In a triangle ABC, if a:b:c = 4:5:6, then fra | vedclass

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(A) Given, $a:b:c = 4:5:6$. Let $a=4x, b=5x, c=6x$. Semi-perimeter $s = \frac{4x+5x+6x}{2} = \frac{15x}{2}$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} =

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$\frac{71}{64}$

Vedclass · Answer

(A) Given, $a:b:c = 4:5:6$. Let $a=4x, b=5x, c=6x$. Semi-perimeter $s = \frac{4x+5x+6x}{2} = \frac{15x}{2}$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15x}{2} \cdot \frac{7x}{2} \cdot \frac{5x}{2} \cdot \frac{3x}{2}} = \frac{15x^2\sqrt{7}}{4}$. Circumradius $R = \frac{abc}{4\Delta} = \frac{4x \cdot 5x \cdot 6x}{4 \cdot \frac{15x^2\sqrt{7}}{4}} = \frac{8x}{\sqrt{7}}$. We know that $r_1+r_2+r_3 = 4R+r$. Also, $r = \frac{\Delta}{s} = \frac{15x^2\sqrt{7}/4}{15x/2} = \frac{\sqrt{7}x}{2}$. Thus, $r_1+r_2+r_3 = 4R + \frac{\sqrt{7}x}{2} = 4(\frac{8x}{\sqrt{7}}) + \frac{\sqrt{7}x}{2} = \frac{32x}{\sqrt{7}} + \frac{\sqrt{7}x}{2} = \frac{64x+7x}{2\sqrt{7}} = \frac{71x}{2\sqrt{7}}$. Finally, $\frac{1}{4R}[r_1+r_2+r_3] = \frac{1}{4(8x/\sqrt{7})} \cdot \frac{71x}{2\sqrt{7}} = \frac{\sqrt{7}}{32x} \cdot \frac{71x}{2\sqrt{7}} = \frac{71}{64}$.

In a triangle $ABC$, if $a:b:c = 4:5:6$, then $\frac{1}{4R}[r_1+r_2+r_3] =$

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