AB = $\displaystyle \sqrt{AC^{2} + BC^{2}}$ = $\displaystyle \sqrt{48^{2} + 36^{2}}$ = $\displaystyle \sqrt{12^{2}(4^{2} + 3^{2})}$ = 60,
OM = OK = r = $\displaystyle {\frac{AC + BC - AB}{2}}$ = $\displaystyle {\frac{48 + 36 - 60}{2}}$ = 12,
CH = AC . $\displaystyle {\frac{BC}{AB}}$ = 48 . $\displaystyle {\textstyle\frac{36}{60}}$ = $\displaystyle {\textstyle\frac{144}{5}}$,
CP = CH - PH = CH - OM = CH - r = $\displaystyle {\textstyle\frac{144}{5}}$ - 12 = $\displaystyle {\textstyle\frac{84}{5}}$,
OC = $\displaystyle {\frac{OK}{\sin \angle OCK}}$ = $\displaystyle {\frac{r}{\sin 45^{\circ}}}$ = r$\displaystyle \sqrt{2}$ = 12$\displaystyle \sqrt{2}$,
Следовательно,
OP = $\displaystyle \sqrt{OC^{2} - CP^{2}}$ = $\displaystyle \sqrt{(12\sqrt{2})^{2} - \left(\frac{84}{5}\right)^{2}}$ = 12$\displaystyle \sqrt{2 - \frac{49}{25}}$ = $\displaystyle {\textstyle\frac{12}{5}}$.
второй
AB = $\displaystyle \sqrt{AC^{2} + BC^{2}}$ = $\displaystyle \sqrt{48^{2} + 36^{2}}$ = $\displaystyle \sqrt{12^{2}(4^{2} + 3^{2})}$ = 60,
OM = OK = r = $\displaystyle {\frac{AC + BC - AB}{2}}$ = $\displaystyle {\frac{48 + 36 - 60}{2}}$ = 12,
BH = $\displaystyle {\frac{BC^{2}}{AB}}$ = $\displaystyle {\frac{36^{2}}{60}}$ = $\displaystyle {\textstyle\frac{108}{5}}$,
BM = BK = BC - CK = BC - r = 36 - 12 = 24,
OP = MH = BM - BH = 24 - $\displaystyle {\textstyle\frac{108}{5}}$ = $\displaystyle {\textstyle\frac{12}{5}}$.
ответ
$ {\frac{12}{5}}$.
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Втреугольнике ace , угол a = углу e. найдите длины сторон треугольника ace, если ae: ac, как 2: 5. периметр треугольника равен 84 см.