گام به گام تمرین صفحه 48 درس 2 هندسه (1) (قضیۀ تالس، تشابه و کاربردهای آن)

1)

\(k = \frac{{15}}{{10}} = \frac{P}{{P'}} = \frac{3}{2} \)

\(\Rightarrow P = 15 + 12 + 10 = 37 \)

\(\Rightarrow \frac{{37}}{{P'}} = \frac{3}{2} \Rightarrow P' = \frac{{74}}{3}\)

2)

\(\frac{{{S_{MNCB}}}}{{{S_{AMN}}}} = \frac{8}{1} \Rightarrow \frac{{{S_{MNCB}} + {S_{AMN}}}}{{{S_{AMN}}}} = \frac{{8 + 1}}{1} \)

\(\Rightarrow \frac{{{S_{ABC}}}}{{{S_{AMN}}}} = 9 = {k^2}\)

\(\Rightarrow k = 3 = \frac{{AB}}{{MA}} \Rightarrow \frac{{MB}}{{MA}} = \frac{{AB - MA}}{{MA}} = \frac{{3 - 1}}{1} \)

\(\Rightarrow \frac{{MB}}{{MA}} = 2\)

3)

\(\frac{{AB}}{{CD}} = \frac{{AE}}{{BC}} = \frac{{BE}}{{BD}} \)

\(\Rightarrow \left\{ \begin{array}{l}2x - 7 = x + y \Rightarrow x - y = 7\\2x - 1 = y + 15 \Rightarrow 2x - y = 16\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}x = 9\\y = 2\end{array} \right.\)

\(\frac{{{S_{BCD}}}}{{{S_{ABE}}}} = {\left( {\frac{{BD}}{{DE}}} \right)^2} = {\left( {\frac{{BE + DE}}{{DE}}} \right)^2} \)

\(= {\left( {\frac{{BE}}{{DE}} + 1} \right)^2} = {\left( {2 + 1} \right)^2} = {3^2} = 9\)

4)

الف)

\(\frac{{{S_{ABH}}}}{{{S_{ABC}}}} = \frac{{\frac{1}{2}AH \cdot BH}}{{\frac{1}{2}AB \cdot AC}} = \frac{{AH \cdot BH}}{{AB \cdot AC}} = \frac{{BH}}{{AB}} \times \frac{{AH}}{{AC}} \)

\(= \frac{{AB}}{{BC}} \times \frac{{AB}}{{BC}} = {\left( {\frac{{AB}}{{BC}}} \right)^2}\)

\(\frac{{{S_{ACH}}}}{{{S_{ABC}}}} = \frac{{\frac{1}{2}AH \cdot CH}}{{\frac{1}{2}AB \cdot AC}} = \frac{{AH \cdot CH}}{{AB \cdot AC}} = \frac{{AH}}{{AB}} \times \frac{{CH}}{{AC}} \)

\(= \frac{{AC}}{{BC}} \times \frac{{AC}}{{BC}} = {\left( {\frac{{AC}}{{BC}}} \right)^2}\)

ب)

\(\frac{{{S_{ABH}}}}{{{S_{ABC}}}} + \frac{{{S_{ACH}}}}{{{S_{ABC}}}} = \frac{{{S_{ABH}} + {S_{ACH}}}}{{{S_{ABC}}}} = \frac{{{S_{ABC}}}}{{{S_{ABC}}}} \)

\(= {\left( {\frac{{AB}}{{BC}}} \right)^2} + {\left( {\frac{{AC}}{{BC}}} \right)^2} = \frac{{A{B^2} + A{C^2}}}{{B{C^2}}}\)

\(\Rightarrow 1 = \frac{{A{B^2} + A{C^2}}}{{B{C^2}}} \Rightarrow B{C^2} = A{B^2} + A{C^2}\)

5)

\(\frac{x}{{1/6}} = \frac{{20 + x}}{6} \Rightarrow 6x = 32 + 1/6x\)

\( \Rightarrow 4/4x = 32 \Rightarrow x = \frac{{32}}{{4/4}} = \frac{{80}}{{11}}\)

\(\frac{x}{{1/6}} = \frac{{80 + x}}{{h + 3/2}} \Rightarrow h.x + 3/2x = 128 + 1/6x \)

\(\Rightarrow h = \frac{{128 - 1/6x}}{x} = \frac{{128}}{x} - 1/6\)

\(h = \frac{{128}}{{\frac{{80}}{{11}}}} - 1/6 = 17/6 - 1/6 = 16\)

6)

\(\frac{{3/{5^{cm}}}}{{4/{2^{cm}}}} = \frac{{{h^m}}}{{{6^m}}} \Rightarrow h = \frac{{3/5 \times 6}}{{4/2}} = {5^m}\)