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Question Number 187901 by Rupesh123 last updated on 23/Feb/23

Commented by Rupesh123 last updated on 23/Feb/23

Find R and r

Commented by BaliramKumar last updated on 23/Feb/23

$$R=\frac{3\sqrt{6}}{4}$$

Answered by mr W last updated on 23/Feb/23

$$s=\frac{7+10+15}{2}=16$$$$\mathrm{\Delta }=\sqrt{16\times 9\times 6\times 1}=12\sqrt{6}$$$$R=\frac{\mathrm{\Delta }}{s}=\frac{12\sqrt{6}}{16}=\frac{3\sqrt{6}}{4}$$$$\cos C=\frac{{15}^2+{10}^2-7^2}{2\times 15\times 10}=\frac{23}{25}$$$$\frac{R-r}{R+r}=\sin \frac{C}{2}=\sqrt{\frac{1-\cos C}{2}}=\frac{1}{5}$$$$\Rightarrow r=\frac{2R}{3}=\frac{\sqrt{6}}{2}✓$$

Commented by Rupesh123 last updated on 23/Feb/23

Very nice solution!

Answered by HeferH last updated on 23/Feb/23

$$A=s\cdot I;I=R$$$$s=\frac{7+10+15}{2}=16$$$$R=\frac{A}{s}=\frac{\sqrt{16(16-10)(16-15)(16-7)}}{16}=\frac{\sqrt{16\cdot 6\cdot 9}}{16}=\frac{3\sqrt{6}}{4}$$$$\frac{12r}{\sqrt{6}}+2\sqrt{\frac{3r\sqrt{6}}{4}}=9$$$$2r\sqrt{6}+\sqrt{3r\sqrt{6}}=9$$$$3r\sqrt{6}={(9-2r\sqrt{6})}^2$$$$3r\sqrt{6}=81+24r^2-36r\sqrt{6}$$$$24r^2-39r\sqrt{6}+81=0$$$$8r^2-13r\sqrt{6}+27=0$$$$r=\frac{13\sqrt{6}\pm \sqrt{169\cdot 6-4\left(8\right)\left(27\right)}}{16}=\frac{13\sqrt{6}\pm \sqrt{6\cdot 169-6\cdot 144}}{16}$$$$=\frac{13\sqrt{6}\pm 5\sqrt{6}}{16}$$$$R>r\Rightarrow r=\frac{\sqrt{6}}{2};$$

Commented by Rupesh123 last updated on 23/Feb/23

Very nice solution!

Commented by HeferH last updated on 23/Feb/23

Commented by Rupesh123 last updated on 23/Feb/23

Very detailed, sir

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