решите 10 и 12плиз очень нужно

Задача 10

[tex]\Delta ABC:\; 6^2 + x^2 = (5 + y)^2.\\CD^2 = 5\cdot y.\\\\\Delta CDB:\; CD^2 = x^2 - 5^2.[/tex]

Решаем систему.

[tex]\begin{cases}6^2 + x^2 = (5 + y)^2,\\5y = x^2 - 5^2;\end{cases}\Longleftrightarrow\;+\begin{cases}36 + x^2 = 25 + y^2 + 10y,\\5y - x^2 + 25 = 0.\end{cases}\\36 + 5y + 25 = 25 + y^2 + 10y;\\y^2 + 5y - 36 = 0;\\y_{1,2} = \dfrac{-5\pm\sqrt{25 + 4\cdot36}}{2} = \left[\begin{array}{c}4,&-9;\end{array}\Longrightarrow \;y = 4.[/tex]

[tex]5\cdot4 - x^2 + 25 = 0;\\x^2 = 45;\\x_{1,2} = \pm\sqrt{45};\\x_{1,2} = \left[\begin{array}{c}3\sqrt{5},&-3\sqrt{5};\end{array}\Longrightarrow \;x = 3\sqrt{5}.[/tex]

Ответ: [tex]3\sqrt5[/tex].

Задача 12

[tex]AE = \dfrac{AD-BC}{2} = \dfrac{9-5}{2} = 2.\\\\BE= \sqrt{BA^2-AE^2} = \sqrt{6^2-2^2} = \sqrt{36-4} = \sqrt{32} = 4\sqrt2.[/tex]

Ответ: [tex]\bf4\sqrt2[/tex].