[tex]
\text{Am atasat desenele notate, iar rezolvarea o voi face aici. } \\
\text{Suntem la capitolul "ASEMANAREA TRIUNGHIURILOR"} \\ \\
a) \displaystyle\\
In~ disenul~~a)~~avem: \\
\Delta ABC ~~si ~~\Delta CDE ~~avem: \\
\widehat{ABC}=\widehat{CDE}=90^o \\
\widehat{ACB}=\widehat{CED} \\
\Longrightarrow ~~~\text{suntem in cazul UU de asemanare } \\
\Longrightarrow ~~~\Delta ABC ~\sim~\Delta CDE \\
\text{Scriem rapoartele de asemanare: } \\ \\
\frac{AB}{CD}= \frac{BC}{DE}
[/tex]
[tex]\displaystyle \\
\frac{AB}{CD}= \frac{BC}{DE} \\ \\
\frac{x-1,4}{10}= \frac{12}{x+5,6} \\ \\
(x-1,4)(x+5,6)=10 \cdot 12 \\
x^2 -1,4x + 5,6x- 1,4\cdot 5,6 = 120 \\
x^2 -4,2x- 7,84 - 120=0 \\
x^2 -4,2x - 127,84=0 \\ \\
x_{12}= \frac{-b \pm \sqrt{b^2 -4ac} }{2a} =\frac{7,84 \pm \sqrt{17,64 +511,36} }{2} = \\ \\
\frac{7,84 \pm \sqrt{529} }{2} =\frac{7,84 \pm 23}{2} \\ \\
x_1 = \frac{7,84 + 23}{2} = \frac{30,84}{2} = \boxed{15,42} [/tex]
[tex]\displaystyle \\ \\
x_2 = \frac{7,84 - 23}{2} =\frac{-15,16}{2}= -7,58 \\ \\
\text{Aceasta solutie va fi eliminata deoarece laturile ar ar fi negative } [/tex]
[tex]b) \displaystyle\\
In~ disenul~~b)~~avem: \\
\Delta ABE ~~si ~~\Delta CBD ~~in~care~ avem: \\
\widehat{ABE}= \widehat{CBD}~~~fiind~opuse~la~varf. \\
\widehat{BAE}=\widehat{BCD} \\
\Longrightarrow ~~~\text{suntem in cazul UU de asemanare } \\
\Longrightarrow ~~~\Delta ABE ~\sim~\Delta CBD \\
\text{Scriem rapoartele de asemanare: } \\ \\
\frac{AB}{CB}= \frac{BE}{BD} \\ \\
\frac{18}{12,6}= \frac{x+2}{x-1} \\ \\
18(x-1)= 12,6(x+2) [/tex]
[tex]\displaystyle \\
18x -12,6x = 25,2+18 \\
5,4x=43,2 \\ \\
x = \frac{43,2}{5,4}=\boxed{8} [/tex]
