In triangle ABC, if a=7, b=8 and c=9, then fr | vedclass

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(B) Given $a=7, b=8, c=9$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{7+8+9}{2} = 12$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(

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$\frac{5}{72}$

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(B) Given $a=7, b=8, c=9$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{7+8+9}{2} = 12$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 	imes 5 	imes 4 	imes 3} = \sqrt{720} = 12\sqrt{5}$. We know $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$. Therefore, $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2} = \frac{(s-a)^2+(s-b)^2+(s-c)^2}{\Delta^2}$. Substituting the values: $= \frac{(12-7)^2+(12-8)^2+(12-9)^2}{720} = \frac{5^2+4^2+3^2}{720} = \frac{25+16+9}{720} = \frac{50}{720} = \frac{5}{72}$.

In $\triangle ABC$, if $a=7, b=8$ and $c=9$, then $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=$

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