and r be the in-radius. Show that:

         r< \(\frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}\).

Solution:

  We know that for a \(\Delta ABC \)

  r =\( \frac{\Delta}{S}\), Where \( \Delta \) is Area of triangle ABC and S is semi-perimeter.

2S = a + b + c  and \( \Delta = \frac{1}{2}bc sin(A)\) =\(\frac{1}{2}ca sin(B)\)=\(\frac{1}{2}ab sin(C)\) .

r=\( \frac{\Delta}{\frac{a+b+c}{2}}\) then r(a+b+c)=\(2\Delta\)

\(2\Delta\)=bc sin(A)

                =ca Sin(B)

                =ab Sin(C)

\(6\Delta\)=bc Sin(A) + ca Sin(B) + ab Sin(C)

3r(a+b+c)=\(6\Delta\)

3r(a+b+c)=bc Sin(A) + ca Sin(B) + ab Sin(C) -------(1)

We know that maximum value of sin is 1

so we get

 bc Sin(A) < bc

 ca Sin(B) < ca

 ab Sin(C) < ab

adding these we get

 bc Sin(A) + ca Sin(B) + ab Sin(C) < bc + ca +ab

From (1) we get ,

3r(a+b+c) < bc + ca + ab ------------(2)

 We know that G.M \( \leq \) A.M

  \(\sqrt{b^{2}c^{2}} \leq \frac{b^{2}+c^{2}}{2}\)

 bc \( \leq \frac{b^{2}+c^{2}}{2}\)

Similarly we get

 ca \( \leq \frac{c^{2}+a^{2}}{2}\)

  ab \( \leq \frac{a^{2}+b^{2}}{2}\)

Adding these we get ,

 \( ( bc + ca + ab) \leq \frac{2(a^{2}+b^{2}+c^{2})}{2} \)

  bc + ca + ab \( \leq (a^{2}+b^{2}+c^{2})\)

From (2) we get,

  3r(a+b+c) < bc + ca + ab \( \leq (a^{2}+b^{2}+c^{2}) \)

  3r(a+b+c) < \( a^{2}+b^{2}+c^{2}\)

   r < \( \frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}\)

Hence it is proved.

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