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Question Number 156184 by cortano last updated on 09/Oct/21

Commented by cortano last updated on 09/Oct/21

$$\mathrm{base}=12,\mathrm{height}=9$$$$\mathrm{find}\mathrm{radius}$$

Answered by mr W last updated on 09/Oct/21

Commented by mr W last updated on 09/Oct/21

$$hypotenuseAC=\sqrt{9^2+{12}^2}=15inch$$$$\Rightarrow \mathrm{\Delta }PQR\sim \mathrm{\Delta }ACB$$$$PR=\frac{9}{15}\times 2r=\frac{6r}{5}$$$$PQ=\frac{12}{15}\times 2r=\frac{8r}{5}$$$$AF=AD=9-\frac{6r}{5}-r=9-\frac{11r}{5}$$$$CG=CE=12-\frac{8r}{5}-r=12-\frac{13r}{5}$$$$AC=AF+FG+GC$$$$=9-\frac{11r}{5}+2r+12-\frac{13r}{5}=15$$$$\frac{14r}{5}=6$$$$r=\frac{15}{7}$$$$\Rightarrow circlediameter=2r=\frac{30}{7}inch$$

Commented by cortano last updated on 09/Oct/21

$$\frac{\mathrm{PR}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{AC}};\frac{\mathrm{RQ}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{BC}}$$

Commented by Tawa11 last updated on 10/Oct/21

$$\mathrm{great}\mathrm{sir}$$

Answered by nikif99 last updated on 09/Oct/21

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