Задание №76346 ЕГЭ Базовой математике

$\mathit{𝐴}\mathit{𝐵}=21\sqrt{2};\mathit{𝐶}\mathit{𝑀}=10;\mathit{𝐵}\mathit{𝑀}=14;\mathit{𝑅}-?$

по сойству биссектрисы $\frac{\mathit{𝐴}\mathit{𝐶}}{\mathit{𝐴}\mathit{𝐵}}=\frac{\mathit{𝐶}\mathit{𝑀}}{\mathit{𝐵}\mathit{𝑀}}$

$\mathit{𝐴}\mathit{𝐶}=\frac{\mathit{𝐴}\mathit{𝐵}·\mathit{𝐶}\mathit{𝑀}}{\mathit{𝐵}\mathit{𝑀}}=\frac{21\sqrt{2}·10}{14}=\frac{3·\sqrt{2}·10}{2}=15\sqrt{2}$

$\mathit{𝑅}=\frac{\mathit{𝑎}\mathit{𝑏}\mathit{𝑐}}{4\mathit{𝑆}}$

Проведем высоту АН; запишем теорему Пифагора в $∆\mathit{𝐴}\mathit{𝐶}\mathit{𝐻}$ и $∆\mathit{𝐴}\mathit{𝐵}\mathit{𝐻}$, обозначив СН=х, тогда $\mathit{𝐵}\mathit{𝐻}=24-\mathit{𝑥}$

$\mathit{𝐴}{\mathit{𝐻}}^2=\mathit{𝐴}{\mathit{𝐶}}^2-\mathit{𝐶}{\mathit{𝐻}}^2={\left(15\sqrt{2}\right)}^2-{\mathit{𝑥}}^2$

$\mathit{𝐴}{\mathit{𝐻}}^2=\mathit{𝐴}{\mathit{𝐵}}^2-\mathit{𝐵}{\mathit{𝐻}}^2={\left(21\sqrt{2}\right)}^2-{\left(24-\mathit{𝑥}\right)}^2$

${\left(15\sqrt{2}\right)}^2-{\mathit{𝑥}}^2={\left(21\sqrt{2}\right)}^2-{\left(24-\mathit{𝑥}\right)}^2$

$450-{\mathit{𝑥}}^2=882-576+48\mathit{𝑥}-{\mathit{𝑥}}^2$

$48\mathit{𝑥}=450+576-882$

$48\mathit{𝑥}=144$

$\mathit{𝑥}=144:48=3;\mathit{𝐶}\mathit{𝐻}=3$

$\mathit{𝐴}{\mathit{𝐻}}^2={\left(15\sqrt{2}\right)}^2-3^2=450-9=441={21}^2;\mathit{𝐴}\mathit{𝐻}=21$

${\mathit{𝑆}}_{∆\mathit{𝐴}\mathit{𝐵}\mathit{𝐶}}=\frac{1}{2}\mathit{𝐵}\mathit{𝐶}·\mathit{𝐴}\mathit{𝐻}=\frac{1}{2}·24·21=12·21=252$

$\mathit{𝑅}=\frac{15\sqrt{2}·24·21\sqrt{2}}{4·252}=15$