گام به گام تمرین صفحه 36 درس قضیۀ تالس، تشابه و کاربردهای آن هندسه (1)

1)

2)

\(\frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} = \frac{{AE}}{{AC}} \Rightarrow \frac{2}{{AB}} = \frac{{DE}}{4} = \frac{1}{{1/5}}\)

\(\Rightarrow \left\{ \begin{array}{l}AB = 3\\DE = \frac{8}{3}\end{array} \right.\)

3)

\(\frac{9}{{9 + x}} = \frac{x}{{x + 4}} = \frac{{2y - 1}}{{x + 2}} \Rightarrow \frac{9}{{9 + x}} = \frac{x}{{x + 4}}\)

\(\Rightarrow 9x + 36 = 9x + {x^2} \Rightarrow {x^2} = 36 \Rightarrow x = 6\)

\(\frac{x}{{x + 4}} = \frac{{2y - 1}}{{x + 2}} \Rightarrow \frac{6}{{10}} = \frac{{2y - 1}}{8}\)

\(\Rightarrow 2y - 1 = \frac{{24}}{5} \Rightarrow y = \frac{{29}}{{10}}\)

4)

\(\left\{ \begin{array}{l}\frac{{OA}}{{OA'}} = \frac{{OB}}{{OB'}}\\\frac{{OC}}{{OC'}} = \frac{{OB}}{{OB'}}\end{array} \right. \Rightarrow \frac{{OA}}{{OA'}} = \frac{{OC}}{{OC'}}\)

(عکس قضیه تالس) \(\Rightarrow AC\parallel A'C'\)

5)

\(\left\{ \begin{array}{l}\frac{{AB}}{{AD}} = \frac{{AC}}{{AE}}\\\frac{{AB}}{{AD}} = \frac{{AE}}{{AF}}\end{array} \right. \Rightarrow A{E^2} = AC.AF\)

6)

\(\frac{3}{{60 + 3}} = \frac{1}{x} \Rightarrow x = \frac{{63}}{3} = 21\)

7)

\(\left\{ \begin{array}{l}\frac{{AM}}{{MD}} = \frac{{AK}}{{KC}}\\\frac{{BN}}{{NC}} = \frac{{AK}}{{KC}}\end{array} \right. \Rightarrow \frac{{AM}}{{MD}} = \frac{{BN}}{{NC}}\)

8)

\(\frac{9}{{9 + 2}} = \frac{{2/43}}{{x + 1/8 + 0/3}} \Rightarrow \frac{9}{{11}} = \frac{{2/43}}{{x + 2/1}}\)

\(\Rightarrow 9x + 18/9 = 26/73 \Rightarrow x = 0/87\)

بازیکن 87 سانتی متر به هوا پریده است.