In triangle ABC, if R = frac{65}{8}, rr_1 = 4 | vedclass

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(D) We know that $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$.Given $rr_1 = 42$, we have $\frac{\Delta^2}{s(s-a)} = 42$. $(i)$Given $r_1 - r

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$168$

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(D) We know that $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$.Given $rr_1 = 42$, we have $\frac{\Delta^2}{s(s-a)} = 42$. $(i)$Given $r_1 - r = 6.5$, we have $\frac{\Delta}{s-a} - \frac{\Delta}{s} = 6.5$.$\frac{\Delta(s - (s-a))}{s(s-a)} = 6.5 \Rightarrow \frac{\Delta a}{s(s-a)} = 6.5$. (ii)Dividing $(i)$ by (ii), we get $\frac{\Delta^2}{s(s-a)} 	imes \frac{s(s-a)}{\Delta a} = \frac{42}{6.5}$.$\frac{\Delta}{a} = \frac{42}{6.5} = \frac{420}{65} = \frac{84}{13}$.Since $R = \frac{abc}{4\Delta}$, we have $\frac{65}{8} = \frac{abc}{4\Delta} \Rightarrow \frac{bc}{\Delta} = \frac{65}{8} 	imes \frac{4}{a} = \frac{65}{2a}$.Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, we know $r_1 = \frac{\Delta}{s-a} = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$.From $rr_1 = 42$, we have $\frac{\Delta^2}{s(s-a)} = 42$. Since $\Delta^2 = s(s-a)(s-b)(s-c)$, this simplifies to $(s-b)(s-c) = 42$.From $r_1 - r = 6.5$, we have $\frac{\Delta a}{s(s-a)} = 6.5$. Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, $\frac{a\sqrt{s(s-a)(s-b)(s-c)}}{s(s-a)} = 6.5 \Rightarrow \frac{a\sqrt{42}}{s(s-a)} = 6.5$.Actually, using $r_1 - r = 4R \sin^2(A/2)$ is not needed here. Simply note $rr_1 = (s-b)(s-c) = 42$.Also $r_1 - r = 2R \sin A 	an(A/2) = 2R \sin^2(A/2) / \cos(A/2) ...$ actually, $r_1 - r = 4R \sin^2(A/2)$.Given $r_1 - r = 6.5$, $4 	imes \frac{65}{8} \sin^2(A/2) = 6.5$ $\Rightarrow \frac{65}{2} \sin^2(A/2) = 6.5$ $\Rightarrow \sin^2(A/2) = \frac{6.5 	imes 2}{65} = 0.2 = \frac{1}{5}$.Then $\cos A = 1 - 2\sin^2(A/2) = 1 - 2(1/5) = 3/5$.$s(s-a) = \frac{\Delta^2}{42}$. Since $\Delta = rs$, and $r = \frac{\Delta}{s}$, we have $s(s-a) = 168$.

In $\triangle ABC$, if $R = \frac{65}{8}$, $rr_1 = 42$ and $r_1 - r = 6.5$, then $s(s-a) = $

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