Длины катетов прямоугольно треугольника равны 24 и 10. Найдите расстояния от центра вписанной окружности до вершин этого треугольника

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}}$.