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| 1. |
If ABC-AORP, arRE.and BC -15 cm, then find PR. |
| Answer» {tex}\\frac{\\operatorname{arc}(\\triangle \\mathrm{ABC})}{\\operatorname{ar}(\\triangle \\mathrm{QRP})}{/tex}\xa0=\xa0{tex}\\left(\\frac{B C}{R P}\\right)^{2}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{9}{4}{/tex}\xa0=\xa0{tex}\\left(\\frac{15}{P R}\\right)^{2}{/tex}{tex}\\Rightarrow{/tex} PR = 10 cm | |