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Question Number 101686 by bemath last updated on 04/Jul/20

Commented by bemath last updated on 05/Jul/20

$$\mathrm{R}=\frac{\mathrm{abc}}{{4\mathrm{L}}_{\mathrm{\Delta }\mathrm{ABC}}},\mathrm{s}=\frac{13+14+15}{2}=\frac{42}{2}=21$$$$\mathrm{area}\mathrm{\Delta }\mathrm{ABC}=\sqrt{21\times 8\times 7\times 6}$$$$=7\times 3\times 4=84$$$$\mathrm{R}=\frac{13\times 14\times 15}{4\times 84}=\frac{13\times 15}{24}$$

Commented by bobhans last updated on 04/Jul/20

$$\mathrm{translate}\mathrm{in}\mathrm{english}$$

Commented by bemath last updated on 04/Jul/20

In the figure the outer circle of the triangle abc is drawn if AB = 13, BC = 14 and AC = 15. Find the shortest distance from the center of the circle to AB

Commented by bobhans last updated on 04/Jul/20

$$\mathrm{d}=\sqrt{{(8.125)}^2-{(6.5)}^2}=\sqrt{(8.125+6.5)(8.125-6.5)}$$$$\mathrm{d}=\sqrt{14.625\times 1.625}=4.875=\frac{39}{8}$$$$\mathrm{answer}\mathrm{D}.\left(★\right)$$

Commented by 1549442205 last updated on 05/Jul/20

$$\mathrm{By}\mathrm{above}\mathrm{the}\mathrm{solution}\mathrm{the}\mathrm{question}\mathrm{mus}\mathrm{be}$$$${}^{″}\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circumcircle}\mathrm{of}$$$$\mathrm{the}\mathrm{triangle}{\mathrm{ABC}}^{″}.\mathrm{R}=\frac{13\times 15}{24}$$$$\mathrm{d}=\sqrt{{\mathrm{R}}^2-{\left(\frac{\mathrm{AB}}{2}\right)}^2}=\sqrt{{\left(\frac{13\times 15}{24}\right)}^2-\frac{{13}^2}{4}}$$$$=\sqrt{\frac{{13}^2}{4}\times (\frac{{15}^2}{{12}^2}-1)}=\frac{13}{2}\sqrt{\frac{{15}^2-{12}^2}{{12}^2}}$$$$=\frac{13}{24}\sqrt{81}=\frac{39}{8}$$

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